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English Pages 159 [148] Year 2021
MINISTRY OF EDUCATION AND SCIENCE
OF THE REPUBLIC OF KAZAKHSTAN KAZAKHBRITISH TECHNICAL UNIVERSITY
B.Sh. Kulpeshov
AN INTRODUCTION TO
PROBABILITY THEORY AND
MATHEMATICAL STATISTICS
ALMATY 2021
UDC 510 LBC 22.1 K91
Kulpeshov B. Sh. An Introduction to Probability Theory and Mathematical Statistics.  Almaty, 2021  159 pages.
ISBN 9786012690859
The present manual has been composed by using the lectures read by the author at KazakhBritish Technical University within last two academic years for students of technical specialties of the English branch. The present course studies elements of probability theory and mathematical statistics: random events, geometric probabilities, combinatorial analysis, independence of events, conditional probability, theorems of addition and multiplication, random variables, mathematical expectation, dispersion, basic laws of distribution, the law of large numbers and limit theorems, variation series, sampling, methods of estimation, and testing of statistical hypotheses. The course has been prepared for teaching Probability Theory and Mathematical Statistics during one semester (15 weeks).
ISBN 9786012690859
© Kulpeshov B.Sh., 2021
INTRODUCTION
Events (phenomena) observed by us could be subdivided on the following three types: reliable, impossible and random. A reliable (universal) event is an event that necessarily will happen if a certain set of conditions S holds. For example, if a vessel contains water with a normal atmospheric pressure and temperature 20 degrees, the event «water in a vessel is in a liquid state» is reliable. In this example the given atmospheric pressure and temperature of water make the set of conditions S. An impossible (null) event is an event that certainly will not happen if the set of conditions S holds. For example, the event «water in a vessel is in a rigid state» certainly will not happen if the set of conditions of the previous example holds. A random event is an event that can either occur, or not to occur for holding the set of conditions S. For example, if a coin is tossed it can land on one of two sides: heads or tails. Therefore, the event «the coin lands on heads» is random. Each random event, in particular an appearance of heads, is the consequence of a functioning very many random reasons (in our example: the force with which the coin tossing, the form of the coin and many others). It is impossible to take into account an influence on result of all these reasons, as their number is very great and laws of their functioning are unknown. Therefore, probability theory does not pose the problem to predict that a single event whether or not will happen  it simply is unable to solve this problem. We have another picture if we consider random events that can multiply be observed for holding the same conditions S, i.e. if the speech goes on mass homogeneous random events. It appears that a rather large number of homogeneous random events independently from their concrete nature are subordinated to definite regularities, namely probability regularities. Probability theory studies these regularities. Thus, the subject of probability theory is studying probability regularities of mass homogeneous random events. Methods of probability theory are widely applied in various branches of natural sciences and techniques: theory of reliability, theory of mass service, theoretical physics, geodesy, astronomy, theory of shooting, theory of mistakes of observation, theory of automatic control, general theory of communication and many others theoretical and applied sciences. Probability theory serves also to substantiate Mathematical Statistics that in turn is used at planning and organizing a manufacture, at analysis of technological processes, and for many other purposes. All life a person has to make decisions: in personal sphere (in which university or college to enter, with whom to communicate; how to study); in public (to attend evenings, theatres, meetings, assemblies, elections); in
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industrial (determining factors essentially influencing on productivity, quality of materials and etc.); in scientific (promotion and checking scientific hypotheses). Decisionmaking usually pursues one of the following purposes: forecasting of a future state of a process (an object); control (i.e. how should to change certain parameters of an object in order that other parameters have taken on a desirable value); an explanation of internal structure of an object. Usually decisionmaking is preceded with an analysis of known data (based on previous experience, common sense, intuition, and etc.). Aspiring to see and prove regularities in uncertain processes, the humanity has developed the whole arsenal of methods that refer to as mathematical statistics (applied statistics or data analysis'). Mathematical statistics is a section of mathematics in which mathematical methods of ordering, processing and using statistical data for scientific and practical conclusions are studied. Abraham Wald (19021950) spoke that «mathematical statistics is theory of decisionmaking in conditions of uncertainty». In essence, mathematical statistics gives a unique, mathematically proved apparatus for solving problems of control and forecasting at absence of obvious regularities (presence of randomness) in investigated processes.
It is wonderful that the science begun with consideration of gambles was fated to become the major object of human knowledge ... PierreSimon Laplace (17491827)
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LECTURE 1 Basic concepts of probability theory
Hereinafter, instead of speaking «the set of conditions S holds» we shall speak briefly: «the trial has been made». Thus, an event will be considered as a result of a trial.
Example. A shooter shoots in a target subdivided into four areas. One shot is the trial. Hit in a certain area of the target is an event. Example. There are colour balls in an um. One takes at random one ball from the um. An extracting a ball from the um is the trial. An appearance of a ball of a certain colour is an event. Events are incompatible if an appearance of one of them excludes an appearance of other events in the same trial. Otherwise, they are compatible.
Example. An item is extracted at random from a box with items. An appearance of a standard item excludes an appearance of a nonstandard item. The events «a standard item has appeared» and «a nonstandard item has appeared» are incompatible. Example. A coin is tossed. An appearance of «heads» excludes an appearance of «tails». The events «heads have appeared» («the coin lands on heads») and «tails have appeared» («the coin lands on tails») are incompatible. For example, a landing two different prizes under only one ticket of a lottery are incompatible events, and a landing the same prizes under two tickets are compatible events. Obtaining marks «excellent», «good» and «satisfactory» by a student at an exam in one discipline are incompatible events and an obtaining the same marks at exams in three disciplines are compatible events.
Some events form a complete group if in result of a trial at least one of them will appear. In other words, the appearance of at least one of events of a complete group is a reliable event. In particular, if events forming a complete group are pairwise incompatible then in result of a trial one and only one of these events will appear. Example. Two tickets of a moneything lottery have been bought. One necessarily will happen one and only one from the following events: «a landing a prize on the first ticket and a nonlanding a prize on the second», «a landing a prize on both tickets», «a nonlanding a prize on the first ticket and a landing a
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prize on the second», «a nonlanding a prize on both tickets». These events form a complete group of pairwise incompatible events.
Example. A shooter has made one shot in a target. One necessarily will happen one from the following two events: hit, miss. These two incompatible events form a complete group.
Events are equally possible if there is reason to consider that none of them is more possible (probable) than other. Example. An appearance of heads and an appearance of tails at tossing a coin are equally possible events. Appearances of «one», «two», «three», «four», «five» or «six» on a tossed die are equally possible events.
Several events are uniquely possible if at least one of them will necessarily happen as a result of a trial. For example, the events consisting in that a family with two children has: A  «two boys», B  «one boy and one girl» and C  «two girls» are uniquely possible. Classical definition of probability
Example. Let an um contain 6 identical, carefully shuffled balls, and 2 of them are red, 3  blue and 1  white. Obviously, the possibility to take out at random from the um a colour ball (i.e. red or blue) is more than the possibility to extract a white ball. Whether it is possible to describe this possibility by number? It appears it is possible. This number is said to be the probability of an event (appearance of a colour ball). Thus, the probability is the number describing the degree of possibility of an appearance of an event. Let the event A be an appearance of a colour ball. We call each of possible results of a trial (the trial is an extracting a ball from the um) by elementary event. We denote elementary events by CDγ, CD2, CD3 and et cetera. In our example the following 6 elementary events are possible: ω1  the white ball has appeared; ω2,ω3  a red ball has appeared; ω4,ω5,ω6  a blue ball has appeared. These events form a complete group of pairwise incompatible events (it necessarily will be appeared only one ball) and they are equally possible (a ball is randomly extracted; the balls are identical and carefully shuffled). We call those elementary events in which the event interesting for us occurs, as favorable to this event. In our example, the following 5 events favor to the event A (appearance of a colour ball): ω2,ω3,ω4,ω5,ω6. In this sense the event A is subdivided on some elementary events; an elementary event is not subdivided into other events. It is the distinction between the event A and an elementary event.
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The ratio of the number of favorable to the event A elementary events to their total number is said to be the probability of the event A and it is denoted by P(A). In the considered example we have 6 elementary events; 5 of them favor to the event A. Therefore, the probability that the taken ball will be colour is equal to P(A) = 5/6. This number gives such a quantitative estimation of the degree of possibility of an appearance of a colour ball which we wanted to find.
The probability of the event A is the ratio of the number of favorable elementary events for this event to their total number of all equally possible incompatible elementary events forming a complete group.
Thus, the probability of the event A is determined by the formula:
m P(A)=n
where m is the number of elementary events favorable to Л; n is the number of all possible elementary events of a trial. Here we suppose that elementary events are incompatible, equally possible and form a complete group.
The definition of probability implies the following its properties: Property 1. The probability of a reliable event is equal to 1. In fact, if an event is reliable then each elementary event of a trial favors to the event. In this case m = n and consequently P(A) = m/n = n/n = 1.
Property 2. The probability of an impossible event is equal to 0. Indeed, if an event is impossible then none of elementary events of a trial favors to the event. In this case m = 0 and consequently P(A) = m/n = 0/n = 0. Property 3. The probability of a random event is the positive number between 0 and 1. In fact, a random event is favored only part of the total number of elementary events of a trial. In this case 0 < m < и; then 0 < m/n < 1 and consequently 0 < P(A) < 1. Thus, the probability of an arbitrary event A satisfies the double inequality:
I O≤∕W≤1 I
Relative frequency
The relative frequency (statistical probability) of an event is the ratio of the number of trials, in which the event has appeared, to the total number of actually made trials. Thus, the relative frequency of the event A is defined by the formula:
m W(A)=n 7
where m is the number of appearances of the event, n is the total number of trials. Comparing the definitions of probability and relative frequency, we conclude: the definition of probability does not demand that the trials should be made actually; the definition of relative frequency assumes that the trials were made actually. In other words, the probability is calculated before an experiment, and the relative frequency — after an experiment.
Example. The quality department has detected 3 nonstandard items in a group consisting of 80 randomly selected items. The relative frequency of appearance of nonstandard items is IV(A) = 3/80. Example. There have been made 24 shots in a target, and 19 hits were registered. The relative frequency of hit in the target is IV(A) = 19/24. The long observations have shown that if experiments are made in identical conditions, in each of which the number of trials is rather great, the relative frequency has a stability property. This property is that for different experiments the relative frequency is changed a little (the less changes, the more trials were made), oscillating about some constant number. There was found out that this constant number is the probability of appearance of the event. Thus, if the relative frequency is established by a practical experiment, the obtained number can be accepted for approximate value of probability. Geometric probabilities
To overcome defect of the classical definition of probability consisting that it is inapplicable to trials with infinite number of events (outcomes) enter geometric probabilities  the probability of hit of a point in area (segment, part of a plane and etc.). Let a segment I be a part of a segment L. A point is set (thrown) at random in the segment L. It means that the following suppositions hold: the thrown point can appear in any point of the segment L, the probability of hit of the point in the segment l is proportional to the length of this segment and does not depend on its disposition concerning the segment L. In these suppositions the probability of hit of the point in the segment l is determined by the equality P = the length of l / the length of L
Example. A point B(x) is thrown at random in a segment OA of the length L of the numeric axis Ox. Find the probability that the smaller of the segments OB and BA has the length more than L/3. It is assumed that the probability of hit of a
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point in the segment is proportional to the length of the segment and does not depend on its disposition on the numeric axis. Solution'. Let's divide the segment OA by points C and D on three equal parts. The request of the problem will be executed if the point B(x) will hit in the segment CD of the length L/3. The required probability P = (L/3)/L = 1/3. Let a flat figure g be a part of a flat figure G. A point is thrown at random in the figure G. It means that the following suppositions hold: the thrown point can appear in any point of the figure G, the probability of hit of the thrown point in the figure g is proportional to the area of this figure and does not depend on both its disposition concerning the figure G and the form of g. In these suppositions the probability of hit of the point in the figure g is determined by the equality P = the area of g / the area of G
Example. Two concentric circles of which the radiuses are 5 and 10 cm respectively are drawn on the plane. Find the probability that the point thrown at random in the large circle will hit in the ring formed by the constructed circles. It is assumed that the probability of hit of a point in a flat figure is proportional to the area of this figure and does not depend on its disposition concerning the large circle. Solution '. The area of the ring (the figure g) Sg =TtR2 πr2 =π(102 52) = 75π. The area of the large circle (the figure G) Sg =π∙R2 = π ∙ IO2 = 100л. _ 73π _ The required probability ʃ = ɪθθ^ =0,75. Glossary probability theory  теория вероятностей reliable event  достоверное событие random event  случайное событие vessel  сосуд; trial (experiment)  испытание (опыт, эксперимент) urn  урна; heads or tails?  орел или решка? at random  наудачу; to land a prize  получить приз complete group of events  полная группа событий equally possible events  равновозможные события uniquely possible events  единственно возможные события асе  очко (при игре в кости); die  кость (игральная) dice  игра в кости, кости; hit  попадание; miss  промах to shuffle  перемешивать; relative frequency  относительная частота
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favorable case  благоприятствующий (благоприятный) случай mass homogeneous events  массовые однородные события Exercises for Seminar 1
1.1. There are 50 identical items (and 5 of them are painted) in a box. Find the probability that the first randomly taken item will be painted. 1.2. A die is tossed. Find the probability that an even number of aces will appear. 1.3. Participants of a tossup pull a ticket with numbers from 1 up to 100 from a box. Find the probability that the number of the first randomly taken ticket does not contain the digit 5 (tossup  жеребьевка; to pull  тянуть; ticket  жетон). 1.4. In a batch of 100 items the quality department has found out 5 nonstandard items. What is the relative frequency of appearance of nonstandard items equal to? (batch  партия) 1.5. At shooting by a rifle the relative frequency of hit in a target has appeared equal to 0,85. Find the number of hits if 120 shots were made (a rifle винтовка). 1.6. One die is randomly taken from a carefully hashed full set of 28 dice of domino. Find the probability that the second randomly taken die can be put to the first if the first die: a) is a double; b) is not double (to hash  перемешивать; double  дубль). The answer: a) 2/9; b) 4/9. 1.7. A point B(x) is randomly put in a segment OA of the length L of the numeric axis Ox. Find the probability that the smaller of the segments OB and BA has the length which is less than LA. It is supposed that the probability of hit of a point in the segment is proportional to its length and does not depend on its location on the numeric axis.
1.8. Two persons have agreed to meet in a certain place between 18 and 19 o'clock and have agreed that a person come the first waits for another person within 15 minutes then leaves. Find the probability of their meeting if arrival of everyone within the specified hour can take place at any time and the moments of arrival are independent. The answer. 7/16. 1.9. Two dice are tossed. Find the probability that: a) the same number of aces will appear on both dice; b) two aces will appear at least on one die; c) the sum of aces will not exceed 6 (to exceed  превышать).
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1.10. A point is randomly taken inside a circle of radius 5. Find the probability that the point will be inside a proper (equilateral) triangle entered in the circle (a proper triangle  правильный треугольник; equilateral  равносторонний). 3√3 The answer: ~— . 4π Exercises for Homework 1
1.11. A die is tossed. Find the probability that the upper side of the die shows: a) six aces; b) an odd number of aces; c) no less than four aces; d) no more than two aces. 1.12. An um contains 12 balls: 3 white, 4 black and 5 red. Find the probability that a randomly taken ball will be black. 1.13. The first box contains 5 balls with numbers from 1 up to 5, and the second  5 balls with numbers from 6 up to 10. It has been randomly extracted on one ball from each box. Find the probability that the sum of numbers of the extracted balls will be: 1) no less than 7; 2) equal to 11; 3) no more than 11. The answer: 2) 0,2; 3) 0,6. 1.14. The relative frequency of workers of an enterprise having a higher education is equal to 0,18. Determine the number of workers having a higher education if the total number of workers of the enterprise is 350. 1.15. 78 seeds have germinated of 100 planted seeds. Find the relative frequency of germination of seeds (seed  семя; to germinate  прорастать, to plant сажать). 1.16. Two dice are tossed. Find the probability that: a) the sum of aces equals 5, and the product equals 6; b) the product of aces doesn’t exceed 6; c) the product of aces is divided on 6. 1.17. A point C is randomly appeared in a segment AB of the length 3. Determine the probability that the distance between C and B exceeds 1. The answer: 2/3. 1.18. A point is randomly thrown inside of a circle of the radius R. Find the probability that the point will be inside the square entered in the circle. It is supposed that the probability of hit of a point in the square is proportional to its area and does not depend on its location regarding the circle. The answer. Ик. 1.19. A coin is tossed twice. Find the probability that the coin lands on heads in both times. The answer: 1/4. 11
1.20. A point B(x) is randomly put in a segment OA of the length 6 of the numeric axis Ox. Find the probability that both the segments OB and BA have the length which is greater than 2. The answer: 1/3.
LECTURE 2 Basic formulas of combinatorial analysis
Here is a typical problem of interest involving probability. A communication system is to consist of n seemingly identical antennas that are to be lined up in a linear order. The resulting system will then be able to receive all incoming signals  and will be called functional  as long as no two consecutive antennas are defective. If it turns out that exactly m of the n antennas are defective, what is the probability that the resulting system will be functional? For instance, in the special case where n = 4 and m = 2 there are 6 possible system configurations  namely,
0110,0101, 1010,0011, 1001, 1100 where 1 means that the antenna is working and 0 that it is defective. As the resulting system will be functional in the first 3 arrangements and not functional in the remaining 3, it seems reasonable to take 3/6 = 1/2 as the desired probability. In the case of general n and m, we could compute the probability that the system is functional in a similar fashion. That is, we could count the number of configurations that result in the system being functional and then divide by the total number of all possible configurations. From the preceding we see that it would be useful to have an effective method for counting the number of ways that things can occur. In fact, many problems in probability theory can be solved simply by counting the number of different ways that a determinate event can occur. The mathematical theory of counting is formally known as combinatorial analysis. Combinatorial analysis studies quantities of ordered sets subordinate to determinate conditions, which can be made of elements, indifferent of a nature, of a given finite set. Permutations are ordered sets consisting of the same n different elements and distinguishing only by the order of their disposition (location).
The number of all possible permutations Pn = n!
Where
w!=l∙2∙3∙...∙w
Observe that it is convenient to consider 0!, assuming by definition that 12
O! = 1
Example. How many threeplace numbers can be made of the digits 1, 2, 3 if each digit is included into the image of a number only once? Solution: P3 = 3! = 1 ■ 2 ■ 3 = 6. Allocations are ordered sets composed of n different elements on m elements, which are differed by either structure of elements or their order. The number of all possible allocations ∕√l C = «(« 1)(«  2)...(π  да +1) =  — s ι.e. (nm)∖
дт _
rι∙ (nm)∖
Example. How many signals is it possible to make of 6 flags of different colour taken on 2? Solution Λ62 = 6 ∙ (6  2 +1) = ɪ = 30. (62)! Combinations are ordered sets composed of n different elements on m elements that are differed by at least one element. The number of all possible combinations
m
= —— —
m∖(nm)∖
Example. How many ways are there to choose two items from a box containing 10 items? 2 _ 10! Solution'. The required number of ways ɛɪo  45 ∙ Observe that the numbers of allocations, permutations and combinations are connected by the equality Af = Pm ■ Cf.
Remark. It was above supposed that all n elements are different. If some elements are repeated then in this case ordered sets with repetitions are calculated by other formulas. For example, if there are n1 elements of one kind, n2 elements of other kind and et cetera among n elements then the number of permutations with repetitions
Pn(Pl,n2
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where n1 + n2 + ... = n.
Example. How many sevenplace numbers consisting of the digits 4, 5 and 6 are there in which the digit 4 is repeated 3 times, and the digits 5 and 62 times? Solution: Each sevenplace number differs from other by the order of consecution of the digits (so that n↑ = 3, n2 = 2, n2 = 2, and their sum is equal to 7), i.e. is the permutation with repetitions of 7 elements: 7! P7 (3; 2; 2)== 210. 7 3!2!2! If in allocations (combinations) of n elements on m some of elements (or all) can appear identical, such allocations (combinations) are said to be allocations (combinations) with repetitions of n elements on m. For example, allocations with repetitions of 5 elements a, b, c, d, e on 3 are abc, cba, bed, cdb, bbe, ebb, beb, ddd and et cetera; combinations with repetitions  abc, bed, bbe, ddd and et cetera. The number of allocations with repetitions of n elements on m is
The number of combinations with repetitions of n elements on m is
Example. 10 films participate in a competition on 5 nominations. How many variants of distribution of prizes are there, if on each nomination are established: a) different prizes; b) identical prizes? Solution: a) Each of variants of distribution of prizes represents a combination of 5 films from 10 that differs from other combinations by both the structure of elements and their order on nominations so that the same films can be repeated some times (a film can receive prizes on both one nomination and some nominations), i.e. it represents the allocations with repetitions of 10 elements on 5: ⅛ =IO5 =100000. b) If identical prizes are established on each nomination then the order of following the films in a combination of 5 prizewinners doesn’t have any significance. Therefore, the number of variants of distribution of prizes represents the number of combinations with repetitions of 10 elements on 5: 14 13 12 11 10 = 2002. 10+51 1∙2∙3∙4∙5
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The following rules are used for solving problems of combinatorial analysis: Sum rule. If some object A can be chosen from the set of objects by m ways, and another object B can be chosen by n ways, then we can choose either A or B by m + n ways.
Example. There are 300 items in a box. It is known that 150 of them are items of the first kind, 120  the second kind, and the rest  the third kind. How many ways of extracting an item of the first or the second kind from the box are there? Solution: An item of the first kind can be extracted by n↑ = 150 ways, of the second kind  by n2 = 120 ways. By the sum rule there are n1 + n2 = 150 + 120 = 270 ways of extracting an item of the first or the second kind. Product rule. If an object A can be chosen from the set of objects by m ways and after every such choice an object B can be chosen by n ways then the pair of the objects (A, B) in this order can be chosen by mn ways.
Example. There are 30 students in a group. It is necessary to choose a leader, its deputy and head of professional committee. How many ways of choosing them are there? Solution: A leader can be chosen from any of 30 students, its deputy  from any of the rest 29 students, and head of professional committee  from any of the rest 28 students, i.e. n↑ = 30, n2 = 29, n3 = 28. By the product rule the total number of ways is n↑ ∙ n2 ∙ n3 = 30 ∙ 29 ∙ 28 = 24360. Operations over events
We introduce the following operations over events: sum, product and negation. The sum of two events A and B is such third event A + B which consists in appearance of at least one of these events, i.e. A or B. If A and B are compatible events then their sum A + B means appearance of either the event A, or the event B, or both A and B. If A and B are incompatible events then their sum A + B means appearance of either the event A, or the event B. For example, if two shots are made by a gun and A is hit at the first shot, B is hit at the second shot then A + B is either hit at the first shot or hit at the second shot, or two hits. The event A + B + C consists of appearance of one of the following events: A; B; C; both A and B; both A and C; both B and C; all the events A, B and C. The product of two events A and B is such third event AB which consists in simultaneous appearance of the events A and B. If A and B are incompatible events then their product AB is an impossible event. The negation of an event A is the event A (not A) which consists in nonappearance of the event A. Observe that A + A is a reliable event, and A ∙ A is an impossible event.
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Example. A winner of a competition is rewarded: by a prize (the event A), a money premium (the event B), a medal (the event C). What do the following events represent: a) A + B, b) ABC’, c) ACB ? Solution: a) The event A + B consists in rewarding the winner by a prize, or a money premium, or simultaneously both a prize and a money premium. b) The event ABC consists in rewarding the winner by a prize, a money premium and a medal simultaneously. c) The event ACB consists in rewarding the winner by both a prize and a medal simultaneously without giving a money premium. Glossary seemingly на вид, повидимому; arrangement  расположение allocation  размещение; combination  сочетание permutation  перестановка; threeplace number  трехзначное число consecution  следование; repetition  повторение to extract  извлекать; simultaneous  одновременный negation  отрицание Exercises for Seminar 2
2.1. A college planning committee consists of 3 freshmen, 4 sophomores, 5 juniors, and 2 seniors. A subcommittee of 4, consisting of 1 person from each class, is to be chosen (a freshman  первокурсник; a sophomore второкурсник). How many different subcommittees are possible? 2.2. How many outcome sequences are possible when a die is rolled four times, where we say, for instance, that the outcome is 3, 4, 3, 1 if the first roll landed on 3, the second on 4, the third on 3, and the fourth on 1? 2.3. There are five disks on the general axis of a lock. Each disk is subdivided into six sectors on which different letters are written. The lock opens only in the event that each disk occupies one certain position regarding the case of the lock. Find the probability that the lock can be opened at any installation of disks (the case of a lock  корпус замка). The answer. 0,0001286. 2.4. The order of performance of 7 participants of a competition is determined by a tossup. How many different variants of the tossup are possible? 2.5. There are 10 cards each of which contains one letter: 3 cards with letter A, 2 cards with letter S and 5 cards with letters D, R, O, M, B. A child takes cards in a random order and puts one to another. Find the probability that the word «AMBASSADOR» will be turned out (to turn out  оказываться). 2.6. By the conditions of the lottery «Sportloto 6 of 45» a participant of the lottery who have guessed 4, 5 or 6 numbers from 6 randomly selected numbers 16
of 45 receives a monetary prize. Find the probability that the participant will guess: a) all 6 numbers; b) 4 numbers. 2.7. 10 of 30 students have sport categories. What is the probability that 3 randomly chosen students have sport categories? The answer: 0,03.
2.8. A group consists of 12 students, and 8 of them are pupils with honor. 9 students are randomly selected. Find the probability that 5 pupils with honor will be among the selected. The answer: 0,255. 2.9. Eight different books are randomly placed on one shelf. Find the probability that two certain books will be put beside (a shelf  полка, beside  радом). The answer: 0,25. 2.10. A box contains 5 red, 3 green and 2 blue pencils. 3 pencils are randomly extracted from the box. Find the probabilities of the following events: A  all the extracted pencils are different color; B  all the extracted pencils are the same color; C  one blue pencil among the extracted; D  exactly two pencils of the same color among the extracted. The answer: P(A) = 0,25; P(B) = 0,092; P(C) = 0,467; P(D) = 0,658. 2.11. It has been sold 21 of 25 refrigerators of three marks available in quantities of 5, 7 and 13 units in a shop. Assuming that the probability to be sold for a refrigerator of each mark is the same, find the probability of the following events: a) refrigerators of one mark have been unsold; b) refrigerators of three different marks have been unsold. The answer: a) 0,06; b) 0,396. 2.12. A shooter has made three shots in a target. Let A1 be the event «hit by the shooter at the /th shot» (z = 1, 2, 3). Express by Al, A2, A2 and their negations the following events: A  «only one hit»; B  «three misses»; C  «three hits»; D  «at least one miss»; E  «no less than two hits»; F «no more than one hit». Exercises for Homework 2
2.13. How many different 7place codes for license plates are possible if the first 3 places are to be occupied by letters of Latin alphabet and the final 4 by numbers? The answer: 175760000. 2.14. In Ex. 2.13, how many codes for license plates would be possible if repetition among letters or numbers were prohibited?
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2.15. 10 persons participate in competitions, and three of them will take the first, second and third places. How many different variants are possible? The answer: 720. 2.16. How many ways of choosing 3 persons of 10 are possible? The answer: 120. 2.17. A randomly taken phone number consists of 5 digits. What is the probability that all digits of the phone number are: a) identical; b) odd? It is known that any phone number does not begin with the digit zero. The answer, a) 0,0001; b) 0,0347. 2.18. There are 3 cards with letter S, 3 cards with letter T, 2 cards with letter I, 1 card with letter A and 1 card with letter C. Cards are mixed and randomly taken out without replacement by one. Find the probability that cards with letters are taken out by the way of consecution of letters of the word «STATISTICS». The answer. 0,0000198. 2.19. A box contains 15 items, and 10 of them are painted. A collector chooses at random 3 items. Find the probability that the chosen items are painted (collector  сборщик). The answer: 0,264. 2.20. Find the probability that from 10 books located in a random order, 3 certain books will be beside. The answer. 0,0667. 2.21. Four tickets are distributed among 25 students (15 of them are girls). Everyone can take only one ticket. What is the probability that owners of these tickets will be: a) four girls; b) four young men; c) three young men and one girl? The answer, a) 0,108; b) 0,017; c) 0,142. 2.22. There are 100 products (including 4 defective) in a batch. The batch is arbitrarily divided into two equal parts which are sent to two consumers. What is the probability that all defective products will be got: a) by one consumer; b) by both consumers fiftyfifty? The answer, a) 0,117; b) 0,383. 2.23. A library consists of ten different books, and five books cost on 4 thousands of tenghe each, three books  on one thousand of tenghe and two
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books  on 3 thousands of tenghe. Find the probability that two randomly taken books cost 5 thousands of tenghe. The answer. 1/3. 2.24. A coin is tossed three times. Let A1be the event «an appearance of heads at the /th tossing» (z = 1, 2, 3). Express by Al, A2, A2 and their negations the following events: A  «three heads»; B  «three tails»; C  «at least one heads»; D  «at least one tails»; E  «only one heads»; F«only one tails».
LECTURE 3 Theorem of addition of probabilities of incompatible events
Let events A and B be incompatible and let the probabilities of these events be known. How can we find the probability of A + Br> Theorem. The probability of appearance of any of two incompatible events is equal to the sum of the probabilities of these events: P(A + B) = P(A) + P(B)
Corollary. The probability of appearance of any of several pairwise incompatible events is equal to the sum of the probabilities of these events:
P(A1 + A2 + ...+ Ar) =P(A1) +P(A2)+ ... + P(Ar).
Example. There are 30 balls in an um: 10 red, 5 blue and 15 white. Find the probability of appearance of a colour ball. Solution: An appearance of a colour ball is an appearance of either red or blue ball. The probability of appearance of a red ball (the event A) is equal to P(A) = 10/30 = 1/3. The probability of appearance of a blue ball (the event B) is equal to: P(B) = 5/30 = 1/6. The events A and B are incompatible (an appearance of a ball of one colour excludes an appearance of a ball of other colour), therefore the theorem of addition is applicable. The required probability is: P(A +B) = P(A) + P(B) = 1/3 + 1/6 = 1/2.
Example. A shooter shoots in a target subdivided into three areas. The probability of hit in the first area is 0,45 and in the second  0,35. Find the probability that the shooter will hit at one shot either in the first area or in the second area. Solution: The events A  «the shooter hit in the first area» and B  «the shooter hit in the second area» are incompatible (hit in one area excludes hit in other
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area). Therefore, the theorem of addition is applicable. The required probability is: P(A + B)= P(A) + P(B) = 0,45 + 0,35 = 0,80. Complete group of events Theorem. The sum of the probabilities of pairwise incompatible events A1, A2,
A3, ..., An which form a complete group is equal to 1: P(A1) + P(A2) + P(A3) + ... + P(An) = 1
Example. A consulting point of an institute receives packages with control works from the cities A, B and C. The probability of receiving a package from the city A is equal 0.7: from the city B  0,2. Find the probability that next package will be received from the city C. Solution: The events «a package has been received from А», «a package has been received from В» and «a package has been received from С» form a complete group. Therefore, the sum of probabilities of these events is equal to 1: 0,7 + 0,2 +p = 1 Then the required probability is equal to p= 10,9 = 0,1. Opposite events
Two uniquely possible events forming a complete group are opposite. If A denotes one of two opposite events, then the opposite to A event is denoted by A. Example. Hit and miss at a shot in a target are opposite events. Example. An item is randomly taken from a box. The events «a standard item has appeared» and «a nonstandard item has appeared» are opposite. Theorem. The sum of the probabilities of opposite events is equal to 1:
P(A) +P(A) = 1
Example. The probability that a day will be rainy is p = 0,7. Find the probability that a day will be clear. Solution: The events «a day is rainy» and «a day is clear» are opposite, therefore the required probability is q = 1 p = 1  0,7 = 0,3.
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Example. There are n items in a box, and m of them are standard. Find the probability that there is at least one standard item among к randomly extracted items. Solution: The events «there is at least one standard item among the extracted items» and «there is no standard item among the extracted items» are opposite. Denote the first event by A, and the second  by A . Obviously, P(A) = 1 P(A). Find P(A). The total number of ways by which one can extract к items from n items is Ckn. The number of nonstandard items is n  m. One can extract
к nonstandard items from n  m nonstandard items by Ckn_m ways. Therefore, the probability that there is no standard item among к extracted items is P(A) = Ck_m /Ck. The required probability is: P(A) = (P(A) = (Ck_ml Ck. Conditional probability
A random event has been before determined as an event that can take place or not to take place for holding the set of conditions S'. If there are no any other restrictions except for the conditions S for a calculation of the probability of an event then such a probability is unconditional', if there are other auxiliary conditions then the probability of an event is said to be conditional. For example, the probability of an event B is very often calculated with an auxiliary condition that an event A was happened. Observe that unconditional probability also, strictly speaking, is conditional since it is supposed that the conditions S hold. The conditional probability Pa(B) is the probability of the event B calculated in assumption that the event A has already happened. Example. There are 3 white and 3 black balls in an um. One takes out twice on one ball from the um without replacement. Find the probability of appearance of a white ball at the second trial (the event B) if a black ball was extracted at the first trial (the event A). Solution: There are 5 balls after the first trial, and 3 of them are white. The required probability is Pa(B) = 3/5.
The conditional probability of an event B with the condition that an event A has already happened is equal to:
where P(A) > 0.
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Indeed, return to our example. The probability of appearance of a white ball at the first trial P(A) = 3/6 = 1/2. Find the probability P(AB) that a black ball will be appeared at the first trial and a white ball  at the second trial. The total number of events  a joint appearance of two balls (indifferently from colour) is equal to the number of allocations ʌ2 = 6 ∙ 5 = 30. The event AB is favored 3 ∙ 3 = 9 events from the total number. Consequently, P(AB) = 9/30 = 3/10. Thus,
Pa(B) = P(AB)ZP(A) = (3/10)/(l/2) = 3/5. Theorem of multiplication of probabilities Theorem. The probability of joint appearance of two events is equal to the
product of the probability of one of them on the conditional probability of another event calculated in assumption that the first event has already happened: P(AB) = P(A) Pa(B) Remark. P(BA) =P(B) ■ Pb(A) .
Since the event BA does not differ from the event AB, P(AB) =P(B) ∙ Pb(A). Consequently, P(A) Pa(B) =P(B) ■ Pb(A). Corollary. The probability of joint appearance of several events is equal to the product of the probability of one of them on the conditional probabilities of all the rest events so that the probability of each subsequent event is calculated in assumption that all preceding events have already happened:
P(AlA2A3..An) = P(A1) ∙ P1 (A2) ∙ P4λ (A3)∙... ∙Pλa^i (An)
where PAlA2..A„ ɪ (An) is the probability of the event An calculated in assumption that the events A↑, A2, ..., ʌɪ have happened. In particular, for three events P(ABC) = P(A) ∙ Pa (B) ∙ Pab (C). Observe that the order of location of the events can be chosen any, i.e. it is indifferently which event is assumed the first, the second and et cetera.
Example. There are 3 conic and 7 elliptic cylinders at a collector. The collector has taken one cylinder, and then he has taken the second cylinder. Find the probability that the first taken cylinder is conic, and the second  elliptic. Solution: The probability that the first cylinder will be conic (the event A) P(A) = 3/10. The probability that the second cylinder will be elliptic (the event B) calculated in assumption that the first cylinder is conic, i.e. the conditional probability Pa(B) = 7/9. By theorem of multiplication, the required probability is P(AB) = P(A) ■ Pa(B) = (3/10) ∙ (7/9) = 7/30.
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Example. There are 5 white, 4 black and 3 blue balls in an um. Each trial consists in extracting at random one ball without replacement. Find the probability that a white ball will appear at the first trial (the event A), a black ball will appear at the second trial (the event B), and a blue ball will appear at the third trial (the event C). Solution: The probability of appearance of a white ball at the first trial P(A) = 5/12. The probability of appearance of a black ball at the second trial calculated in assumption that a white ball has appeared at the first trial, i.e. the conditional probability Pa(B) =4/11. The probability of appearance of a blue ball at the third trial calculated in assumption that a white ball has appeared at the first trial, and a black ball has appeared at the second trial, i.e. the conditional probability Pab(C) = 3/10. The required probability is 5 4 3 1
P(^C) = PU).λw.^(C) = .π. = 
Glossary opposite events  противоположные события to extract  извлекать; without replacement  без возвращения conditional probability  условная вероятность preceding  предшествующий conic  конический; elliptic  эллиптический cylinder  валик; collector  сборщик Exercises for Seminar 3
3.1. In a cashprize lottery 150 prizes and 50 monetary winnings are played on every 10000 tickets. What is the probability of a winning indifferently monetary or prize for an owner of one lottery ticket equal to? 3.2. The events A, B, C and D form a complete group. The probabilities of the events are those: P(A) =0,1; P(B) = 0,4; P(C) = 0,3. What is the probability of the event D equal to? 3.3. The probability that a shooter will beat out 10 aces at one shot is equal to 0,1 and the probability to beat out 9 aces is equal to 0,3. Find the probabilities of the following events: A  the shooter will beat out 8 or less aces; B  the shooter will beat out no less than 9 aces. 3.4. There are 10 items in a box, and 2 of them are nonstandard. Find the probability that in randomly selected 6 items appears no more than one non standard item. Direction: If A is «there is no nonstandard items» and B is «there is one nonstandard item» then P(A + B) = P(A) + P(B) =...
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3.5. An enterprise produces 95% standard products, and 86% of them have the first grade. Find the probability that a randomly taken product made at the enterprise will be standard and the first grade (grade  сорт). 3.6. Two dice are tossing. Find the conditional probability that each die lands on 5 if it is known that the sum of aces is divided on 5. 3.7. If two dice are rolled, what is the conditional probability that the first one lands on 4 given that the sum of the dice is 8? 3.8. In a certain community, 36 percent of the families own a dog, and 22 percent of the families that own a dog also own a cat. In addition, 30 percent of the families own a cat. What is (a) the probability that a randomly selected family owns both a dog and a cat; (b) the conditional probability that a randomly selected family owns a dog given that it owns a cat? 3.9. An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 ace (a pile стопка; a deck  колода; an ace  туз). 3.10. A coin is tossed until it will not land on the same side 2 times in succession. Find the probability that the experiment will terminate before the sixth tossing (in succession  подряд). Exercises for Homework 3
3.11. By the statistical data of a repair shop 20 stops of a lathe are on the average: 10  for change of a cutter; 3  because of malfunction of a drive; 2 because of delayed submission of items. The rest stops occur for other reasons. Find the probability of stop of the lathe for other reasons (repair shop ремонтная мастерская; lathe  токарный станок; cutter  резец; malfunction неисправность; drive  привод). The answer. 0,25. 3.12. There are 30 TVs in a shop, and 20 of them are import. Find the probability that no less than 3 import TVs will be among 5 TVs sold for one day, assuming that the probabilities of purchase of TVs of different marks are identical. The answer. 0,81. 3.13. If two dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is 11? The answer. 0,5.
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3.14. An um contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black? The answer. 0,066. 3.15. Fiftytwo percent of the students at a certain university are females. Five percent of the students in this university are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, find the conditional probability that (a) this student is female, given that the student is majoring in computer science; (b) this student is majoring in computer science, given that the student is female. The answer, a) 0,4; b) 0,038. 3.16. Celine is undecided as to whether to take a French course or a chemistry course. She estimates that her probability of receiving an A grade would be 1/2 in a French course and 2/3 in a chemistry course. If Celine decides to base her decision on the flip of a coin, what is the probability that she gets an A in chemistry (grade  оценка)? The answer. 0,33. 3.17. Suppose that an um contains 8 red and 4 white balls. We draw 2 balls from the um without replacement. If we assume that at each draw each ball in the um is equally likely to be chosen, what is the probability that both balls drawn are red (to draw  тянуть)? The answer. 0,424. 3.18. An um contains 10 white, 15 black, 20 blue and 25 red balls. A ball is taken at random from the um. Find the probability that the taken ball is: a) white or black; b) blue or red. 3.19. Two shooters shoot in a target. Find the probability that the target will be stmck at least one of the shooters if it is known that the probability of miss by both shooters is equal to 0,27. 3.20. Two cards are randomly selected from a pack of 36 playing cards. Find the probability that both cards are the same color (a pack  колода). The answer: 0,486.
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LECTURE 4 Independent events
An event B is said to be independent from an event A if appearance of the event A does not change the probability of the event B, i.e. if the conditional probability of the event B is equal to its unconditional probability:
Pa(B)=P(B) (*) Since P(A) ■ Pa(B) = P(B) ■ Pb(A), by using (*) we have P(A) ■ P(B) = P(B) ■ Pb(A) and consequently Pb(A) = P(A), i.e. the event A does not depend from the event B. Thus, if an event B does not depend from an event A then the event A does not depend from the event B; this means that the property of independence is mutual.
For independent events theorem of multiplication P(AB) = P(A) ■ Pa(B) has the following form: ___________________ P(AB) = P(A) P(B) (**)
i.e. the probability of joint appearance of two independent events is equal to the product of the probabilities of these events. The equality (**) is accepted as a definition of independent events.
Two events are independent if the probability of their joint appearance is equal to the product of the probabilities of these events; otherwise they are dependent. At practice one concludes on independence of events on the sense of a problem. For example, the probability of hit in a target by each of two guns does not depend on that whether another gun has hit in the target, therefore the events «the first gun has hit in the target» and «the second gun has hit in the target» are independent. Example. Find the probability of joint hit in a target by two guns if the probability of hit in the target by the first gun (the event A) is equal to 0,8; and by the second gun (the event B)  0,7. Solution: The events A and B are independent, therefore by theorem of multiplication, the required probability P(AB) = P(A) ■ P(B) = 0,8 ∙ 0,7 = 0,56. Remark. If events A and B are independent, the events A and B , A and B, A
and B are also independent. Proof ofRemark: A =AB +AB^> P(A) = P(AB) + P(AB) = P(AB) + P(A) ■ P(B).
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Then we have P(AB) = P(A) ∙ [1  P(B)] = P(A) ∙ P(B), i.e. the events A and B are independent. □ Several events are pairwise independent if each two of them are independent. For example, events A, B and C are pairwise independent if the events A and B, A and C, B and C are independent. Several events are independent in union (or just independent) if each two of them are independent and each event and all possible products of the rest events are independent. For example, if events A1, A2 and A3 are independent in union then the events A1 and A2, A1 and A3, A2 and A3, Al and .L.13. A2 and Aμ43, A3 and A1A2 are independent. Let's underline that a pairwise independence of several events does not imply their independence in union in general, i.e. the demand of independence of events in union is stronger than the demand of their pairwise independence.
Example. There are 4 coloured balls in an um: one ball is coloured in red colour (A), one ball is coloured in blue colour (B), one ball is coloured in black colour (C) and one ball  in all these three colours (ABC). Find the probability that a ball extracted at random from the um has red colour. Since two of four balls have red colour, P(A) = 2/4 = 1/2. Reasoning analogously, we have P(B) = 1/2, P(C) = 1/2. Assume now that the taken ball has blue colour, i.e. the event B has already happened. Is the probability that the extracted ball has red colour changed, i.e. is the probability of the event A changed? One ball of two balls having blue colour has also red colour, therefore the probability of the event A is still equal to 1/2. In other words, the conditional probability of the event A calculated in assumption that the event B has happened is equal to its unconditional probability. Consequently, the events A and B are independent. By analogy we have that the events A and C, B and C are independent. Thus, the events A, B and C are pairwise independent. Are these events independent in union? Let the extracted ball have two colours, for example, blue and black. What is the probability that this ball has also red colour? Since only one ball is coloured in all three colours, therefore the taken ball has also red colour. Thus, assuming that the events B and C have happened, we have the event A will necessarily happen. Consequently, this event is reliable and its probability is equal to 1. In other words, the conditional probability Pbc(A) = 1 of the event A is not equal to its unconditional probability P(A) = 1/2. Thus, the pairwise independent events A, B and C are not independent in union. Corollary. The probability of joint appearance of several events that are independent in union is equal to the product of the probabilities of these events:
P(A1A2...An) =P(A1) P(A2) ■... ■ P(An).
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Remark. If events A1, A2, ..., An are independent in union, then the opposite to
them events ∠41, H2 , ..., An are also independent in union.
Example. Find the probability of a joint appearance of heads at tossing two coins. Solution: The probability of appearance of heads on the first coin (the event A) P(A) = 1/2. The probability of appearance of heads on the second coin (the event B) P(B) = 1/2. The events A and B are independent; therefore the required probability by theorem of multiplication is equal to: P(AB) = P(A)P(B) = 1/2 ∙ 1/2 = 1/4. Example. There are 3 boxes containing 10 items each. There are 8 standard items in the first box, 7  in the second and 9  in the third box. One takes at random on one item from each box. Find the probability that all three taken items will be standard. Solution: The probability that a standard item has been taken from the first box (the event A) P(A) = 8/10 = 0,8. The probability that a standard item has been taken from the second box (the event B) P(B) = 7/10 = 0,7. The probability that a standard item has been taken from the third box (the event C) P(C) = 9/10 = 0,9. Since the events A, B and C are independent in union, the required probability (by theorem of multiplication) is equal to: P(ABC) = P(A) P(B) P(C) = 0,8 ∙ 0,7 ∙ 0,9 = 0,504. Probability of appearance of at least one event
Let as a result of a trial n events independent in union or some of them (in particular, only one or none) can appear, so that the probabilities of appearance of each of the events are known. How can we find the probability that at least one of these events will happen? For example, if as a result of a trial three events can appear, then an appearance of at least one of these events means an appearance of either one, or two, or three events. The answer on the posed question is given by the following theorem. Theorem. The probability of appearance of at least one of the events A1, A2, ..., An independent in union is equal to the difference between 1 and the product of the probabilities of the opposite events Al, A2,..., An : P(A) = lq1q2...qn
where A is the appearance of at least one of the events A1, A2, ..., An, P(Ai) = qi, i = l,n. Partial case. If the events A1, A2, ..., An have the same probability which is equal to p then the probability of appearance of at least one of these events:
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P(A) = 1qn
where A is the appearance of at least one of the events A1, A2, ..., An, P(A1) = q = lp, i=l,n.
Example. The probabilities of hit in a target at shooting by three guns are the following: p1 = 0,8; p2 = 0,7; p3 = 0,9. Find the probability of at least one hit (the event A) at one shot by all three guns. Solution: The probability of hit in the target by each of the guns doesn’t depend on results of shooting by other guns, therefore the considered events A1 (hit by the first gun), A2 (hit by the second gun) and A3 (hit by the third gun) are independent in union. The probabilities of events which are opposite to the events A1, A2 andA3 (i.e. the probabilities of misses) are equal respectively: q1 = 1 p1 = 1 — 0,8 = 0,2; q2 = 1 ~p2 = 1 — 0,7 = 0,3; q3 = 1 — p3 = 1 — 0,9 = 0,1. The required probability P(A) = 1  q1q2q3 = 1  0,2 ∙ 0,3 ∙ 0,1 = 0,994. Example. There are 4 flatprinting machines at typography. For each machine the probability that it works at the present time is equal to 0,9. Find the probability that at least one machine works at the present time (the event A). Solution: The events «a machine works» and «a machine doesn’t work» (at the present time) are opposite, therefore the sum of their probabilities is equal to 1: p + q = 1. Consequently, the probability that a machine doesn’t work at the present time is equal to q = 1 p = 1  0,9 = 0,1. The required probability P(A) = 1  q4 = 1  (0,l)4 = 0, 9999. Example. A student looks for one formula necessary to him in three directories. The probability that the formula is contained in the first, second and third directories, is equal to 0.6: 0,7 and 0,8 respectively. Find the probability that the formula is contained: a) only in one directory (the event A); b) only in two directories (the event B); c) in all the directories (the event C); d) at least in one directory (the event D); e) neither of the directories (the event E). Solution: Consider elementary events and their probabilities: A1  the formula is in the first directory, P(A1) = 0,6; P(Ai) = 1  0,6 = 0,4; A2  the formula is in the second directory, P(√42) = 0,7; P(A2) = 10,7 = 0,3; A3 the formula is in the third directory, P(A3) = 0,8; P(A3) = 1  0,8 = 0,2.
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Express all the events AE by the elementary events and their negations, and apply the abovestated theorems: a) A = /11 ∙ A2 ∙ ∕13 + ∕ll ∙ A2 ∙ Λ3 + /11 ∙ A2 ∙ ∕13,
P(A) = P(A1A2A3) + P(A1A2A3) + P(A1 A2A3) = = P(A1) ∙ P(A2) ∙ P(A3) + P(A1) ∙ P(A2) ∙ P(A3) + P(A1) ∙ P(A2) ∙ P(A3) = = 0,6 ∙ 0,3 ∙ 0,2 + 0,4 ∙ 0,7 ∙ 0,2 + 0,4 ∙ 0,3 ∙ 0,8 = 0,188. b)
B = /11 ∙ /12 ∙ ∕13 + /11 ∙ /12 ∙ Λ3 + /11 ∙ /12 ∙ ∕13, P(B) = P(A1A2A3) + P(A1A2A3) + P(A1A2A3) =
= P(A1) ∙ P(A2) ∙ P(A3) + P(A1) ∙ P(A2) ∙ P(A3) + P(A1) ∙ P(A2) ∙ P(A3) = = 0,6 ∙ 0,7 ∙ 0,2 + 0,6 ∙ 0,3 ∙ 0,8 + 0,4 ∙ 0,7 ∙ 0,8 = 0,452. c) C = √41 ∙ √42 ∙ √43, P(C) = P(A1 A2A3)= P(A1) ∙ P(Λ2) ∙ P(A3) = 0,6 ∙ 0,7 ∙ 0,8 = 0,336.
d)
P(D) = 1  P(A1) ∙ P(A2) ∙ P(A3) = 1  0,4 ∙ 0,3 ∙ 0,2 = 0,976.
e)
E = A1A2A3, P(E) = 0,4∙ 0,3∙0,2 = 0,024.
Glossary independent events  независимые события mutual  взаимный; independent in union  независимые в совокупности flatprinting machine  плоскопечатная машина directory  справочник Exercises for Seminar 4
4.1. The probability that a shooter hit in a target at one shot is equal to 0,9. The shooter has made 3 shots. Find the probability that all 3 shots will strike the target.
4.2. A coin and a die are tossed. Find the probability of joint appearance of the following events: «the coin lands on heads» and «the die lands on 6». 4.3. What is the probability that at tossing three dice 6 aces will appear at least on one of the dice (the event A)? The answer: 0,421. 4.4. There are 8 standard items in a batch of 10 items. Find the probability that there is at least one standard item among two randomly taken items. 4.5. Two dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers?
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4.6. The probability of hit in a target by the first shooter at one shot is equal to 0,8, and by the second shooter  0,6. Find the probability that the target will be struck only with one shooter. The answer. 0,44. 4.7. The probability to receive high dividends under shares at the first enterprise  0.2: on the second  0.35: on the third  0,15. Determine the probability that a shareholder having shares of all the enterprises will receive high dividends: a) only at one enterprise; b) at least on one enterprise (a share  акция). The answer: a) 0,4265; b) 0,564. 4.8. The first brigade has 6 tractors, and the second  9. One tractor demands repair in each brigade. A tractor is chosen at random from each brigade. What is the probability that: a) both chosen tractors are serviceable; b) one of the chosen tractors demands repair (serviceable  исправный). The answer: a) IQTlI; b) 13/54. Exercises for Homework 4
4.9. There are items in two boxes: in the first  10 (3 of them are standard), in the second  15 (6 of them are standard). One takes out at random on one item from each box. Find the probability that both items will be standard. The answer. 0,12.
4.10. There are 3 television cameras in a TV studio. For each camera the probability that it is turned on at present, is equal to p = 0,6. Find the probability that at least one camera is turned on at present (the event A). The answer. 0,936. 4.11. What is the probability that at least one of a pair of dice lands on 6, given that the sum of the dice is 8? The answer. 0,4. 4.12. 10 of 20 savings banks are located behind a city boundary. 5 savings banks are randomly selected for an inspection. What is the probability that among the selected banks appears inside the city: a) 3 savings banks; b) at least one? The answer, a) 0,348; b) 0,984. 4.13. There are 16 items made by the factory № 1 and 4 items of the factory № 2 at a collector. Two items are randomly taken. Find the probability that at least one of them has been made by the factory № 1. The answer. 92/95.
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4.14. Three buyers went in a shop. The probability that each buyer makes purchases is equal to 0,3. Find the probability that: a) two of them will make purchases; b) all three will make purchases; c) only one of them will make purchases. The answer: a) 0,189; b) 0,027; c) 0,441. 4.15. Three students pass an exam. The probability that the exam will be passed on "excellent" by the first student is equal to 0,7; by the second  0,6; and by the third  0,2. What is the probability that the exam will be passed on "excellent" by: a) only one student; b) two students; c) at least one; d) neither of the students? The answer: a) 0,392; b) 0,428; c) 0,904; d) 0,096. 4.16. Three shots are made in a target. The probability of hit at each shot is equal to 0,6. Find the probability that only one hit will be in result of these shots. The answer: 0,288. LECTURE 5 Theorem of addition of probabilities of compatible events
Two events are compatible if appearance of one of them doesn’t exclude appearance of another event at the same trial. Example. A  appearance of four aces at tossing a die; B  appearance of an even number of aces. The events A and B are compatible. Let events A and B be compatible, and the probabilities of these events and the probability of their joint appearance be given. How can we find the probability of the event A + B consisting in that at least one of the events A and B will appear? Theorem. The probability of appearance of at least one of two compatible
events is equal to the sum of the probabilities of these events without the probability of their joint appearance: P (A +B) = P(A) + P(B) P(AB) Remark 1. Using the obtained formula one should remember that the events A
and B can be both independent and dependent. Forindependentevents: P (A + B) =P(A) +P(B)P(A) P(B) For dependent events: P (A + B) = P(A) + P(B)  P(A) ■ Pa(B) 32
Remark 2. If the events A and B are incompatible then their joint appearance is
an impossible event and consequently, P(AB) = 0. Then for incompatible events A and B, P (A + B) = P(A) + P(B).
Example. The probabilities of hit in a target at shooting by the first and the second guns are equal to: p1 = 0,7; p2 = 0,8 respectively. Find the probability of hit at one shot (by two guns) by at least one of the guns. Solution: The probability of hit in the target by each of guns doesn’t depend on result of shooting by another gun, therefore the events A (hit by the first fun) and B (hit by the second gun) are independent. The probability of the event AB (both the first and the second guns gave hit) P(AB) = P(A) ■ P(B) = 0,7 ∙ 0,8 = 0,56. The required probability is: P (A +B) = P(A) + P(B)  P(AB) = 0,7 + 0,8  0,56 = 0,94. Remark 3. Since in this example the events A and B are independent, we can use the formula P(A + B) = 1  P(A) ∙ P(B) (the probability of appearance of at least one of the events). In fact, the probabilities of the events which are opposite to the events A and B, i.e. the probabilities of misses are: P(A) = 1  P(A) = 1  0,7 = 0,3; P(B) = 1  P(B) = 1  0,8 = 0,2. The required probability that at least one of guns gives hit at one shot is equal to P(A + B) = 1  P(A) ■ P(B) = 1  0,3 ∙ 0,2 = 0,94.
Formula of total probability
Let an event A can be happened only in case of appearance of one of incompatible events B1, B2, ..., Bn which form a complete group. Let both the probabilities of these events and the conditional probabilities 1’,. (A), Pb^ (A),..., P,. (√4)of the event A be known. How can we find the probability of the event Л? The answer on this question gives the following: Theorem. The probability of an event A which can be happened only in case of appearance of one of incompatible events B1, B2, ..., Bn forming a complete group is equal to the sum of products of the probabilities of each of these events on corresponding conditional probability of the event A:
P(A) = P(B1) ∙Pbi (A) + P(B2) ∙Pb> (A) +... + P(Bn) ∙Pb^ (A) This formula is «the formula of total probability».
Example. There are two sets of items. The probability that an item of the first set is standard is equal to 0,8; and of the second set  0,9. Find the probability that a randomly taken item (from a randomly taken set) is standard. 33
Solution: Denote by A the event «an extracted item is standard». An item can be extracted from either the first set (the event Bi) or the second set (the event B2). The probability that an item has been extracted from the first set P(B1) = 1/2. The probability that an item has been extracted from the second set P(B2) = 1/2. The conditional probability that a standard item will be extracted from the first set PB1 (A) = 0,8. The conditional probability that a standard item will be extracted from the first set PBi (A) = 0,9. The required probability that a randomly extracted item is standard is equal by the formula of total probability to P(A) = P(Bi) ∙ Pbi (A) + P(B2) ∙ P⅛ (A) = 0,5 ∙ 0,8 + 0,5 ∙ 0,9 = 0,85.
Example. There are 20 radio lamps (including 18 standard ones) in the first box, and 10 radio lamps (including 9 standard ones) in the second box. A lamp has been taken randomly from the second box and placed to the first box. Find the probability that a lamp randomly extracted from the first box is standard. Solution: Denote by A the event «a standard item has been extracted from the first box». One could be extracted from the second box either a standard item (the event B↑) or a nonstandard item (the event B2). The probability that a standard item has been extracted from the second box P(B1) = 9/10. The probability that a nonstandard item has been extracted from the second box P(B2) = 1/10. The conditional probability that a standard item will be extracted from the first box provided that a standard lamp was placed from the second box to the first one PB1 (A) = 19/21. The conditional probability that a standard item will be extracted from the first box provided that a nonstandard lamp was placed from the second box to the first one PB1 (A) = 18/21. The required probability that a standard lamp will be extracted from the first box is equal by the formula of total probability to 9 19 1 18 P(A) = P(B1) ∙ PBi (A) + P(B2) . Pb (A) =■ + ■ = 0,9. Probability of hypotheses. Bayes’s formulas.
Let an event A can happen only in case of appearance of one of incompatible events B1, B2, ..., Bn forming a complete group. Since it isn’t known beforehand which of these events will happen, we call them by hypotheses. The probability of appearance of the event A is defined by the formula of total probability: P(A) = P(B1). Pbi (A) + P(B2) ■ Pbi (A) +... + P(Bn) ∙ P,, (A) (*)
Assume that a trial has been made in result of which the event A was appeared. Pose the problem to determine how the probabilities of the hypotheses have been changed (in connection with that the event A has already happened). In
34
other words, we will look for the conditional probabilities Pa(Bi), Pa(B2), , Pa(Bπ). Find firstly the conditional probability Pa(Bi). By theorem of multiplication we have P(AB1) = P(A) ∙ Pλ (B1) = P(B1) ∙ Pb^ (A).
Consequently,
Pa(B1) =
P(B1)Pb (A)
л ɪ Replacing P(A) by (*), we obtain
P(A)
P(B1)Pb (A) Pa (Bl) =1 . P(B1) ∙PBi (A) + P(B2) ∙РВг (A) +... +P(Bn) ∙P11 (A)
Formulas determining the conditional probabilities of the rest hypotheses are deduced analogously, i.e. the conditional probability of any hypothesis B1 (i = 1, 2, ..., n) can be calculated by the formula
P(Bi)Pb (A) Pa (Bi) =■ P(B1) ∙ Pbi (A) + P(B2) ∙ Pb (A) +... + P(B ) ∙ Pb (A) The obtained formulas are the Bayes ,s formulas (on name of British mathematician who deduced them; published in 1764). The Bayes’s formulas allow overestimating the probabilities Ofhypotheses after that a trial has been made in result of which the required event has appeared. Example. Items produced by a factory shop are given for checking them on standard to one of two controllers. The probability that an item will be given to the first controller is equal to 0,6; and to the second  0,4. The probability that a suitable item will be recognized standard by the first controller is equal to 0,94; and by the second controller  0,98. A randomly chosen item that is suitable has been recognized standard at checking. Find the probability that this item was checked by the first controller. Solution: Denote by A the event consisting in that a suitable item has been recognized standard. One can make two assumptions: 1) The item has been checked by the first controller (the hypothesis IP)'. 2) The item has been checked by the second controller (the hypothesis B2). Find the required probability that the item has been checked by the first P(B1) ■ PbfA) controller by the Bayes’s formula: Pa (Pi ) P(B1)Pb (A)+ P(B2)Pb (A)
We have: P(B1) = 0,6) P(B2) = 0,4) P131 (A) = 0,94; PBi (A) = 0,98.
p.R._ 0,6∙0,94 The required probability Pa(B1)  0,6.0,94 + 0,4.0,98 ≈ °’59'
35
We see the probability of the hypothesis B↑ equals 0,6 before the trial, and after that a trial result has become known the probability of this hypothesis (more precise, the conditional probability) has been changed and become 0,59. Thus, using the Bayes’s formula allowed overestimating the probability of the considered hypothesis. Glossary compatible events  совместные события; hypothesis  гипотеза beforehand  заранее; to recognize  признавать, распознавать a factory shop  цех завода; suitable item  годная деталь to overestimate  переоценить Exercises for Seminar 5
5.1. There are 20 skiers, 6 bicyclists and 4 runners in a group of sportsmen. The probability to fulfil the corresponding qualifying norm is: for a skier  0,9, for a bicyclist  0,8, and for a runner  0,75. Find the probability that a randomly chosen sportsman will fulfil the norm (to fulfil  выполнить). The answer. 0,86. 5.2. The first box contains 20 items and 15 of them are standard; the second  30 items and 24 of them are standard; the third  10 items and 6 of them are standard. Find the probability that a randomly extracted item from a randomly taken box is standard. The answer. 43/60. 5.3. There are radio lamps in two boxes. The first box contains 12 lamps, and 1 of them is nonstandard; the second box contains 10 lamps, and 1 of them is nonstandard. A lamp is randomly taken from the first box and placed in the second. Find the probability that a randomly extracted lamp from the second box will be nonstandard. Theanswer 13/132. 5.4. At a deviation of an automatic device from the normal operating mode the signaling device CI acts with the probability 0,8, and the signaling device C11 acts with the probability 1. The probabilities that the automatic device is supplied with CI or CIl are equal to 0,6 and 0,4 respectively. A signal about cutting the automatic device has been received. What is more probable: the automatic device is supplied with the signaling device CI or C11? The answer. The probability that the automatic device is supplied with CI, is equal to 6/11, and CIl 5/11. 5.5. The probability for products of a certain factory to satisfy the standard is equal to 0,96. A simplified system of checking on standardness gives positive result with the probability 0,98 for products satisfying the standard, and with the probability 0,05  for products nonsatisfying the standard. A randomly taken 36
product has been recognized as standard at checking. Find the probability that it really satisfies the standard. The answer. 0,998.
5.6. A digit is firstly randomly chosen from the digits {1, 2, 3, 4, 5}, and then the second digit  from the rest four digits. Find the probability that an odd digit will be chosen: a) for the first time; b) for the second time; c) in both times. The answer, a) 3/5; b) 3/5; c) 3/10. 5.7. Three cards are randomly selected without replacement from an ordinary deck of 52 playing cards. Compute the conditional probability that the first card selected is a spade, given that the second and third cards are spades. The answer. 0,22. 5.8. Um A contains 2 white balls and 1 black ball, whereas um B contains Iwhite ball and 5 black balls. A ball is drawn at random from um A and placed in um B. A ball is then drawn from um B. It happens to be white. What is the probability that the ball transferred was white? The answer: 0,8. Exercises for Homework 5
5.9. A collector has received 3 boxes of items made by the factory № 1, and 2 boxes of items made by the factory № 2. The probability that an item of the factory № 1 is standard is equal to 0,8, and the factory № 2  0,9. The collector has randomly extracted an item from a randomly taken box. Find the probability that a standard item has been extracted (a collector  сборщик). The answer. 0,84. 5.10. There are 4 kinescopes in a television studio. The probabilities that the kinescope will sustain the warranty period of service are equal to 0,8; 0,85; 0,9; 0,95 respectively. Find the probability that a randomly taken kinescope will sustain the warranty period of service (to sustain  выдержать). The answer. 0,875. 5.11. A die has been randomly extracted from the full set of 28 dice of domino. Find the probability that the second randomly extracted die can be put to the first. The answer. 7/18. 5.12. For participation in student selective sport competitions 4 students has been directed from the first group, 6  from the second, 5  from the third group. The probabilities that a student of the first, second and third group gets in the combined team of institute, are equal to 0,9; 0,7 and 0,8 respectively. A randomly chosen student as a result of competition has got in the combined
37
team. Which of groups is this student most likely belonged to (a combined team  сборная)? The answer. The probabilities that the student has been chosen from the first, second and third group are equal to 18/59, 21/59, 20/59 respectively.
5.13. English and American spellings are rigour and rigor, respectively. A man staying at a Parisian hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If 40 percent of the Englishspeaking men at the hotel are English and 60 percent are Americans, what is the probability that the writer is an Englishman (rigour (rigor)  суровость; a vowel  гласная)? The answer: 5/11. 5.14. Um A has 5 white and 7 black balls. Um B has 3 white and 12 black balls. We flip a coin. If the outcome is heads, then a ball from um A is selected, whereas if the outcome is tails, then a ball from um B is selected. Suppose that a white ball is selected. What is the probability that the coin landed on tails (to flip  подбросить)? The answer: 12/37. 5.15. There are four urns. The first um contains 1 white and 1 black ball, the second  2 white and 3 black balls, the third  3 white and 5 black balls, and the fourth  4 white and 7 black balls. The event Hi is the choosing the /th um (i = 1, 2, 3, 4). It is known that the probability of choosing the /th um is equal to /710. A ball is randomly extracted from a randomly chosen um. Find the probability that a randomly extracted ball is white. The answer: 0,388.
LECTURE 6 Repetition (recurrence) of trials. The Bernoulli formula
If several trials are made and the probability of an event A for each trial doesn’t depend on outcomes of other trials, such trials are called independent from the event A. At various independent trials an event A can have either different probabilities or the same probability. We will further consider only such independent trials in which the event A has the same probability. We use below the notion of complex event meaning by that overlapping of several events, which are called simple. Let n independent trials will be made in each of which an event A can either appear or not to appear. Assume that the probability of the event A for each trial is the same, namely equals p. Consequently, the probability of nonhappening the event A in each trial is also constant and equals q = 1 p.
38
Let’s pose the problem of calculating the probability that for n trials the event A will happen exactly к times and consequently will not happen n  к times. It is important to underline that it is not required the event A repeated exactly к times in a certain sequence. For example, if the speech is about appearance of the event A three times in four trials then the following complex events are possible:
AAAA, AAAA, AAAA, AAAA. The entry AAAA denotes that the event A happened at the first, second and third trials and it didn’t happen at the fourth trial, i.e. the opposite event A happened; the rest entries have the corresponding sense. Denote the required probability by Pn(k). For example, the symbol P$(3) denotes the probability that the event will happen exactly 3 times for 5 trials and consequently it will not happen 2 times. One can solve the posed problem by means of suchcalled Bernoulli formula. Deduction of the Bernoulli formula: The probability of one complex event consisting in that for n trials the event A will happen к times and will not happen n  к times is equal by the theorem of multiplication of probabilities of independent events to pkqn ^^ k. There can be such complex events as much as combinations of n elements on к elements can be composed, i.e. Ck . Since these complex events are incompatible, by the theorem of addition of probabilities of incompatible events the required probability is equal to the sum of the probabilities of all possible complex events. Since the probabilities of all these complex events are the same, the required probability (of appearance of the event A к times for n trials) is equal to the probability of one complex event multiplied on their number:
PnCO = Ck pkqk
or
ft) LP (.P) = :— p я " k∖(nkE4
Example. The probability that the expense of electric power during one day will not exceed the established norm is equal to p = 0,75. Find the probability that at the closest 6 days the expense of electric power will not exceed the norm for 4 days. Solution: The probability of normal expense of electric power during each of 6 days is constant and equals 0,75. Consequently, the probability of overexpenditure of electric power for each day is also constant and equals q = 1  p = 10,75 = 0,25. The required probability by the Bernoulli formula is equal to P6 (4) = C6W =^.(0,15Y ∙(O,25)2 = 0,30.
39
The most probable number k$ of occurrences of an event in independent trials is determined from the double inequality:
npq=1
Property 1. The mathematical expectation of a constant is equal to the constant: M(C) = C Property 2. A constant multiplier can be taken out for a sign of mathematical expectation, i.e. M(kX) = IM(X) Property 3. The mathematical expectation of the algebraic sum of finitely many random variables is equal to the sum of their mathematical expectations, i.e.
M(∑Xi)=∑M(Xi). /=1
/=1
Property 4. The mathematical expectation of the product of finitely many mutually independent random variables is equal to the product of their mathematical expectations: A∕χl⅛)=∏wυ /=1
/=1
Property 5. The mathematical expectation of deviation of a random variable from its mathematical expectation is equal to zero: M[XM(X)] = 0 Dispersion of a discrete random variable
Although M(X) yields the weighted average of the possible values of X, it does not tell us anything about the variation, or spread, of these values. For instance, although random variables W, Y, and Z, having probability mass functions determined by W=O with probability 1 with probability ɪ 2 + 1 with probability —
1 F=
6.
Polygon and histogram
Variation series are presented graphically by a polygon and (or) a histogram. Polygon offrequencies is a broken line of which segments connect points (x1; П1), (x2; n2), ..., (xk; nk). Polygon of relative frequencies is a broken line of which segments connect points (x1; n1∕n), (x2; n2∕n), ..., (xk; nk∕n). Construct a polygon of frequencies for Example 1:
The figure consisting of rectangles with the basis к and heights ni is called a histogram of frequencies. For a histogram of relative frequencies ni∕n is considered as a height. Construct a histogram of frequencies for Example 2.
Numerical characteristics of variation series
Variation series allow receiving a first representation on a studied distribution. Further it is necessary to investigate numerical characteristics of distribution. Arithmetic mean of a variation series is the sum of products of all variants on the corresponding frequencies divided on the sum of frequencies: m
where xi are variants of a discrete variation series or middles of intervals of an interval variation series; ni are the corresponding to them frequencies; 96
m n = "∑nd m is the number of intervals or nonrepeating variants, wi are the >=1 relative frequencies of variants or intervals. Properties of arithmetic mean 1. If all the results of observations multiply on the same number then the following holds: z =Cx = Cx. 2. If we add to (subtract from) all the results of observations the same number then z = x + C = x + C. 3. If all the frequencies of variants multiply on the same number then arithmetic mean is not changed. Mean linear (absolute) deviation of a variation series is the arithmetic mean of absolute quantities of deviations of variants from their arithmetic mean:
jr∣χ.χ∣w. d = . n For example, if we have the following series: xi 13 6 16 ni 4 10 5 1 then _ _ 4∙l + 10∙3 + 5∙6 + l∙16 _ 80 4 + 10 + 5 + 1 20
4∙ 114  +10∙  341 +5∙  641 +1∙ 1164 _ 20
The mean linear deviation serves to characterize a variance of a variation series.
Median Me of a variation series is the value of the attribute falling on the middle of ranked series of observations. If a discrete variation series has 2n + 1 members in a ranked set: x1, x2, ..., x„, xti+ι, ..., x2κ+ι then Me = xκ+1. If a discrete variation series has 2n members in a ranked set: x1, x2, ..., x„, xti+ι, ..., x2κ then Me = (xn + ‰+ι)∕2. Mode Mo of a variation series is the variant to which the greatest frequency is corresponded. The elementary parameter of a variation is the variation scope R which is equal to the difference between greatest and least variants of the series: R = xmax  xmin.
Example 1. Let the following distribution be given: x1 0 1 2 3 4 ni 6 7 3 5 3 We have:
_
0∙6 + l∙7 + 2∙3+3∙5 + 4∙3 24
97
— = 1,67; 24
M =1; M = —= 1; A = 40 = 4. ‘
°
2
We have the following formulas for median and mode of an interval variation series: lvl e
cum
r∖ r
+ k^^±
M =X
λMιr^ nMe
where xMe is the beginning of the median interval, к is the length of partial interval, n is the set size, ≤,"1 is the cumulative frequency of the interval preceding to the median interval, nMe is the frequency of the median interval. J ʃ _ γ + r^rtMo ~ nMoI ) (nMo  nMoI ) + (nMo  nMM )
where xMo is the beginning of the modal interval, к is the length of partial interval, nMo is the frequency of the modal interval, πmo.1 is the frequency of the premodal interval, and nMo+1 is the frequency of the postmodal interval.
Example 2. Let the following interval series be given: Intervals with length к = Frequency ni 1,6 45,6 5 5,67,2 17 7,28,8 9 8,810,4 15 10,412,0 10 12,013,6 1 13,615,2 3 60 Σ We have: 0,56022 Me =7,2 + 1,6 ∙ = 8,62; 9 (175) Mo =5,6 + 1,6 ∙ = 6,56. (175)+ (179) Mode and median are used as a characteristic of average position in case when the boundaries of a series are fuzzy or a series is not symmetrical. Dispersion D of a variation series is the arithmetic mean of squares of deviations of variants from their arithmetic mean: m
_
Σ(*,*)4 D = —
98
m = ∑(x,  x)2 wi■
Mean square deviation σ of a variation series is the arithmetic value of square root of its dispersion:
Coefficient ofvariation'. V  — ■ 100% (x ≠ 0).
Coefficient of variation characterizes the relative value of the mean square deviation and usually serves for comparing variance quantities according to sample mean of two variation series: one of these series has greater variance if its coefficient of variance is greater. Properties of dispersion 1. If we add to (subtract from) all the results of observations the same number C then dispersion is not changed. 2. If all the results of observations multiply on the same number C then dispersion is increased in C2 times. 3. If all the frequencies of variants multiply on the same number then dispersion is not changed.
The rule of addition of dispersions If we combine several distributions in one “new” distribution, the general dispersion of the new distribution is equal to the sum of the intragroup dispersion and the intergroup dispersion:
Xi
X2
Xm
Σ
«11
«12
^lm
«1
«21
«22
^2m
«2
«31
«32
^3m
«3
«И
«к2
W km
nk
N1
N2
Nm
N
Z 1 2 3
к Σ
Σ(⅞⅛)2^∙ Dgen
= Ant™ + Antrr
yχ2w
or Dgen = — = Nl  (xgen
99
where nij is the frequency of the уth variant of the zth partial distribution (/ = 1, ...,m; i= 1,2, ...Д);
xij is the уth variant of the i th partial distribution (j = 1, ..., m; i = 1, 2, ..., k)', ni is the size of the zth partial distribution;
N = ∑nlj is the frequency of the уth variant of the new distribution; N = ∑nl is the size of the new distribution;
Σ" x xl = — is the arithmetic mean of the zth partial distribution (i = 1,2, ...,ky, nl V xiN . x = j j is the anthmetιc mean ot the new distribution; g" N
∑x(,nv
D1 = 1 (χz )2 is the dispersion of the zth partial distribution; tlι Dintra =
ɪ8 ɑ16 intragroup dispersion;
Dinter = ———l is the intergroup dispersion; Example. Let be given the following distribution consisting of two groups: The first group The second group xi
xi
ni
ni
1 3 2 4 7 8 3 5 2 Find the group dispersions, the intragroup dispersion, the intergroup dispersion and the general dispersion of the distribution. Solution: We have: jV1=∑zzi =10, N2 =∑ni =5, N = N1 + N2 = 10 + 5 = 15. Find the group means: ¾ = {∑ntxt')∣γnt = (1∙2+ 7 ∙4+ 2∙ 5)/10 = 4; x2 = (2∙3 + 3∙8)∕5 = 6. Find the required group dispersions: D1 =f∑nEχl∙M')v∣ = (l∙(24)2 +7∙(44)2+2∙(54)2)∕10 = 0,6; D2 = (2∙ (36)2+3∙ (86)2)∕5 = 6. Findthe intragroup dispersion: Dintra =(N1D1 +N2D2)∕N = (10∙0,6 + 5∙6)∕15 = 12/5. 1
∑n.x.
l∙2 + 7∙4 + 2∙5+2∙3 + 3∙8 14 15 ^ 3 ' Find the required intergroup dispersion:
Find the general mean: xgen
100
= = 1Q.(414∕3)2 + 5(614∕3)2 = 8 """ n 15 9' Find the general dispersion: Dgen = Diatra + Diater = 12/5 + 8/9 = 148/45. m
^Xtni
Initial moment Mt of the кth order of a variation series is Mk = —m
∑xM
In particular, M1 = λl— = x.
Central moment mt of the кth order of a variation series is ∑(χi x)kni ¾ =—∙ m
_
∑(xi x)2ni
Obviously, ∞ι = θ= m2 = ~ “ = ʃɔWhen studying distributions other than normal, there is a need to quantify this difference. For this purpose special characteristics are entered, in particular, asymmetry and excess. These characteristics are equal to zero for the normal distribution. Therefore, if for a studying distribution the asymmetry and excess have small values, we can assume the closeness of this distribution to normal. On the contrary, great values of asymmetry and excess indicate a significant deviation from the normal. m
_
∑(xixfni
Coefficient of asymmetry of a variation series is as = —=
Excess
(or
coefficient m _ Y(xix)4ni
of
excess')
of
a
Consider Example 2. Construct the following table: Groups of Average Frequency xini О, *У", farms on value of (the amount of interval number of workers (a) farms in on 100 the group) hectares ni 45,6 4,8 5 24 72,708 101
 ɜ ∙
is
variation
series
z —∖ 3 I X1  X 1 nι ∖ σ )
( X, nι ∖ σ J
18,954
29,554
Λ,Y
5,617,2 7,218,8 8,81 10,4 10,41 12,0 12,01 13,6 13,61 15,2 Σ
6,4 8 9,6
17 9 15
108,8 83,28 72 3,386 144 14,603
12,601 0,142 0,985
11,404 0,036 0,397
11,2
10
112
66,908
11,832
12,514
12,8
1
12,8
17,528
5,017
8,588
14,4
3
43,2
100,457
39,74
94,03

60
516,8 358,869
25,876
156,523
Wehave: j . ⅛ .≡.8.6,3; θ.⅛⅛.≡.⅞,>t
и
60
и
60
V(x^)2wi ,σ 2 446  — = √5,981 = 2,446; V =  • 100% =   100% = 28,4%. п х 8,613
Thus, the average number of employees per 100 ha of farmland on analyzing set of economies amounted to 8,61 persons. The density of employees on average varied in the interval *±σ = 8,61±2,45, i.e. from 6,16 to 11,06 persons per 100 ha of farmland. This interval and the coefficient of variation indicate that there are big differences in the availability of economies by labor. m _ (xi x)3ni
V ' '_ 25,876 Coefficient of asymmetry: as  : —~—  0,43. nσ 60 V (x x)4n 156,523 l eek  Excess: 3 3  —— 3  u,39. nσ 60
The found value of coefficient of asymmetry (not sufficiently close to zero) specifies that the distribution is not symmetrical. The excess is also different from zero indicating a possible difference from normal distribution. Glossary gathering  сбор; gathering data  сбор данных; nominating  выдвижение ordering, systematization  систематизация; processing  обработка data processing  обработка данных; observation  наблюдение revealing  выявление; attribute  признак; sampling  выборочный метод sample  выборка; census  перепись parent population  генеральная совокупность; essence  сущность judgment  суждение; destruction  разрушение inevitable  неизбежный; representative  представительный 102
properly  собственно; occurring  происходящий; wheat  пшеница farm  хозяйство; shift  смена; to rank  ранжировать frequency  частота; cumulative frequency  накопленная частота inexpedient  нецелесообразный arithmetic mean  среднее арифметическое; disparate  несоизмеримый variance  рассеяние; excess  эксцесс farmland  сельскохозяйственные угодья economy  хозяйство; labor  рабочая сила Exercises for Seminar 12
12.1. The sample is given as a distribution of frequencies: xi 2 5 7 ni 13 6 Find the distribution of relative frequencies. 12.2. Find the empirical function of the following distribution: xi 2 5 7 8 ni 13 2 4 12.3. Construct the polygon of frequencies for the following distribution: xi 2 35 6 ni 10 15 5 20 12.4. Construct the polygon of relative frequencies for the following distribution: x, 2 4 5 7 10 wi 0,15 0,2 0,1 0,1 0,45 12.5. Construct the histogram of frequencies for the following distribution: The number of interval The partial interval Frequency i ∏i XiXi + 1 1 15 10 2 59 20 3 913 50 4 1317 12 5 1721 8 12.6. Construct the histogram of relative frequencies for the following distribution: The number of interval The partial interval Frequency i ∏i XiXi + 1 2 1 1015 2 1520 4 3 2025 8 4 2530 4 5 3035 2 103
12.7. Find the group means of a population consisting of two groups: the first group ... Xi 0,4 0,6 0,1 ni 3 2 5 the second group .. . Xi 0,3 0,4 0,1 10 4 6 ni 12.8. Find the intragroup, intergroup and general dispersions of a population consisting of three groups: the first group ... Xi 1 2 8 ni 30 15 5 the second group .. . Xi 1 6 ni 10 15 the third group ... Xi 3 8 20 5 ni 12.9. There are 100 workers at an enterprise according to the list who have the following categories: 1, 5, 2, 4, 3, 4, 6, 4, 5, 1, 2, 2, 3, 4, 5, 3, 4, 5, 2, 1, 4, 5, 5, 4, 3, 4, 6, 1, 2, 4, 4, 3, 5, 6, 4, 3, 3, 1, 3, 4, 3, 1, 2, 4, 4, 5, 6, 1, 3, 4, 5, 3, 4, 4, 3, 2, 6, 1, 2, 4, 5, 3, 3, 2, 3, 6, 4, 3, 4, 5, 4, 3, 3, 2, 6, 3, 3, 4, 5, 4, 4, 3, 3, 2, 1, 2, 1, 6, 5, 4, 3, 2, 3, 4, 4, 3, 5, 6, 1,5. Compose the series of distribution of workers on categories. Find cumulative and relative frequencies. Determine the average category of a worker, the modal and median category, the dispersion and the mean square deviation.
12.10. There are the following conditional data on size of quotas in millions dollars of 100 countries  members of the International currency fund: 353 326 344 324 339 332 324 344 349 352 348 316 329 354 358 302 325 324 351 333 341 312 331 351 304 345 332 382 342 351 396 341 353 318 325 354 338 321 398 359 376 355 382 342 374 354 358 332 368 343 344 376 324 339 372 366 381 334 369 332 371 312 334 361 304 362 354 366 378 348 352 362 356 364 372 342 344 346 353 334 336 364 352 348 347 368 329 335 363 312 378 342 354 363 361 366 354 364 348 351 It is required: a) Compose the interval variation series taking the beginning of the first interval equal 300, and the width of each interval equal 10: b) Construct the histogram and the polygon of relative frequencies of distribution; c) Find the mode and the median of the variation series; 104
d) Find empirical function of distribution of the variation series and construct its graph.
12.11. Let the following distribution of employers by an experience of their work be given: Distribution of employers on the experience of work The experience of Up to 1 15 510 1020 2040 work, years Number of employers 8 12 16 14 10
Total 60
Find the average experience of work, the mean square deviation and the coefficient of variation (experience  стаж). Exercises for Homework 12
12.12. The sample is given as a distribution of frequencies: xi 4 7 8 12 «,523 10 Findthedistributionofrelativefrequencies. 12.13. Find the empirical function of the following distribution: xi 4 7 8 ni 5 2 3 12.14. Construct the polygon of frequencies for the following distribution: xi 15 20 25 30 35 ni 10 15 30 20 25 12.15. Construct the polygon of relative frequencies for the following distribution: x1 1 4 5 8 9 wi 0,15 0,25 0,3 0,2 0,1 12.16. Construct the histogram of frequencies for the following distribution: The number of The partial interval Frequency interval ∏i XiXi + 1 i 1 27 5 2 712 10 3 1217 25 4 1722 6 5 2227 4 12.17. Construct the histogram of relative frequencies for the following distribution: The number of The partial interval Frequency interval ∏i XiXi + 1 i 1 25 6 105
2 3 4
10 4 5
58 811 1114
12.18. Find the intragroup, intergroup and general dispersions of a population consisting of two groups: the first group ... xt 2 7 ni 6 4 the second group... xi 2 7 ni 2 8 12.19. By oral interrogation the quality of production released by a firm and sold in a shop of this firm was studied. Visitors were estimating the quality by a tenmark scale. The summary data have been received. Mark estimation of production of the enterprise Estimation of quality of 12 34 56 78 production, point Number of cases 3 8 36 89
910
45
Determine the average mark of quality of production, the mean square deviation, the coefficient of variation, the parameters of asymmetry and excess (interrogation  опрос; to release  выпускать; visitor  посетитель).
12.20. Carry out the analysis of the data of annual levels of profit of three companies: Year 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992
«Lemon Motors»
«Cherry Computers» 14,2 12,3 16,2 15,4 17,2 10,3 6,3 7,8 3,4 12,2
6,2 13,3 8,4 27,3 28,2 14,5 2,4 3,1 15,6 18,2
«Orange Electronics» 37,5 10,6 40,3 5,4 6,2 10,2 13,8 11,5 6,2 27,5
Find the average value and the mean square deviation of the profit for each of the companies. Compare the received results of their activity for 10 years. Which of the companies, in your opinion, is the activity more successful for? 106
12.21. There are the following data on a number of industrial subdivisions on each of 100 agricultural enterprises: 2, 4, 5, 3, 4, 6, 7, 4, 5, 3, 3, 4, 2,6, 5, 4, 7, 2, 3, 4, 4, 5, 4, 3, 4, 6, 6, 5, 2, 3, 4, 3, 5, 6, 7, 2, 4, 3, 4, 5, 4, 6, 7, 2, 5,3, 5, 4, 3, 7, 2, 4, 3, 4, 5, 4, 3, 2, 6, 7, 6, 4, 3, 2, 3, 4, 5, 4, 3, 5, 4, 3, 2, 6, 4, 5, 7,5, 4, 3, 4, 5, 7, 4, 3, 4, 5, 6, 5, 3, 4, 2, 2, 4, 3, 7, 5,6, 4,5. Compose the series of distribution of the agricultural enterprises by number of industrial subdivisions for one economy. Find cumulative and relative frequencies. Determine the average number of industrial subdivisions for one economy, the modal and median values of the number of subdivisions, the dispersion and the mean square deviation (economy  хозяйство). 12.22. The administration of a supermarket is interested in the optimal level of stocks of products in a trading hall, and also the monthly average volume of purchases of the goods which are not subjects of daily consumption in a family (for example such as soda). For findingout of this question the manager of a supermarket within January was registering the frequency of purchases of 100gramme packages with soda and has collected the following data xi∖ 4, 4, 9, 3, 3, 1, 2, 0, 4, 2, 3, 5, 7, 10, 6, 5, 7, 3, 2, 9, 8, 1, 4, 6, 5, 4, 2, 1, 0, 8 Construct the variation series, determine its numerical characteristics. Which recommendations would you give to the administration of the supermarket (stock  запас; to find out  выяснить)?
LECTURE 13 Statistical estimations of parameters of distribution
Mathematical theory of sampling is based on properly random sample. Arithmetic means of distribution of an attribute in a parent and sample populations are parent and sample means respectively, and dispersions of these distributions parent and sample dispersions. The major problem of sampling is estimating parameters (characteristics) of a parent population by sample data. Let a quantitative attribute of a parent population be required to study. Assume we have established that the attribute has a certain distribution. The task of estimation of parameters of the distribution arises naturally. For example if it is known that a studied attribute is normally distributed, it is necessary to estimate (to find approximately) mathematical expectation and mean square deviation since these parameters completely determine a normal distribution. And if there is reason to consider that the attribute has Poisson distribution, it is necessary to estimate the parameter λ which determines this distribution.
107
A researcher usually has only data of a sample, for example the values of a quantitative attribute x1, x2, ..., x„ received in result of n observations (hereinafter we assume that observations are independent). An estimated parameter is expressed by these data. Considering x1, x2, ...,xn as independent random variables A1, X2, ..., Xn we can say that to find statistical estimation of an unknown parameter of a theoretical distribution is to find function of observed random variables which gives an approximate value of the estimated parameter. Formulate the problem of estimation of parameters in a general form: Let distribution of an attribute X of a parent population be given by function of probabilities φ(xi, θ) = P(X = xi) (for a discrete random variable X) or probability density φ(x, θ) (for a continuous random variable X) which contains an unknown parameter θ. For example, it is the parameter λ for Poisson distribution or the parameters a and σ2 for the normal law of distribution and etc. One tries to judge on a parameter θ by a sample consisting of values (variants) x1, x2, ..., xn. These values can consider as partial values (realizations) of n independent random variables X1, X2, ..., Xn each of which has the same law of distribution with the random variable X. An estimator θ* of a parameter θ is a function of results of observations over a random variable X by means of which one judges on value of the parameter θ:
θ* = θ*(JXl,X1,...,Xn). A function of results of observations (i.e. a sample function) is called statistic. It can be said that an estimator θ* of a parameter θ is statistic which is close in certain sense to true value of θ. In order to statistical estimators gave “good” approximations of estimated parameters it is necessary that these estimators satisfy the certain requirements. Let θ* be a statistical estimator of an unknown parameter θ of a theoretical distribution. Assume that an estimator 0’ has been found on a sample of size n. Repeat the experience, i.e. extract another sample of the same size from the parent population and find an estimator 02* on its data. Repeating the experience many times, we obtain numbers θ2 ,θ,2,...,θk that are in general different between them. Thus, the estimator θ* can be considered as a random variable, and the numbers θ,1 ,θ2,...,θ,k as its possible values. Suppose that the estimator provides an approximate value θ with a surplus. Then each number θ'(i = ∖,2,...,k) found on the data of samples is greater than the true value θ. It is clear that the mathematical expectation (the mean value) of the random variable θ* is greater than θ, i.e. M(θ*) > θ. Obviously, if θ* provides an estimation with a lack then M(θ*) < θ. Thus, an using a statistical estimator of which mathematical expectation is not equal to an estimated parameter provides systematical mistakes (In theory of mistakes of measuring 108
nonrandom mistakes distorting results of measuring to one certain direction are called systematical}. Therefore it is naturally to require the following: M(θ*) = θ. An estimator θ* of a parameter θ is unbiased if its mathematical expectation is equal to the estimated parameter, i.e. M(θ*) = θ. Otherwise it is biased. The requirement of unbiasedness guarantees an absence of regular mistakes at estimating. Also this requirement is important for a small number of observations (experiences). If for a finite volume of sample n, M(θ*) ≠ θ, i.e. the bias (displacement) of an estimator b(θ*')=M(θ*')θ ≠0, but Iim 6(6>*) = 0, then such an estimator θ* n→∞ is said to be asymptotically unbiased. However, it would be a mistake to believe that an unbiased estimator always provides a good approximation of an estimated parameter. Indeed, the possible values of 6* can be strongly dispersed around of its average value, i.e. dispersion D(θ*) can be big. In this case an estimator 0’ can be too distant from the average value θ *, consequently from the estimated parameter θ, therefore if we accept 61, as an approximate value of θ then we obtain a big mistake. If we require that dispersion of θ* be small then the possibility to obtain a big mistake will be eliminated. Therefore, the requirement of efficiency is added to a statistical estimator. A statistical estimator θ* of a parameter θ is efficient if it has the least dispersion among all possible estimators of the parameter θ calculated on samples of the same volume n. At considering samples of a big size (n is very great) the requirement of consistency is presented to statistical estimators. An estimator θ* of a parameter θ is consistent if it satisfies the law of large numbers, i.e. it converges on probability to the estimated parameter: θ * θ
< v) = 1.
Iim P^
In case of use of consistent estimators an increasing of sample volume is justified since for this significant mistakes become improbable at estimating. Therefore, only consistent estimators have a practical sense. It is desirable as statistical estimators ofparameters of a parent population to use the estimators satisfying simultaneously requirements of unbiasedness, consistency and efficiency. However to reach it is not always possible. One can appear that for simplicity of calculations it is expedient to use insignificantly biased estimators or estimators possessing the greater dispersion in comparison with effective estimators, etc. Pointwise estimators of mathematical expectation and dispersion
Let a random variable X with mathematical expectation a = M(X) and dispersion D(X) be studied. Statistic used as an approximate value of unknown parameter
109
of parent population is its pointwise estimator, i.e. a pointwise estimator of characteristic of parent population is a number defined by a sample. Let x1, x2, ..., x„ be a sample obtained in result of conducting n independent observations for the random variable X. To underline a random character of quantities x1, x2, ...,xn rewrite them as Xλ, X2, ...,Xn, i.e. by X1 we mean the value of the random variable X in the /th experience. The random variables Xλ, X2, ..., Xn can be considered as n independent “exemplars” of X. Therefore M(X) = M(X) = a, D(X) = D(X). Theorem. Let Xl, X2, ..., Xn be a sample from a parent population and M(X) =
TxX M(X) = a, D(X) = D(X) (i =1, ■■■, n). Then the sample mean xs =— is an n unbiased and consistent estimator of mathematical expectation (parent mean). Remark 1. If original variants xi are large numbers then for simplification of
counting it is expedient to subtract the same number C from each variant, i.e. go to conditional variants ui = xiC (it is profitable to take as C the number close to the sample mean; because the sample mean is unknown, the number C is picked approximately). Then 1) ¾ = C+^fnlul∖n.
2) Dispersion is not changed: Ds(x) = Ds(u) = u2 ∖uf =
ʌ , ,
TfVf
n n Remark 2. If original variants are decimal fractions with к decimal sings after comma, then in order to avoid any actions with fractions, we multiply the original variants on a constant number C = 10*. i.e. go to conditional variants ui = Cxi. For this the dispersion is increased in C2 times. Therefore at finding the dispersion of conditional variants it needs divide this dispersion on C2: Ds(x)=Ds(u)IC2. ∑(¾ ¾)4 Theorem. The sample dispersion Ds = — is a biased estimator of the
n _ɪ parent dispersion D since M(Ds) = D . p n p Therefore we revise the sample dispersion by multiplying on nl(n  1) and
obtaining the formula: s1 = —— ∙ Ds. и1
Corollary. The revised sample dispersion s2 = D=^——;—г— is an и1
unbiased estimator of the parent dispersion.
110
и1
X ' γι χ½ _ \ И Х
/И
It is more convenient the following formula: s2=^———‘ ,j—. In и1 ∖ ’ yi — Γ∖ ’ 77 и F / 77 conditional variants it has the following form: s2u = ʌ , ,—lz" , ,j—, and for и1 this if ui = xiC then s2x = s2u,if ui = Cxi then s2x = s2u IC2.
Example. A sample of size 50 is extracted from a parent population: variant xi 2 5 7 10 frequency ni 16 12 8 14 Find an unbiased estimator of the parent mean. Solution: An unbiased estimator of the parent mean is the sample mean: ∑*Λ xs = ≡1 = (16∙2 + 12∙5 + 8∙7+14∙10)∕50 = 5,76. n
Example. Find the sample mean for the following distribution of sample of size n = 10: xi 1250 1270 1280 ni 2 5 3 Solution: The original variants are large numbers; therefore we go to conditional variants: ui =xi  1270 Thus, we obtain the following distribution of conditional variants: ui 20 0 10 ni 2 5 3 Find the required sample mean: V u,nl xs=C + sɪ = 1270 + n
10
= 1270 1 = 1269.
Example. For a sample of size и = 41 a biased estimator Ds = 3 of the parent dispersion has been found. Find an unbiased estimator of dispersion of parent population. n
41
Solution', s2 = D, = 3 = 3,075. иl 40
Example. As a result of five measurements of length of a rod by one device (without systematical mistakes) the following results have been obtained (in mm): 92; 94; 103; 105; 106. Find: a) sample mean of length of the rod; b) sample and revised dispersions of mistakes of the device (rod  стержень). Solution: a) Find sample mean: xs = 92 + (0 + 2 +11 +13 +14) / 5 = 92 + 8 = 100. b) Find sample dispersion:
111
Ds =
∑(x,xj2
 — = [(92 IOO)2 + (94 IOO)2 + (103 IOO)2 + n + (105100)2 +(106100)2]∕5 = 34. 5
и
Find the revised dispersion: 52 = D =  ∙ 34 = 42,5. f и1 ‘ 4 Example. Find sample dispersion for the following distribution of sample of size n = 10: xi O5Ol 0,04 0,08 ni 5 3 2 Solution: In order to avoid any actions with fractions go to conditional variants ui = 100x,. Thus, we obtain the following distribution: ui 14 8 ni 5 3 2 Find the sample dispersion of conditional variants: ∑w,⅝~∣2 _ 5∙l2+3∙42+2∙82 _(5∙l + 3∙4 + 2∙8Y ∑ntr Ds(u)=^ l l и _ 10 I4 10 J n
Find the required sample dispersion of the original variants: Ds(x) = Ds (и) ∕1002 = 7,21/10000 = 0,0007. Example. Find the revised sample dispersion for the following distribution of Sampleofsizew = IO: xi 0,01 0,05 0,09 ni 1 3 5 Solution: In order to avoid any actions with fractions go to conditional variants ui = 100x,. Thus, we obtain the following distribution: ui 15 9 ni 2 3 5 Find the revised sample dispersion of conditional variants: 2
∑w,w,2[∑",w,]2∕w
su =   ■
и1
Substituting the data of the example in this formula we get: 52 = 10,844. Find the required revised dispersion of the original variants: s2 = S2
∕1002 = 10,844/10000 ≈ 0,0010844.
Glossary sampling  выборочный метод; sample  выборка census  перепись; parent population  генеральная совокупность essence  сущность; judgment  суждение; inevitable  неизбежный representative  представительный; estimator  оценка surplus  избыток; lack  недостаток; to distort  искажать unbiased  несмещенная; unbiasedness  несмещенность consistent  состоятельная
112
Exercises for Seminar 13
13.1 A sample of size 60 is extracted from a parent population: variant x, 1 3 6 26 frequency ni 8 40 10 2 Find an unbiased estimator of the parent mean. The answer: 4 13.2 Find the sample mean for the following distribution of sample of size « = 20: xi 2560 2600 2620 2650 2700 ni 2 3 10 4 1 Direction: Go to conditional variants. The answer: 2621.
13.3. For a sample of size n = 51 a biased estimator Ds = 5 of the parent dispersion has been found. Find an unbiased estimator of dispersion of the parent population. The answer. 5,1. 13.4. As a result of four measurements of some physical quantity by one device (without systematical mistakes) the following results have been obtained: 8; 9: 11; 12. Find: a) the sample mean of results of measurements; b) the sample and revised dispersions of mistakes of the device. The answer: a) xs =10; b) Ds =2,5; s2 =10/3. 13.5. The results of measuring the height (in centimeters) of randomly selected 100 students have been obtained: Height 154158162166170174178158 162 166 170 174 178 182 The number 10 14 26 28 12 8 2 of students Find the sample mean and sample dispersion of height of these students. Direction: Find the middles of intervals and take them as variants. The answer: xs = 166; Ds = 33,44. 13.6. Find the n = 10: xi 186 ni 2 The answer: Ds 13.7. Find the n = 50: xi 0,1 ni 5 The answer: Ds
sample dispersion for the following distribution of sample of size 194 3
192 5 =8,04.
sample dispersion for the following distribution of sample of size
0,5 15
0,6 20
0,8 10
= 0,0344.
113
13.8. Find the revised sample dispersion for the following distribution of sample of size и = 20: xi O5I 0,5 0,7 0,9 ni 6 12 1 1 The answer: s1 2 = 0,0525. Exercises for Homework 13
13.9. Find the «=100: xi 340 ni 20 The answer: D,
sample dispersion for the following distribution of sample of size
360 50
375 18
380 12
= 167,29.
13.10. Find the sample dispersion for the following distribution of sample of size n = 50: xi 18,4 18,9 19,3 19,6 ni 5 10 20 15 Direction: Go to the conditional variants ui= 10x, 195. The answer: Ds = 0,1336. 13.11. Find the revised sample dispersion for the following distribution of sample of size n = 10: xi 23,5 26,1 28,2 30,4 ni 2 3 4 1 The answer: s2 = 4,89. 13.12. Find the revised sample dispersion for the following distribution of sample of size n = 100: xi 1250 1275 1280 1300 ni 20 25 50 5 The answer: s2 = 168,88.
LECTURE 14 Methods of finding of estimations
1. Method of moments. According to the method of moments offered by K. Pearson for finding pointwise estimators of unknown parameters of a given distribution is to equate a certain amount of the sampling moments (initial Mk
either central mk, or those and others) to the corresponding theoretical moments of distribution (v k or μk) ofa random variable ofX.
114
Remind that the sampling moments Mk and mk are determined by the formulas: m m ∑xM ∑(xix)kni
Mk = — , mk =
 ,
n corresponding
n theoretical
and the them vk =M(Xk), μk =M[X M(X))1}.
moments

vk =∑xi P Pt =∑(χ∣ аУр< /=1
/=1
for a discrete random variable with the function of probabilities pi = φ(xi, θ); vk = ʃ xkφ(x, θ)dx,
μk = ʃ (x  a)k φ(x, θ)dx
for a continuous random variable with density of probabilities φ(x, θ)), where a = M(X). If a distribution is determined by only one parameter then for its finding one theoretical moment is equated to one empirical moment of the same order. For example, we can equate an initial theoretical moment of the first order to the initial empirical moment of the first order: v1 =MvTaking in account that v1 =M(X) and M1 =xs, we obtain: M(X) = χ, (*) Mathematical expectation is a function of an unknown parameter of the given distribution, therefore solving the equation (*) with respect to the unknown parameter we obtain its pointwise estimator. If a distribution is determined by two parameters then we equate two theoretical moments to the corresponding empirical moments of the same order. For example, we can equate the initial theoretical moment of the first order to the initial empirical moment of the first order and the central theoretical moment of the second order to the central empirical moment of the second order: v1 = M1, μ2 =m2. Taking in account that v1 = M(X), M1=xs, μ2= D(X), m2 =Ds, we have: [M(X) =x , J v 7 /**ʌ [D(X) = Dv The left parts of these equalities are functions of unknown parameters, therefore solving the system (**) with respect to the unknown parameters we obtain their pointwise estimators. Of course, to calculate sample mean xsand sample dispersion Ds we need a sample x1, x2, ..., x„. Example. A random variable X is distributed under the law of Poisson: PmDO = IMDxJ
115
where m is the number of trials made in one experience; xi is the number of appearances of an event in the /th experience. Find by method of moments on a sample xb x2, ..., xn the pointwise estimator of the unknown parameter λ determining the Poisson distribution. Solution: It is required to estimate one parameter; therefore it is sufficiently to have one equation with respect to this parameter. Equate the initial theoretical moment of the first order vɪ to the initial empirical moment of the first order Λ∕1: V1 =Mr Taking in account that rl = M(X), M1 = ¾, we obtain M(X) = ¾. Also taking in account that the mathematical expectation of the Poisson distribution is equal to the parameter λ of this distribution, we finally have: λ = xs. Thus, the pointwise estimator of the parameter λ of the Poisson distribution is the sample mean: 2* = ¾.
Estimators of the method of moments are usually consistent', however they are not "best" on efficiency, their efficiencies e(θ*) are often less than 1. Nevertheless, the method of moments is frequently used in practice since it results in rather simple calculations. 2. Method of maximal (the greatest) plausibility. The basic method of
obtaining estimations of parameters of parent population on sample data is the method of maximal plausibility offered by R. Fisher. The method of maximal plausibility of pointwise estimator of unknown parameters of a given distribution is reduced to finding maximum of a function of one or several estimated parameters. A. Discrete random variables. Let X be a discrete random variable which in result of n experiences has taken on possible values x1, x2, ..., xn. Assume that the form of law of distribution is given, but a parameter θ determining this law is unknown; it is required to find its pointwise estimator θ* = θ*(xl,x2,...,xn). Denote the probability that in result of trial the variable X take on a value xi by p(x; θ). The function of plausibility of a discrete random variable X is called the function of argument ft L(xγ,x2,...,xi,...,xn',θ) = p(xγ,θ)∙ p(x2,θ)∙ ...■ p(xi,θ)∙ ...■ p(xn,θ), or L(xl,x2,...,xi,...,xn',θ) = ∏Xx,,ft). /=1
An estimator of the greatest plausibility of parameter θ is called such its value ft * at which the function of plausibility reaches a maximum.
116
Functions L and In L reach a maximum for the same value θ, therefore instead of finding a maximum of the function L we find (it is frequently convenient) a maximum of the function In L. The logarithmic function ofplausibility is called the function In L. A point of maximum of a function In L of argument θ can be found, for example, as follows:
JlnL
1. Find the derivative —ττ ∙ dθ 2. Equate the derivative to zero and find the critical point θ *  the root of the obtained equation (it is called an equation ofplausibility}. d1 3. Find the second derivative —~~— ∙1Γ the second derivative for θ = θ*is Ciu negative then θ * is a point of maximum. The found point of maximum θ * is taken as an estimator of the greatest plausibility of the parameter θ.
InL
B. Continuous random variables. Let X be a continuous random variable which in result of n trials has taken on the values x1, x2, ..., xn. Assume that the form of density of distribution  the function f(x)  is given, but the parameter θ determining this function is unknown. The function of plausibility of a continuous random variable X is called the function of argument ft L(x1,x2,...,xi,...,xn;&) = f{xλ,θ')∙ f(x2,θ)...∙ f(xiβ}...f{xn,θ'), An estimator of the greatest plausibility of an unknown parameter of distribution of a continuous random variable is found analogously as in case of a discrete random variable. If density of distribution f(x) of a continuous random variable is determined by two unknown parameters θλ and θ2 then the function of plausibility is a function of two independent arguments θλ and 02. L(pcλ,x2,...,xi,...,xnff,θ2f =f(x1,θ1,θ2>fCc2,θ1,θ2)∙f^M∖ Further the logarithmic function of plausibility is found, and for finding its maximum the following system is composed and solved:
⅛ = 0, ∂θl '^ = 0.
∂θ1
Example. Find by method of maximal plausibility a pointwise estimator of unknown parameter p (probability of appearance of an event in one trial) of a binomial distribution:
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Pm(χi) = Cxmpx∙dpΓx∙ where xi is the number of appearances of an event in the ith experience (i = 1, 2, ..., ri), m is the number of trials in one experience, n is the number of experiences. Solution: Compose the function of plausibility: L=p(x1',θ)p(x2',θ)...p(xn',θ) Taking in account that θ = p and Pm (X = x,) = Cm px,(i p)m x, we obtain: L = [c>4 (1  p)^ ]∙ &p* (1  pΓ' ]∙... ∙ [cx"px" (1  pyχ∙ ]
or P =(CxiCx2 (Jx^y pxl+x2++x .Q_ pym 30, the critical point χ2r (a;k)
can be found from the WilsonGilferti equality: X2 (α; к) = к ∙ [1  21 ,)k + za √2∕(9⅛) ]3 where zα is found using Laplace function (Appendix 1) from the equality Φ(zα) = (l2α)∕2.
Example. A sample of size n = 21 has been extracted from a normal parent population and the revised sample dispersion s2 = 16,2 has been found. It is required to test the null hypothesis H0. σ2 =σ2 =15, at the significance level
0,01 taking H1'. σ2 >15 as the competing hypothesis.
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Solution’. Find the observed value of criterion: χ°bs = fL_J_L_ = σ0
= 21д 15
By the hypothesis of the example the competing hypothesis has the following form: σ2>15. Therefore the critical area is righthanded (Rule 1). By Table (Appendix 4), the significance level 0,01 and the number of degrees of freedom & = «1=211= 20 find the critical point χ2r (0.01: 20) = 37,6. Since χ2bs < XH , there is no reason to reject the null hypothesis on equality of parent
dispersion to the hypothetical value σ2 = 15. In other words, the difference between the revised dispersion (16,2) and hypothetical parent dispersion (15) is not significant.
Example. A batch of items is accepted if the dispersion of controllable size does not exceed 0,2 significantly. The revised sample dispersion found on a sample of size n = 121 is ʌɪ =0,3. Can we accept the batch at the significance level 0,01? Solution: The null hypothesis H0. σ2 = σ0 = 0,2. The competing hypothesis H1: σ2 > 0,2.
π ∙ ,ʌ, л value ι otг criterion: t ■ v2b, = ;— =(121—1)∙0,3 = 180. F md the observed Z„ σ0 0,2 The competing hypothesis has the form: σ2 > 0,2. Consequently the critical area is righthanded. Since there is no the number of degrees of freedom к = 120 in Table (Appendix 4), find the critical point approximately from the WilsonGilferti equality. Find firstly (taking in account that α = 0,01) that za = z00l from the equality: Φ(z0 01) = (l2α)∕2 = (l2∙0,01)∕2 = 0,49. By Table of Laplace function (Appendix 1) using linear interpolation we find z0 0 = 2,326. Replacing к = 120, za = 2,326 in the WilsonGilferti formula, we
obtain ɪ2 (0,01; 120) = 158,85. Since χ2bs> χ2r, the null hypothesis is rejected, i.e. the batch can’t be accepted. Glossary discrepancy  расхождение; reliability  надежность confidence interval  доверительный интервал accuracy  точность; slot machine  игровой автомат competing hypothesis  конкурирующая гипотеза; batch  партия significance level  уровень значимости Exercises for Seminar 15
15.1. Find the confidence interval for an estimator with reliability 0,99 of unknown mathematical expectation a of a normally distributed attribute X of a
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parent population if the parent mean square deviation σ, sample mean ʌ and sample size n are known: σ = 4, xs = 10,2 ,n = 16. The answer'. 7,63 < a < 12,77.
15.2. Five equally exact measurements of distance between a gun and a target have been made by the same device with the mean square deviation of random mistakes of measurements σ = 40 m. Find the confidence interval for an estimator of true distance a to the target with reliability γ = 0,95, knowing that the arithmetic mean of results of measurements xs = 2000 m. It is supposed that results of measurements are normally distributed. The answer. 1964,94 < a < 2035,06. 15.3. An automatic machine stamps cylinders. The sample mean of diameters of made cylinders has been calculated on a sample n = 100. Find the accuracy δ with reliability 0,95 on which the sample mean estimates the mathematical expectation of diameters of making cylinders, knowing that their mean square deviation σ = 2 mm. It is supposed that diameters of cylinders are normally distributed. The answer, δ = 0,392 mm. 15.4. Find minimal size of a sample at which with reliability 0,925 the accuracy of estimator of mathematical expectation of a normally distributed parent population on sample mean is equal to 0,2 if the mean square deviation of the parent population σ = 1,5. The answer: n= 179. 15.5. A sample of size n = 12 has been extracted from a parent population: variant xi 0,5 0,4 0,2 0 0,2 0,6 0,8 1 1,2 1,5 frequency n, 1 2 111 1 1121 Estimate with reliability 0,95 the mathematical expectation a of a normally distributed attribute of a parent population on sample mean by the confidence interval. The answer.  0,04 < a < 0,88. 15.6. By data of nine independent equally exact measurements of some physical quantity the following has been found: the arithmetic mean of results of measurements xs = 30,1 and revised mean square deviation s = 6. Estimate true value of the measured quantity by the confidence interval with reliability γ = 0,99. It is supposed that results of measurements are normally distributed. The answer. 25,38 < a < 34,82.
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15.7. By data of a sample of size n from a parent population of a normally distributed quantitative attribute the following has been found: the revised mean square deviation s. Find the confidence interval covering the parent mean square deviation σ with reliability 0,999 if n = 10, s = 5,1. The answer. 0 < σ < 14,28. 15.8. 10 measurements of some physical quantity have been made by one device (without systematic mistakes), and the revised mean square deviation s of random mistakes appeared equal to 0,8. Find the accuracy of the device with reliability 0,95. It is supposed that results of measurements are normally distributed. The answer. 0,28 < σ < 1,32. 15.9. Independent trials are made with identical but unknown probability p of appearance of an event A in each trial. Find the confidence interval for estimation of probability p with reliability 0,99 if the event A has appeared 60 times in 100 trials. The answer. 0,47 < p < 0,71. 15.10. An event A has appeared 270 times in 360 trials in each of which the probability of appearance of the event is constant and unknown. Find the confidence interval covering the unknown probability p with reliability 0,95. The answer. 0,705
0,18 as the competing hypothesis. Direction: Take conditional variants ut = 10x,  100: calculate firstly Уniu(  [Vniui∖ In s2 u=^÷λ— and then ʌ2 = 52∕102. и1 The answer. The null hypothesis is rejected.
15.15. It has been found in result of long timestudy of assembling a node by different collectors that the dispersion of this time is σ2 = 2 min2 (timestudy хронометраж; node  узел). The results of 20 observations for an inexperienced collector working are the following: the time of assembling of one node in minutes xi 56 58 60 62 64 frequency ni 1 4 10 3 2 Can we consider that the inexperienced collector works rhythmically at the significance level 0,05 (in the sense that the dispersion of time expended by him is not essentially differed from the dispersion of time of the rest collectors)? Direction'. The null hypothesis H0 is σ2 = σ02 = 2; the competing hypothesis H1 is σ2 ≠ σ2. Take ui = xi60 as conditional variants and calculate s]. The answer. The null hypothesis is rejected. The inexperienced collector works nonrhythmically.
Exercises for Homework 15
15.16. Find the confidence interval for an estimator with reliability 0,99 of unknown mathematical expectation a of a normally distributed attribute X of a
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parent population if the parent mean square deviation σ, sample mean ʌ and sample size n are known: σ = 5, xs = 16,8; и = 25. The answer. 14,23 < a < 19,37.
15.17. A sample of big butch of electro lamps contains 100 lamps. The average durability of burning a lamp of the sample is 1000 hours. Find the confidence interval with reliability 0,95 for average durability a of burning a lamp of all the butch if it is known that the mean square deviation of durability of burning a lamp σ = 40 h. it is supposed that the durability of burning lamps is normally distributed. The answer. 992,16 < a < 1007,84. 15.18. By data of 16 independent equally exact measurements of some physical quantity the following has been found: the arithmetic mean of results of measurements xs =42,8 and revised mean square deviation s = 8. Estimate true value of the measured quantity with reliability γ = 0,999. The answer. 34,66 < a < 50,94. 15.19. By data of a sample of size n from a parent population of a normally distributed quantitative attribute the following has been found: the revised mean square deviation s. Find the confidence interval covering the parent mean square deviation σ with reliability 0,999 if n = 50, s = 14. The answer. 7,98 < σ < 20,02. 15.20. 300 trials have been made in each of which an unknown probability p of appearance of an event A is constant. The event A has appeared 250 times in these trials. Find the confidence interval covering the unknown probability p with reliability 0,95. The answer. 0,78