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I N D E X UNIT-1 » ROOTS OF NON-LINEAR EQUATION.................................................. 1 1).

METHOD – 1: BISECTION METHOD (BOLZANO METHOD) ...................................................... 2

2).

METHOD – 2: REGULA-FALSI METHOD............................................................................................ 3

3).

METHOD – 3: SECANT METHOD.......................................................................................................... 5

4).

METHOD – 3: NEWTON-RAPHSON METHOD................................................................................. 9

5).

METHOD – 4: ITERATION METHOD.................................................................................................11

UNIT-2 » INTERPOLATION .............................................................................13 6).

METHOD – 1: NEWTON’S FORWARD DIFFERENCE FORMULA ............................................16

7).

METHOD – 2: NEWTON’S BACKWARD DIFFERENCE FORMULA .........................................18

8).

METHOD – 3: NEWTON’S DIVIDED DIFFERENCE FORMULA ................................................22

9).

METHOD – 4: LAGRANGE’S INTERPOLATION FORMULA .......................................................24

UNIT-3 » NUMERICAL INTEGRATION .............................................................27 10).

METHOD – 1: TRAPEZOIDAL RULE ..................................................................................................27

11).

METHOD – 2: SIMPSON’S 1/3 RULE .................................................................................................29

12).

METHOD – 3: SIMPSON’S 3/8 RULE .................................................................................................32

13).

METHOD – 4: GAUSSIAN QUADRATURE (GAUSSIAN INTEGRATION) ...............................35

UNIT- 4 » NUMERICAL METHODS FOR O.D.E. ................................................37 14).

METHOD – 1: PICARD METHOD ........................................................................................................37

15).

METHOD – 2: TAYLOR’S SERIES METHOD ....................................................................................38

16).

METHOD – 3: EULER’S METHOD .......................................................................................................39

17).

METHOD – 4: IMPROVED EULER’S METHOD ...............................................................................41

18).

METHOD – 5: RK 4TH ORDER METHOD ...........................................................................................42

UNIT-5 » CURVE FITTING ..............................................................................45 19).

METHOD – 1: FITTING A STRAIGHT LINE 𝐲 = 𝐚 + 𝐛 𝐱.............................................................47

20).

METHOD – 2: FITTING A PARABOLA 𝐲 = 𝐚 + 𝐛𝐱 + 𝐜𝐱𝟐 ..........................................................49

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AEMM - 3130908

I N D E X 21).

METHOD – 3: NON-LINEAR REGRESSION ..................................................................................... 51

UNIT 6 – BASIC PROBABILITY .......................................................................53 22).

METHOD – 1: BASIC EXAMPLES ON PROBABILITY................................................................... 55

23).

METHOD – 2: PROBABILITY OF EVENT ......................................................................................... 59

24).

METHOD – 3: CONDITIONAL PROBABILITY ................................................................................ 62

25).

METHOD-6: EXAMPLES ON TWO-DIMENSIONAL RANDOM VARIABLE........................... 67

26).

METHOD – 4: TOTAL PROBABILITY AND BAYE’S THEOREM ............................................... 72

27).

METHOD – 5: PROBABILITY FUNCTION, EXPECTATION, VARIANCE ................................ 76

28).

METHOD – 6: BINOMIAL DISTRIBUTION ...................................................................................... 81

29).

METHOD – 7: CHEBYSHEV’S INEQUALITY .................................................................................... 84

UNIT-7 » BASIC STATISTICS ..........................................................................85 30).

METHOD – 1: EXAMPLES ON CENTRAL TENDENCY, DISPERSION AND SKEWNESS .. 91

31).

METHOD – 2: EXAMPLES ON MOMEMNT (SKEWNESS AND KURTOSIS) ......................... 99

SYLLABUS OF AMEE – 3130908 ……………………………….........………………..*** GTU QUESTION PAPER …………..……………………………….........………………..***

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-1 » Roots of Non – Linear Equation

[1]

UNIT-1 » ROOTS OF NON-LINEAR EQUATION ❖ INTRODUCTION: ✓ An expression of the form f(x) = a0 x n + a1 x n−1 + a2 x n−2 + ⋯ + an−1 x + an , where a0 , a1 , a2 … an are constants and n is a positive integer , is called an algebraic polynomial of degree n if a0 ≠ 0. ✓ The equation f(x) = 0 is called an algebraic equation if f(x) is an algebraic polynomial, e.g. x 3 − 4x − 9 = 0. ✓ If f(x) contains functions such as trigonometric, logarithmic, exponential, etc., then f(x) is called a transcendental equation, e.g., 2x 3 − log(x + 3) tanx + ex = 0. ✓ In general, an equation is solved by factorization. But in many cases, the method of factorization fails. In such cases, numerical methods are used. There are some methods to solve the equation f(x) = 0, which are given as follows. (1). Bisection Method (2). Regula Falsi Method (3). Secant Method (4). Newton Raphson Method (5). Fixed Point Iteration Method ❖ BISECTION METHOD (BOLZANO METHOD): ✓ This is the method which narrow down the interval which contains the root by bisecting the interval. ❖ PROCEDURE TO USE BISECTION METHOD: ✓ Let f(x) = 0 be an equation. (1). Find a and b such that f(a) < 0 & f(b) > 0. (2). Find c such that, c =

a+b 2

i.e. Root lies in interval [a, b].

. Check f(c).

(3). If f(c) < 0, then replace a by c. New Interval will be [c, b]. (4). If f(c) > 0, then replace a by c. New Interval will be [a, c]. (5). Repeat same steps to find new value of "c" till f(c) is equal or nearer to zero.

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-1 » Roots of Non – Linear Equation

[2]

METHOD – 1: BISECTION METHOD (BOLZANO METHOD) C

1

Explain bisection method for solving an equation f(x) = 0. Find the real root of equation x 2 − 4x − 10 = 0 by using this method correct to three decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓. 𝟕𝟒𝟏

H

2

Explain bisection method for solution of equation. Using this method find the approximate solution of x 3 + x − 1 = 0 correct up to three decimal points. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟔𝟖𝟑

T

3

Use bisection method to find a root of equation x 3 + 4x 2 − 10 = 0 in the interval [1,2].Use four iteration. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟑𝟏𝟐𝟓

C

4

Find one root of ex − 3x = 0 correct up to two decimal places using bisection method. Use 1.5 and 1.52 as initial. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟓𝟏

H

5

Perform three iterations of Bisection method to obtain root of the equation 2 sin x − x = 0. Take (1,2) as initial interval. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟖𝟕𝟓

H

6

Find a positive root of the equation x − cos x = 0 using bisection method correct to two places of decimals.

W-19 (4)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕𝟓 C

7

Perform the five iterations of the bisection method to obtain a root of the equation f(x) = cos x − xex = 0. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟓𝟑𝟏𝟐

❖ METHOD OF FALSE POSITION OR REGULA FALSI METHOD: ✓ This is the oldest method to finding the real root of an equation f(x) = 0 and closely resembles the bisection method.

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UNIT-1 » Roots of Non – Linear Equation

[3]

❖ PROCEDURE TO USE FALSE POSITION OR REGULA FALSI METHOD: ✓ Let f(x) = 0 be an equation. (1). Find a and b such that f(a) < 0 & f(b) > 0.

i.e. Root lies in interval [a, b].

b−a

(2). Find c such that, c = a − f(b)−f(a) f(a) . Check f(c). (3). If f(c) < 0, then replace a by c. New Interval will be [c, b]. (4). If f(c) > 0, then replace a by c. New Interval will be [a, c]. ✓ Repeat same steps to find new value of "c" till f(c) is equal or nearer to zero.

METHOD – 2: REGULA-FALSI METHOD 1

C

Find a real root of the equation x 3 − x − 1 = 0 using Regula Falsi method correct up to two decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟑𝟐

2

C

Apply False Position method to find the negative root of the equation x 3 − 2x + 5 = 0 correct to four decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟐. 𝟎𝟗𝟒𝟓

3

T

Find a root of the equation x 3 − 4x − 9 = 0 using False position method correct up to three decimals. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟕𝟎𝟕

H

4

Use the method of Regula Falsi to find the root of an equation x = e−x correct up to three decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟓𝟔𝟕

5

C

Find a real root of cos x − 3x + 5 = 0 corect upto four decimal places using false position method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟔𝟒𝟐𝟕

H

6

Using method of False position, compute the real root of the equation xlog10 x − 1.2 = 0 correct to four decimals. Where (x > 0). 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟕𝟒𝟎𝟔

T

7

Using Regula Falsi method determine the root of the equation x log x = 1.2. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟕𝟒𝟎𝟐

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

W-19 (3)

UNIT-1 » Roots of Non – Linear Equation

[4]

❖ SECANT METHOD: ✓ This method is an improvement over the method of false position as it does not require the condition f(x0 ) f(x1 ) < 0 of that method. Here also the graph of the function y = f(x) is approximated by a secant line but at each iteration, two most recent approximations to the root are used to find the next approximation. Also, it is not necessary that the interval must contain the root. ✓ Taking x0 , x1 as the initial limits of the interval, we write the equation of the chord joining these as y − f(x1 ) =

f(x1 ) − f(x0 ) (x − x1 ) x1 − x0

✓ Then the abscissa of the point where it crosses the X −axis (y = 0) is given by x2 = x1 −

x1 − x0 f(x ) f(x1 ) − f(x0 ) 1

✓ Which is an approximation to the root. The general formula for successive approximations is, therefore, given by xn+1 = xn −

xn − xn−1 f(x ), n ≥ 1 f(xn ) − f(xn−1 ) n

✓ Note: If at any iteration f(xn ) = f(xn−1 ), this method fails and shows that it does not converge necessarily. This is drawback of secant method over the method of false position which always converges.

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UNIT-1 » Roots of Non – Linear Equation

[5]

METHOD – 3: SECANT METHOD H

1

Find the positive solution of f(x) = x − 2sinx = 0 by the secant method, starting form x0 = 2, x1 = 1.9. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟖𝟗𝟓𝟓

H

2

Derive Secant method and solve xex − 1 = 0 correct up to three decimal places between 0 and 1. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟓𝟔𝟖

3

C

Find the positive solution of f(x) = e−x − x by the secant method starting from x0 = 0, x1 = 1.0.

W-19 (3)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟓𝟔𝟕𝟏 H

4

Find the real root of the following by secant method. x 2 − 4x − 10 = 0 (using x0 = 4, x1 = 2,upto six iteration) x 3 − 2x − 5 = 0 (using x0 = 2, x1 = 3, upto four iteration) 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓. 𝟕𝟒𝟏𝟏, 𝟐. 𝟎𝟗𝟒𝟔

C

5

Use Secant method to find the roots of cos x − xex = 0 correct upto 3 decimal places of decimal. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟓𝟏𝟖

❖ NEWTON-RAPHSON METHOD (NEWTON’S METHOD): ✓ This is the method to quickly find a good approximation for the root of real valued function. ❖ PROCEDURE TO USE NEWTON-RAPHSON METHOD: ✓ Let f(x) = 0 be an equation. (1). If f(a) ∙ f(b) < 0, take x0 = a when |f(a)| < |f(b)| OR x0 = b when |f(b)| < |f(a)|. (2). Find xn+1 such that, xn+1 = xn −

f(xn ) ; n = 0,1,2,3 … f ′ (xn )

✓ Where, f ′ (xn ) ≠ 0 ✓ Repeat the steps to find new value of " xn+1 " till f(xn+1 ) is equal or nearer to zero.

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UNIT-1 » Roots of Non – Linear Equation

[6]

❖ EXPLANATION: ✓ Let x1 be the root of f(x) = 0 and x0 be an approximation to x1 . If h = x1 − x0 , then by Taylor’s Series, f(x0 + h) = f(x0 ) + h f

′ (x

h2 ′′ f (x0 ) + ⋯ 0) + 2!

✓ Since, x1 = x0 + h is root. f(x1 ) = f(x0 + h) = 0 ✓ If h is chosen too small enough, then we can neglect 2nd , 3rd and higher powers of h. ✓ We have, 0 = f(x0 ) + h f ′ (x0 ) ⟹h=−

f(x0 ) ; f ′ (x0 ) ≠ 0. f ′ (x0 )

✓ Suppose that, x1 = x0 + h be the better approximation. ⟹ x1 = x0 −

f(x0 ) f ′ (x0 )

✓ By repeating the process, ⟹ x2 = x1 −

f(x1 ) f ′ (x1 )

⟹ x3 = x2 −

f(x2 ) f ′ (x2 )

✓ In general, f(x )

xn+1 = xn − f′ (xn ) ; n = 0,1,2,3 … Where, f ′ (xn ) ≠ 0 n

✓ This is called Newton-Raphson Formula. ❖ NOTE: ✓ If the function is linear, ( f ′ (xn ) = 0), then N-R method has to be failed.

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-1 » Roots of Non – Linear Equation

[7]

❖ THEOREM: ✓ Prove that NR procedure is second order convergent. OR Rate of convergence of the NR methods. ✓ Proof: ✓ Let α be exact root of f(x) = 0 and let xn, xn+1 be two successive approximations to the actual root. If ∈n and ∈n+1 are the corresponding errors then xn= α +∈n xn+1= α +∈n+1 ✓ Substituting in equation of ‘NR METHOD FORMULA’ α +∈n+1 = α +∈n − ∈n+1 =∈n −

f(α +∈n ) f′(α +∈n )

f(α +∈n ) f′(α +∈n )

∈n 2 ′′ f(α) +∈n f ′ (α) + 2! f (α) + ⋯ = ∈n − f ′ (α) +∈n f ′′ (α) + ⋯ ∈n 2 ′′ ∈n f ′ (α) + 2! f (α) + ⋯ = ∈n − (∵ f(α) = 0) f ′ (α) +∈n f ′′ (α) ✓ Neglecting the derivative of order higher then two ∈n 2 ′′ ∈n f ′ (α) + 2! f (α) + ⋯ ∈n+1 = ∈n − f ′ (α) +∈n f ′′ (α) + ⋯ 1 ∈n 2 f ′′ (α) = [ ′ ] 2 f (α) +∈n f ′′ (α) f ′′ (α) ∈n 2 f ′ (α) = [ ] f ′′ (α) 2 1 +∈n ′ f (α) =

∈n 2 f ′′ (α) 2 f ′ (α)

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UNIT-1 » Roots of Non – Linear Equation

[8]

✓ The last equation shows that the error at stage is proportional to the squares of the error in the previous stage. Hence, the NR method has a quadratic convergence and the convergence is of the order 2. ❖ FORMULA FOR 𝐪𝐭𝐡 ROOT: 1

✓ Let, x = N q , where q = 1, 2, 3, … … n and N is an natural number. ⟹ x q − N = 0 ⟹ f(x) = x q − N ⟹ f ′ (x) = qx q−1 ✓ By the general formula we get, xn+1 =

1 N [(q − 1)xn + q−1 ] q xn

✓ Where n = 0, 1, 2 … … ❖ FIND THE ITERATIVE FORMULA FOR √𝐍 BY N-R METHOD: ✓ Let, x = √N ⟹ x 2 − N = 0 ⟹ f(x) = x 2 − N ⟹ f ′ (x) = 2x ✓ By N-R formula, f(x )

xn+1 = xn − f′ (xn )n

xn 2 − N 2xn 2 − xn 2 + N = xn − = 2xn 2xn xn 2 + N 1 N = = (xn + ) 2xn 2 xn 𝟏

❖ FIND THE ITERATIVE FORMULA OF 𝐍 BY N-R METHOD: 1

✓ Let, x = N ⟹

1 −N=0 x

⟹ f(x) =

1 −N x

⟹ f ′ (x) = −

1 x2

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-1 » Roots of Non – Linear Equation ✓ By N-R formula, xn+1 = xn −

f(xn ) f′ (xn )

= xn −

1 −N xn 1 − 2 xn

1

= xn + (x − N) xn 2 = 2xn − Nxn 2 n

xn+1 = xn (2 − Nxn ) , where n = 0, 1, 2, … …

METHOD – 3: NEWTON-RAPHSON METHOD H

1

Explain Newton’s method for solving equation f(x) = 0. Apply this method to find the approximate solution of x 3 + x − 1 = 0 correct up to three decimals. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟔𝟖𝟐

H

2

Find the real root of the equation x 4 − x − 9 = 0 by Newton-Raphson method, correct to three decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟖𝟏𝟑

3

T

Use Newton-Raphson method to find smallest positive root of f(x) = x 3 − 5x + 1 = 0 correct to four decimals. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟐𝟎𝟏𝟔𝟒

4

C

Find the positive root of x = cos x correct up to three decimal places by NR method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕𝟑𝟗

H

5

Find to four decimal places, the smallest root of the equation sin x = e−x Using the N-R starting with x0 = 0.6. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟓𝟖𝟖𝟓

6

T

Using Newton-Raphson method find a root of the equation xex = 2 correct to three decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟖𝟓𝟑

7

C

Find a root of x 4 − x 3 + 10x + 7 = 0 correct up to three decimal places between a = −2 & b = −1 by N-R method. 𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟏. 𝟒𝟓𝟒

H

8

Find a zero of function f(x) = x 3 − cos x with starting point x0 = 1 by NR Method could x0 = 0 be used for this problem? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟖𝟔𝟓𝟓

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

[9]

UNIT-1 » Roots of Non – Linear Equation H

9

[10]

Find the real root of the equation 3x = cos x + 1, taking x0 = 1 by Newton Raphson method correct to four decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟔𝟎𝟕𝟏

C

10

Find a real positive root of equation x sin x + cos x = 0 which is near x = π correct to four significant digits using Newton-Raphson method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟕𝟗𝟖𝟒

H 11

State the formula for finding qth root and find the square root of 8 using Newton-Raphson method correct to two decimal places.

W-19 (4)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟖𝟑 C

12

Find the √10 correct to three decimal places by using Newton-Raphson iterative method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑. 𝟏𝟔𝟐𝟑

H 13

Derive an iterative formula to find √N hence find approximate value of √28 , √65 , √3, correct up to three decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓. 𝟐𝟗𝟏𝟓, 𝟖. 𝟎𝟔𝟐, 𝟏. 𝟕𝟑𝟐

C

14

Derive an iterative formula for finding cube root of any positive number 3

using Newton Raphson method and hence find approximate value of √58. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑. 𝟖𝟕𝟎𝟖 H 15

Derive Newton Raphson’s formula for finding the cube root of positive no. 3

N, hence find √12. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟐𝟖𝟗𝟒 T

16

Find cube root of 72 using Newton Raphson method correct up to four decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒. 𝟏𝟔𝟎𝟐

C

17

1

Find an iterative formula to find N(Where N is positive number) and hence evaluate

1 3

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟑𝟑𝟑 H 18

1

Find an iterative formula to find N(Where N is positive number) and hence evaluate

1 19

1

, . 23

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟎𝟓𝟐𝟔, 𝟎. 𝟎𝟒𝟑𝟓

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-1 » Roots of Non – Linear Equation

[11]

❖ ITERATION METHOD OR SUCCESSIVE APPROXIMATION METHOD: ✓ This is the method to use initial guess to generate a sequence of improving approximate solutions. ❖ PROCEDURE TO USE ITEATION OR SUCESSIVE APPROXIMATION METHOD: (1). To find root of f(x) = 0. Rewrite in the form x = g(x) (2). Check |g ′ (x)| < 1 for all x in I (3). Let x = x0 (initial) then x1 = g(x0 ) x2 = g(x1 )….. (4). Proceeding in this way xn = g(xn−1 ) xn+1 = g(xn ) ; n = 0, 1, 2, 3, …

METHOD – 4: ITERATION METHOD H

1

1

Test the convergence condition for the equation x = 3 (cos x + 1) in the π

interval (0, 2 ) and then solve the equation using successive approximation method correct up to 3 decimal places of decimals taking initial guess as x0 = 0.5. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝟎. 𝟔𝟎𝟕 H

2

State the condition of convergence for iterative method x = g(x). Find to 1

four decimal place accuracy a solution to x = 2 + sin x using the method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝟏. 𝟒𝟗𝟕𝟑 C

3

Find possible roots of equation x 3 + x 2 − 1 = 0 by iteration method correct up to four decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝟎. 𝟕𝟓𝟒𝟗

C

4

Using method of successive approximation solve the equation 2x − log10 x = 7 correct up to four decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝟑. 𝟕𝟖𝟗𝟑

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UNIT-1 » Roots of Non – Linear Equation H

5

[12]

Perform three iterations to find the real root of e−x − 10x = 0 correct to two decimal places by fixed point iteration method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝟎. 𝟎𝟗

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

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UNIT-2 » Interpolation

[13]

UNIT-2 » INTERPOLATION ❖ DEFINITION OF OPERATOR: Forward difference[∆]

∆f(x) = f(x + h) − f(x)

Backward difference [∇]

∇f(x) = f(x) − f(x − h)

Shift Operator [E]

E n f(x) = f(x + nh)

Central difference [δ]

1 1 h h δf(x) = f (x + ) − f (x − ) = E 2 − E −2 2 2

Average Operator [μ]

1 h h E 2 + E −2 μf(x) = [f (x + ) + f (x − )] = 2 2 2 2

Differential Operator [D]

Df(x) =

1

1

d f(x) = f ′ (x) dx

❖ RELATION BETWEEN OPERATORS: 1. 𝐄 = 𝟏 + ∆ Proof: (1 + ∆)f(x) = f(x) + ∆f(x) = f(x) + f(x + h) − f(x) = Ef(x) 𝟏

2. 𝐄𝛁 = ∆= 𝛁𝐄 = 𝛅𝐄 𝟐 Proof: E∇(f(x)) = E(∇f(x)) = E(f(x) − f(x − h)) = Ef(x) − Ef(x − h) = f(x + h) − f(x) ⟹ E∇(f(x)) = ∆f(x) ⟹ E∇= ∆ 1

1

1

1

Now, δE 2 = (E 2 − E −2 ) E 2 = (E − 1) = ∆ (∵ E = 1 + ∆)

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UNIT-2 » Interpolation

[14]

1

⟹ E∇= ∆= ∇E = δE 2 3. ∆𝛁 = ∆ − 𝛁 Proof: ∆∇(f(x)) = ∆(∇f(x)) = ∆(f(x) − f(x − h)) = ∆f(x) − ∆f(x − h) = [f(x + h) − f(x)] − [f(x) − f(x − h)] = ∆f(x) − ∇f(x) = (∆ − ∇)f(x) ⟹ ∆∇(f(x)) = (∆ − ∇)f(x) ⟹ ∆∇= ∆ − ∇ 𝟒.

∆ 𝛁 − =∆+𝛁 𝛁 ∆

Proof: ∆ ∇ ∆2 − ∇2 − = ∇ ∆ ∆∙∇ =

(∆ − ∇) ∙ (∆ + ∇) (∵ ∆∇= ∆ − ∇) ∆−∇

= ∆+∇ 5. (𝟏 + ∆)(𝟏 − 𝛁) = 𝟏

W − 19(3)

Proof: (1 + ∆)(1 − ∇) = 1 − ∇ + ∆ − ∆. ∇= 1 − ∇ + ∆ − (∆ − ∇) = 1 6. 𝛁 = 𝟏 − 𝐄−𝟏 Proof: 1 − E −1 = 1 − (1 + ∆)−1 = 1−

(∵ E = 1 + ∆)

1 1+∆−1 ∆ E∇ = = = = ∇ (∵ ∆= ∇E) 1+∆ 1+∆ 1+∆ E

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UNIT-2 » Interpolation

[15]

7. 𝐄 = 𝐞𝐡𝐃 Proof: By Taylor’s expansion Ef(x) = f(x + h) = f(x) + hf ′ (X) +

h2 ′′ f (x) + ⋯ 2!

h2 2 = f(x) + h Df(X) + D f(x) + ⋯ 2! = [1 + hD +

h2 2 D + ⋯ ] f(x) 2!

⟹ Ef(x) = ehD f(x) ⟹ E = ehD 8. ∆= 𝐞𝐡𝐃 − 𝟏 OR 𝐡𝐃 = 𝐥𝐨𝐠 (𝟏 + ∆) Proof: We Know that, E = ehD . Taking E = 1 + ∆⟹ ∆= ehD − 1 ⟹ hD = log (1 + ∆) 𝟏

9. 𝛍𝛅 = 𝟐 (∆ + 𝛁) Proof: μδ(f(x)) = μ(δf(x)) h h = μ [f (x + ) − f (x − )] 2 2

h h (∵ δf(x) = f (x + ) − f (x − )) 2 2

h h = μf (x + ) − μf (x − ) 2 2 1

= 2 [(f(x + h) + f(x)) − (f(x) + f(x − h))] =

1 [(f(x + h) − f(x)) + (f(x) − f(x − h))] 2

=

1 (∆ + ∇) 2

1

h

h

( ∵ μf(x) = 2 [f (x + 2) + f (x − 2)])

1 ⟹ μδ = (∆ + ∇) 2

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UNIT-2 » Interpolation

[16]

❖ INTERPOLATION: ✓ Let the function y = f(x) take the values y0 , y1 , y2 , … , yn corresponding to the values x0 , x1 , x2 , … , xn of x. The process of finding the value of y corresponding to any value of x = xi between x0 and xn is called interpolation. ✓ Thus, interpolation is a technique of finding the value of a function for any intermediate value of the independent variable. The process of computing the value of the function outside the range of given values of the variable is called extrapolation. ❖ NEWTON’S FORWARD DIFFERENCE FORMULA: ✓ If data are (x0 , y0 ), (x1 , y1 )(x2 , y2 ), … , (xn , yn ). Where, x0 , x1 , x2 , … , xn are equally spaced. 𝐟(𝐱) = 𝐲 = 𝐲𝟎 + 𝐩∆𝐲𝟎 +

𝐩(𝐩 − 𝟏) 𝟐 𝐩(𝐩 − 𝟏)(𝐩 − 𝟐) 𝟑 𝐱 − 𝐱𝟎 ∆ 𝐲𝟎 + ∆ 𝐲𝟎 + ⋯ ; 𝐩 = 𝟐! 𝟑! 𝐡

✓ Where x0 is initial point. ❖ DIFFERENCE TABLE 𝐱

𝐟(𝐱) = 𝐲

𝐱𝟎

y0

∆𝐟(𝐱)

∆𝟐 𝐟(𝐱)

∆𝟑 𝐟(𝐱)

∆𝟒 𝐟(𝐱)

∆y0 = y1 − y0 𝐱𝟏

∆2 y0

y1

∆3 y0

∆y1 = y2 − y1 𝐱𝟐

∆2 y1

y2

∆3 y1

∆y2 = y3 − y2 𝐱𝟑

∆4 y0

∆2 y2

y3 ∆y3 = y4 − y3

𝐱𝟒

y4

METHOD – 1: NEWTON’S FORWARD DIFFERENCE FORMULA C

1

Construct forward difference table for following data. (1, 1.1), (2, 4.2), (3, 9.3), (4, 16.4). 𝐀𝐧𝐬𝐰𝐞𝐫: ∆𝟑 𝐲𝟎 = 𝟎

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UNIT-2 » Interpolation 2

C

[17]

Construct Newton’s forward interpolation polynomial for the following data. Use it to find the value of y for x = 5. x

4

6

8

10

y

1

3

8

16

𝟑𝐱 𝟐 − 𝟐𝟐𝐱 + 𝟒𝟖 𝟏𝟑 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐱) = , 𝐲(𝟓) = 𝟖 𝟖 H

3

State Newton’s forward difference formula and use it to find approximate value of f(1.6), if x

1

1.4

1.8

2.2

f(x)

3.49

4.82

5.96

6.5

W-19 (4)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓. 𝟒𝟑𝟗𝟒 T

4

HDetermine the polynomial by Newton’s forward difference formula from the following table. x

0

1

2

3

4

5

y

−10

−8

−8

−4

10

40

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 𝟑 − 𝟒𝐱 𝟐 + 𝟓𝐱 − 𝟏𝟎 C

5

Find sin 520 using the following values. x

45°

50°

55°

60°

y = sin x

0.7071

0.7660

0.8192

0.8660

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕𝟖𝟖𝟎

❖ NEWTON’S BACKWARD DIFFERENCE FORMULA: ✓ If data are (x0 , y0 ), (x1 , y1 )(x2 , y2 ), … , (xn , yn ).Also x0 , x1 , x2 , … , xn are equally spaced. 𝐟(𝐱) = 𝐲 = 𝐲𝐧 + 𝐩𝛁𝐲𝐧 +

𝐩(𝐩 + 𝟏) 𝟐 𝐩(𝐩 + 𝟏)(𝐩 + 𝟐) 𝟑 𝐱 − 𝐱𝐧 𝛁 𝐲𝐧 + 𝛁 𝐲𝐧 + ⋯ ; 𝐩 = 𝟐! 𝟑! 𝐡

✓ Where xn is ending point.

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UNIT-2 » Interpolation

[18]

❖ DIFFERENCE TABLE 𝐱

𝐟(𝐱) = 𝐲

𝐱𝟎

y0

𝛁 𝟐 𝐟(𝐱)

𝛁𝐟(𝐱)

𝛁 𝟑 𝐟(𝐱)

𝛁 𝟒 𝐟(𝐱)

∇y1 = y1 − y0 𝐱𝟏

∇2 y2

y1

∇3 y3

∇y2 = y2 − y1 𝐱𝟐

∇2 y3

y2

∇4 y4 ∇3 y4

∇y3 = y3 − y2 𝐱𝟑

∇2 y4

y3 ∇y4 = y4 − y3

𝐱𝟒

y4

METHOD – 2: NEWTON’S BACKWARD DIFFERENCE FORMULA C

1

The population of the town is given below. Estimate the population for the year 1925. Year Population in thousand

1891

1901

1911

1921

1931

46

66

81

93

101

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟗𝟔. 𝟖𝟑𝟔𝟖 H

2

State Newton’s backward formula and use it to find the approximate value of f(7.5), if x

3

4

5

6

7

8

f(x)

28

65

126

217

344

513

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝟐𝟐. 𝟖𝟕𝟓 H

3

From the following table of half-yearly premium for policies maturing at different ages, estimate the premium at the age of 63. Age Premium (in $)

45

50

55

60

65

114.84

96.16

83.32

74.48

68.48

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟕𝟎. 𝟓𝟖𝟓𝟏

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UNIT-2 » Interpolation 4

T

[19]

The following table gives distance (in nautical miles) of the visible horizon for the heights (in feet) above earth’s surface. Find the values of y when x = 390 feet. Height(x)

100

Distance(y) 10.63

150

200

250

300

350

400

13.03

15.04

16.81

18.42

19.90

21.47

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝟏. 𝟏𝟐𝟎𝟖 C

The area of circle of diameter d is given by D

80

85

90

95

100

A

5026

5674

6361

7088

7854

Use suitable interpolation to find area of circle of diameter 98. Also Calculate the error. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐀 = 𝟕𝟓𝟒𝟑. 𝟎𝟔𝟕𝟐, 𝐄𝐱𝐚𝐜𝐭𝐀 = 𝟕𝟓𝟒𝟐. 𝟗𝟔𝟒𝟎, 𝐄𝐫𝐫𝐨𝐫 = 𝟎. 𝟏𝟎𝟑𝟐 H

5

Compute f(x) = ex at x = 0.02 and x = 0.38 Using suitable interpolation formula for the data given below. x

0.0

0.1

0.2

0.3

0.4

f(x)

1.0000

1.1052

1.2214

1.3499

1.4918

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟎𝟐𝟎𝟐, 𝟏. 𝟒𝟔𝟐𝟑 H

6

From the following table, find P when t = 1420 c and 1750 c using appropriate Newton’s interpolation formula. Temp. t 0 C

140

150

160

170

180

Pressure P

3685

4854

6302

8076

10225

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏𝟏𝟒𝟐° = 𝟑𝟖𝟗𝟖. 𝟔𝟔𝟖𝟖 ; 𝐏𝟏𝟕𝟓° = 𝟗𝟏𝟎𝟎. 𝟒𝟖𝟒𝟒 C

7

The population of the town is given below. Estimate the population for the year 1895 and 1930 using suitable interpolation. Year

1891

1901

1911

1921

1931

Population in thousand

46

66

81

93

101

𝐀𝐧𝐬𝐰𝐞𝐫:

𝐏𝐨𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐲𝐞𝐚𝐫 𝟏𝟖𝟗𝟓 = 𝟓𝟒. 𝟖𝟓𝟐𝟖 𝐏𝐨𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐲𝐞𝐚𝐫 𝟏𝟗𝟑𝟎 = 𝟏𝟎𝟎. 𝟒𝟕𝟎𝟓

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UNIT-2 » Interpolation H

8

[20]

Compute values of f(0.12) and f(0.40) using suitable interpolation formula for the following data: x

0.10

0.15

0.20

0.25

0.30

f(x)

0.1003

0.1511

0.2027

0.2553

0.3093

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝟎. 𝟏𝟐) = 𝟎. 𝟏𝟐𝟎𝟓, 𝐟(𝟎. 𝟒𝟎) = 𝟎. 𝟒𝟐𝟒𝟏 T

9

Compute cosh(0.56) & cosh(0.76) from the following table. x cosh x

0.5

0.6

0.7

0.8

1.127626 1.185465 1.255169 1.337435

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟏𝟔𝟎𝟗𝟒𝟓, 𝟏. 𝟑𝟎𝟐𝟗𝟕𝟕 T

10

Using Newton’s suitable formula, find the value of f(1.6) & f(2),if x

1

1.4

1.8

2.2

f(x)

3.49

4.8

5.96

6.5

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓. 𝟒𝟑𝟗𝟒, 𝟔. 𝟑𝟑𝟎𝟔 H 11

Find the cubic polynomial which takes the following values: y(0) = 1, y(1) = 0, y(2) = 1,and y(3) = 10. Hence, obtain y(4). 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐱) = 𝐱 𝟑 − 𝟐𝐱 𝟐 + 𝟏, 𝐲(𝟒) = 𝟑𝟑

❖ NEWTON’S DIVIDED DIFFERENCE: [𝐱 𝟎 , 𝐱 𝟏 ] = ⍋𝐲𝟎 =

𝐟(𝐱 𝟏 ) − 𝐟(𝐱 𝟎 ) 𝐱𝟏 − 𝐱𝟎

[𝐱 𝟎 , 𝐱 𝟏 , 𝐱 𝟐 ] = ⍋𝟐 𝐲𝟎 =

[𝐱 𝟏 , 𝐱 𝟐 ] − [𝐱 𝟎 , 𝐱 𝟏 ] 𝐱𝟐 − 𝐱𝟎

[𝐱 𝟎 , 𝐱 𝟏 , … , 𝐱 𝐧 ] = ⍋𝐧 𝐲𝟎 =

[𝐱 𝟏 , 𝐱 𝟐 , … , 𝐱 𝐧 ] − [𝐱 𝟎 , 𝐱 𝟏 , … , 𝐱 𝐧−𝟏 ] 𝐱𝐧 − 𝐱𝟎

❖ NEWTON’S DIVIDED DIFFERENCE FORMULA: 𝐟(𝐱) = 𝐲𝟎 + (𝐱 − 𝐱 𝟎 )⍋𝐲𝟎 + (𝐱 − 𝐱 𝟎 )(𝐱 − 𝐱 𝟏 )⍋𝟐 𝐲𝟎 + (𝐱 − 𝐱 𝟎 )(𝐱 − 𝐱 𝟏 )(𝐱 − 𝐱 𝟐 )⍋𝟑 𝐲𝟎 + ⋯

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UNIT-2 » Interpolation

[21]

❖ DIVIDED DIFFERENCE TABLE 𝐱

𝐟(𝐱) = 𝐲

𝐱𝟎

y0

⍋𝟐 𝐟(𝐱)

⍋𝐟(𝐱)

⍋𝟑 𝐟(𝐱)

⍋𝟒 𝐟(𝐱)

⍋y0 =

y1 − y0 x1 − x0 ⍋2 y0

𝐱𝟏

y1

=

⍋y1 − ⍋y0 x2 − x0 ⍋3 y0

⍋y1 =

y2 − y1 x2 − x1

=

⍋2 y1 − ⍋2 y0 x3 − x0 ⍋4 y0

⍋2 y1 𝐱𝟐

y2

=

⍋y2 − ⍋y1 x3 − x1

⍋3 y1 − ⍋3 y0 x4 − x0

⍋3 y1

⍋y2 =

=

y3 − y2 x3 − x2

=

⍋2 y2 − ⍋2 y1 x4 − x1

⍋2 y2 𝐱𝟑

y3

=

⍋y3 − ⍋y2 x4 − x2

⍋y3 =

𝐱𝟒

y4 − y3 x4 − x3

y4

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UNIT-2 » Interpolation

[22]

METHOD – 3: NEWTON’S DIVIDED DIFFERENCE FORMULA C

1

If f(x) =

1 , prepare the table for finite differences and hence find [a, b] x

W-19 (3)

and [a, b, c]. 𝐀𝐧𝐬𝐰𝐞𝐫: − C

2

𝟏 𝟏 𝐚𝐧𝐝 𝐚𝐛 𝐚𝐛𝐜

Use Newton’s Divided Difference method to evaluate f(4) from the below data. x

0

1

2

3

f(x)

2

3

12

147

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝟒) = 𝟓𝟐𝟔 H

3

The Shear stress in kips, per square foot for 5 specimens in a clay stratum are: Depth(m)

1.9

3.1

4.2

5.1

5.8

Stress(ksf)

0.3

0.6

0.4

0.9

0.7

Use Newton’s divided difference formula to compute the stress at 4.5m depth. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟓𝟒𝟏𝟓 C

4

Write a formula for divided difference [x0 , x1 ] & [x0 , x1, x2 ] .Using Newton’s divided difference formula find f(9.5) from following data. x

8

9

9.2

11

f(x)

2.079442

2.197225

2.219203

2.397895

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝟗. 𝟓) = 𝟐. 𝟐𝟓𝟏𝟐𝟕𝟖 C

5

Determine the missing entry in the following table. x

0

1

2

3

4

y

1

4

17

?

97

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟑) = 𝟒𝟔

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UNIT-2 » Interpolation H

6

[23]

Find f(8) from following data using Newton’s Divided difference formula. x

4

5

7

10

11

13

f(x)

48

100

294

900

1210

2028

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝟖) = 𝟒𝟒𝟖 7

T

Find f(9.2) from the following value using Newton’s divided difference. x

8

9

9.5

11

f(x)

2.079442

2.197225

2.251292

2.397895

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝟗. 𝟐) = 𝟐. 𝟐𝟏𝟗𝟐𝟎𝟖 8

T

Using Newton’s divided difference formula find f(3) from the following table. x

−1

2

4

5

f(x)

−5

13

255

625

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝟑) = 𝟕𝟏 H

9

Using Newton’s divided difference interpolation compute the value of f(6) from the following table. x

1

2

7

8

f(x)

1

5

5

4

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = 𝟔. 𝟐𝟑𝟖𝟏

❖ LAGRANGE’S INTERPOLATION FORMULA: If data are (x0 , y0 ), (x1 , y1 )(x2 , y2 ), … , (xn , yn ). Also, x0 , x1 , x2 , … , xn are unequally spaced. 𝐲=

(𝐱 − 𝐱 𝟏 )(𝐱 − 𝐱 𝟐 ) … (𝐱 − 𝐱 𝐧 ) 𝐲 (𝐱 𝟎 − 𝐱 𝟏 )(𝐱 𝟎 − 𝐱 𝟐 ) … (𝐱 𝟎 − 𝐱 𝐧 ) 𝟎 +

(𝐱 − 𝐱 𝟎 )(𝐱 − 𝐱 𝟐 ) … (𝐱 − 𝐱 𝐧 ) (𝐱 − 𝐱 𝟎 )(𝐱 − 𝐱 𝟏 ) … (𝐱 − 𝐱 𝐧−𝟏 ) 𝐲𝟏 + ⋯ + 𝐲 (𝐱 𝟏 − 𝐱 𝟎 )(𝐱𝟏 − 𝐱 𝟐 ) … (𝐱 𝟏 − 𝐱 𝐧 ) (𝐱 𝐧 − 𝐱 𝟎 )(𝐱 𝐧 − 𝐱 𝟏 ) … (𝐱 𝐧 − 𝐱 𝐧−𝟏 ) 𝐧

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UNIT-2 » Interpolation

[24]

❖ LAGRANGE’S INVERSE INTERPOLATION FORMULA: If data are (x0 , y0 ), (x1 , y1 )(x2 , y2 ), … , (xn , yn ). Also x0 , x1 , x2 , … , xn are unequally spaced. 𝐱=

(𝐲 − 𝐲𝟏 )(𝐲 − 𝐲𝟐 ) … (𝐲 − 𝐲𝐧 ) 𝐱 (𝐲𝟎 − 𝐲𝟏 )(𝐲𝟎 − 𝐲𝟐 ) … (𝐲𝟎 − 𝐲𝐧 ) 𝟎 +

(𝐲 − 𝐲𝟎 )(𝐲 − 𝐲𝟐 ) … (𝐲 − 𝐲𝐧 ) (𝐲 − 𝐲𝟎 )(𝐲 − 𝐲𝟏 ) … (𝐲 − 𝐲𝐧−𝟏 ) 𝐱𝟏 + ⋯ + 𝐱 (𝐲𝟏 − 𝐲𝟎 )(𝐲𝟏 − 𝐲𝟐 ) … (𝐲𝟏 − 𝐲𝐧 ) (𝐲𝐧 − 𝐲𝟎 )(𝐲𝐧 − 𝐲𝟏 ) … (𝐲𝐧 − 𝐲𝐧−𝟏 ) 𝐧

METHOD – 4: LAGRANGE’S INTERPOLATION FORMULA C

1

Find f(9.2) using Lagrange’s interpolation method from the following data. x

9

9.5

11

f(x)

2.1972

2.2513

2.3979

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟐𝟏𝟗𝟐 H

2

Use Lagrange’s interpolation formula to find y(2) from the data given below: x

0

1

3

4

y

1

2

10

17

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟐) = 𝟓 H

3

Using quadratic Lagrange interpolation, compute ln 9.2 from ln 9.0 = 2.1972, ln 9.5 = 2.2513, ln 11 = 2.3979.

W-19 (3)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐧 𝟗. 𝟐 = 𝟐. 𝟐𝟏𝟗𝟐 C

4

Find the value of tan 330 by Lagrange’s formula If, tan 300 = 0.5774, tan 320 = 0.6249 , tan 350 = 0.7002, tan 380 = 0.7813. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐭𝐚𝐧 𝟑𝟑𝟎 = 𝟎. 𝟔𝟒𝟗𝟒

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UNIT-2 » Interpolation 5

T

[25]

Find the Lagrange’s interpolation polynomial from the following data. Also find f(2). x

0

1

4

5

f(x)

1

3

24

39

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = 6

C

𝟏 (𝟑𝐱 𝟐 + 𝟏𝟎𝐱 𝟐 + 𝟐𝟕𝐱 + 𝟐𝟎), 𝐟(𝟐) = 𝟔. 𝟗 𝟐𝟎

Determine the interpolating polynomial of degree three and Compute f(2) by using Lagrange’s interpolation method from the following data. x

−1

0

1

3

f(x)

2

1

0

−1

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟎. 𝟎𝟒𝟏𝟕𝐱 𝟑 − 𝟏. 𝟎𝟒𝟏𝟕𝐱 + 𝟏, 𝐟(𝟐) = −𝟎. 𝟕𝟓 H

7

Find the Lagrange’s interpolating polynomial from the following data. Also, Compute f(4). x

2

3

5

7

f(x)

0.1506

0.3001

0.4517

0.6259

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟎. 𝟎𝟎𝟓𝟓𝐱 𝟑 − 𝟎. 𝟎𝟕𝟗𝟖𝐱 𝟐 + 𝟎. 𝟒𝟒𝟒𝟓𝐱 − 𝟎. 𝟒𝟔𝟑𝟖; 𝐟(𝟒) = 𝟎. 𝟑𝟖𝟗𝟔 H

8

Find a second degree polynomial passing through the points (0,0), (1,1) and (2,20) using Lagrange’s interpolation. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐱) = 𝟗𝐱 𝟐 − 𝟖𝐱

C

9

From the following data find value of x when y = 0.39. x

20

25

30

y = f(x)

0.342

0.423

0.500

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝟐𝟐. 𝟗𝟑𝟎𝟖 H 10

Apply Lagrange’s formula to find a root of the equation f(x) = 13.00. x

44

45

46

47

f(x)

13.40

13.16

12.93

12.68

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝟓. 𝟔𝟗

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UNIT-2 » Interpolation C

11

[26]

Apply Lagrange’s formula to find a root of the equation f(x) = 0. X

30

34

38

42

f(x)

−30

−13

3

18

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝟕. 𝟐𝟑 H 12

By Lagrange’s interpolation formula, Obtain the value of f(x) = 85. x

2

5

8

14

y

94.8

97.9

81.3

68.7

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟔. 𝟏𝟎𝟖𝟐

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

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UNIT-3 » Numerical Integration

[27]

UNIT-3 » NUMERICAL INTEGRATION ❖ NEWTON-COTES FORMULA: b

∫ f(x) dx = h [y0 + a

Where, h =

n n(2n − 3) 2 n(n − 2)2 3 Δy0 + Δ y0 + Δ y0 + ⋯ ] 2 12 24

b−a n

❖ TRAPEZOIDAL RULE (IF 𝐧 IS A MULTIPLE OF 𝟏): b

h ∫ f(x)dx = [(y0 + yn ) + 2(y1 + y2 + ⋯ + yn−1 )] 2 a

=

h [(First term + Last term) + 2(Remaining all Terms)] 2

Where, n is number of subinterval & h =

b−a n

❖ ERROR BOUNDS: ✓ If f′′(x) is continuous in [a, b],then the error in the trapezoidal rule is no longer then |𝐞𝐫𝐫𝐨𝐫| ≤

(𝐛 − 𝐚) 𝟐 𝐡 |𝐟′′(𝐌)| 𝟏𝟐

✓ Where |f′′(M)|is the largest value of |f′′(x)| in [a, b].

METHOD – 1: TRAPEZOIDAL RULE 1

C

1

State Trapezoidal rule with n = 10 and evaluate ∫ ex dx. 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟕𝟏𝟗𝟕 H

2

1 2

State Trapezoidal rule with n = 10 and evaluate ∫ e−x dx. 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕𝟒𝟔𝟐

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UNIT-3 » Numerical Integration H

3

[28]

1.3 x2

Use Trapezoidal rule to estimate ∫ e dx using a strip of width 2.

W-19 (3)

0.5

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟏𝟏𝟕𝟏 C

π

4

Evaluate ∫ sin x dx , taking n = 10. 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟗𝟖𝟑𝟓 H

5

6

Evaluate ∫ log x dx using trapezoidal rule. Take h = 1. 2

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟑𝟏𝟕𝟕 H

6

1

Evaluate ∫ 0

dx using trapezoidal rule. 1 + x2

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕𝟖𝟑𝟕 C

7

Given the data below, find the isothermal work done on the gas as it is v2

compressed from v1 = 22L to v2 = 2L use w = − ∫ P dv. v1

V, L P(atm.)

2

7

12

17

22

12.20

3.49

2.04

1.44

1.11

Use Trapezoidal rule. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟔𝟖. 𝟏𝟐𝟓 H

8

Consider the following tabular values. x

25

25.1

25.2

25.3

25.4

25.5

25.6

y = f(x)

3.205

3.217

3.232

3.245

3.256

3.268

3.28

Determine the area bounded by the given curve and X-axis between x = 25 to x = 25.6 by Trapezoidal rule. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟗𝟒𝟔𝟏

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UNIT-3 » Numerical Integration H

9

[29]

A river is 50 feet wide. The depth "d" in feet at a x feet from one bank is given by: x

0

10

20

30

40

50

d

0

4

7

9

12

15

Use Trapezoidal rule to find the area of cross-section of the river 50

A = ∫ d dx. 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟕𝟗𝟎

❖ SIMPSON’S 𝟏⁄𝟑- RULE (IF 𝐧 IS A MULTIPLE OF 𝟐): b

h ∫ f(x)dx = [(y0 + yn ) + 4(y1 + y3 + ⋯ ) + 2(y2 + y4 + ⋯ )] 3 a =

h [(First term + Last term) + 4(Odd Terms) + 2(Even Terms)] 3

Where n is number of subinterval & h =

b−a n

❖ ERROR BOUNDS: ✓ If f 4 (x) is continuous in [a, b],then the error in the Simpson’s rule is no longer then |𝐞𝐫𝐫𝐨𝐫| ≤

(𝐛 − 𝐚) 𝟒 𝟒 𝐡 |𝐟 (𝐌)| 𝟏𝟖𝟎

✓ Where, |f 4 (M)|is the largest value of |f 4 (x)| in [a, b].

METHOD – 2: SIMPSON’S 1/3 RULE C

1

1

Evaluate ∫ 0

dx 1 by simpson’s rule taking eleven ordinates and hence 1+x 3

find the value of log e 2 correct to five significant digits. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟔𝟗𝟑𝟐, 𝟎. 𝟔𝟗𝟑𝟏

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UNIT-3 » Numerical Integration T

2

[30] 1.3

1rd Derive Trapezoidal rule. Evaluate ∫ e dx by using Simpson’s rule. 3 x2

0.5

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟎𝟕𝟓𝟏 H

3

4

Evaluate integral ∫ ex dx by Simpson’s 1/3 rule (take h = 1). 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓𝟑. 𝟖𝟔𝟑𝟗 H

4

6

Evaluate ∫ 0

1 1 dx by using Simpson’s rule taking h = 1. 1 + x2 3

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟑𝟔𝟔𝟐 H

π

5

Evaluate ∫ sin x dx, take n = 10. 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟎𝟎𝟎𝟏 T

6

6

3

Evaluate ∫(1 + x 2 )2 dx using Simpson’s −2

1 rule with taking 6 3

subintervals. Use four digits after decimal point for calculation. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝟔𝟎. 𝟏𝟖𝟑𝟎 H

7

6

Using Simpson’s 1⁄3 rule evaluate ∫ f(x)dx from following data. h = 1. 0

x

0

1

2

3

4

f(x)

1

0.5

0.3333

0.25

0.2

5

6

0.1666 0.1428

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟗𝟓𝟖𝟔 T

8

2.5

1 Using Simpson’s rule evaluate ∫ f(x)dx from the following data. Take 3 1

h = 0.3. x

1

1.3

1.6

1.9

2.2

2.5

f(x)

1

1.69

2.56

3.61

4.84

6.25

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒. 𝟑𝟐𝟓

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UNIT-3 » Numerical Integration H

9

[31]

The speed, v m/s, of a car, t seconds after it starts, is shown in table: t

0

12

24

36

48

60

72

84

96

108

120

v

0

3.6

10.08

18.90

21.60

18.54

10.26

5.40

4.50

5.40

9.00

Using Simpson’s 1/3rd rule find the distance travelled by the car in 2 minutes. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏𝟐𝟐𝟐. 𝟓𝟔 C

10

A river is 80 meter wide. The depth ‘d’ in meters at a distance x meters from one bank is given by the following table, calculate the area of cross-section 1

of the river using Simpson’s 3 rule. x

0

10

20

30

40

50

60

70

80

d

0

4

7

9

12

15

14

8

3

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟖𝟐𝟑. 𝟑𝟑𝟑𝟑 T

11

16

Consider the following tabular values. Find ∫ ydx by Simpson 10

1 rule. 3

x

10

11

12

13

14

15

16

y

1.02

0.94

0.89

0.79

0.71

0.62

0.55

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒. 𝟕𝟐𝟑𝟑 C

12

The velocity ν of a particle at distance s from point on its linear path is given by the Following table. Estimate the time taken by the particle to travel the distance of 20 m using Simpson’s 1/3 rule. Time (t)

0

5

10

15

20

25

30

Speed (ν)

30

24

19

16

13

11

10

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓𝟏𝟑. 𝟑𝟑𝟑𝟑 T

13

Velocity of a car running on a straight road at intervals of 2 minutes are given below: Time

0

2

4

6

8

10

12

Velocity

0

22

30

27

18

7

0

Apply Simpson’s rule to find the distance covered by the car. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝟏𝟑. 𝟑𝟑

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UNIT-3 » Numerical Integration H 14

[32]

A wind force distributed against the side of a skyscraper is measured as Height l, m

0

30

60

90

120

Force F(l), N/m

0

350

1000

1500

2600

1

Use Simpson’s 3 rule to compute the net force. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐍𝐞𝐭 𝐟𝐨𝐫𝐜𝐞 = 𝟏, 𝟐𝟎, 𝟎𝟎𝟎 𝐍/𝐦

❖ SIMPSON’S 𝟑⁄𝟖- RULE ( IF 𝐧 IS A MULTIPLE OF 𝟑.): b

∫ f(x)dx = a

=

3h [(y0 + yn ) + 2(y3 + y6 + ⋯ ) + 3(y1 + y2 + y4 + y5 + ⋯ )] 8

3h [(First term + Last term) + 2(Multiple of 3) + 3(Not Multiple of 3)] 8

Where, n is number of subinterval & h =

b−a n

❖ ERROR BOUNDS: ✓ If f 4 (x) is continuous in [a, b],then the error in the Simpson’s 3/8 rule is no longer then |error| ≤

(b − a) 4 4 h |f (M)| 80

✓ Where, |f 4 (M)| is the largest value of |f 4 (x)| in [a, b].

METHOD – 3: SIMPSON’S 3/8 RULE C

1

3

1 3 Evaluate ∫ dx using Simpson’s rule with n = 6. 1+x 8

W-19 (4)

0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟑𝟖𝟖𝟖 H

2

1

3 1 1 State Simpson’s rule and evaluate ∫ dx taking h = . and also 8 1 + x2 6 0

by Simpson’s

1 1 rule taking h = . 3 4

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕𝟖𝟓𝟒

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UNIT-3 » Numerical Integration 3

C

[33]

5.2

Evaluate the integral ∫ log e x dx using Simpson’s 4

3 rule. 8

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟖𝟐𝟕𝟖 H

π

4

Dividing the range into 10 equal part, evaluate ∫ sin x dx by Simpson’s 0

3 8

rule. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟎𝟎𝟎𝟒 5

T

1

Evaluate ∫ 0

dx 3 using Simpson’s rule. 1+x 8

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟔𝟗𝟑𝟕 H

6

6

ex Find ∫ dx approximately using Simpson’s 3/8th rule with h = 1. 1+x 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏𝟐𝟑𝟔. 𝟗𝟔 H

7

1.4

1 Evaluate ∫ (sin x − log x + ex ) dx with h = 0.2 by Simpson’s rd and 3 0.2

3/8 th rule. 𝐀𝐧𝐬𝐰𝐞𝐫: By Simpson’s 1/3 rd Rule: 𝐲 = 𝟑. 𝟖𝟑𝟑𝟒 By Simpson’s 3/8 th Rule: 𝐲 = 𝟑. 𝟖𝟐𝟏𝟐 H

8

The speed, v m/s, of a car, t seconds after it starts, is shown in table: t

v

0

0

12

3.6

24

10.08

36

18.90

48

21.60

60

18.54

72

10.26

84

5.4

96

4.5

10

12

8

0

5.4

9.0

Using Simpson’s 1/3rd rule find the distance travelled by the car in 2 minutes. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟕𝟎. 𝟏𝟔𝟓𝟐

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UNIT-3 » Numerical Integration C

9

[34]

A train is moving at the speed of 30 m/sec. Suddenly brakes are applied. The speed of the train per second after t seconds is given by Time(t)

0

5

10

15

20

25

30

Speed(v)

30

24

19

16

13

11

10

The following table. Apply Simpson’s three-eight rule to determine the distance moved by the train in 30 sec. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓𝟏𝟏. 𝟖𝟕𝟓𝟎

❖ GAUSSIAN INTEGRATION (GAUSSIAN QUADRATURE): ✓ One Point Gaussian Quadrature Formula (n = 1) 1

∫ f(x)dx = 2 f(0) −1

✓ Two Point Gaussian Quadrature Formula (n = 2) 1

∫ f(x)dx = f (− −1

1

1 ) + f( ) √3 √3

✓ Three Point Gaussian Quadrature Formula (n = 3) 1

8 5 3 3 ∫ f(x)dx = f(0) + (f (−√ ) + f (√ )) 9 9 5 5

−1

✓ You can also use following formula to find Gaussian Quadrature. b

∫ f(x)dx = a

b−a b−a b+a (w1 f(y1 ) + w2 f(y2 )+ ⋯ + wn f(yn )), Where x = y+ 2 2 2

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UNIT-3 » Numerical Integration

[35]

❖ TABLE: 𝐧 1

𝐰𝐢 2.0000

𝐲𝐢 0.0000

2

1.0000 1.0000

−0.57735 0.57735

3

0.55555 0.88889 0.55555

−0.77460 0.00000 0.77460

𝐍 4

5

𝐰𝐢 0.34785 0.65214 0.65214 0.34785

𝐲𝐢 −0.86114 −0.33998 0.33998 0.86114

0.23693 0.47863 0.56889 0.47863 0.23693

−0.90618 −0.53847 0.00000 0.53847 0.90618

METHOD – 4: GAUSSIAN QUADRATURE (GAUSSIAN INTEGRATION) H

1

1

Evaluate ∫ exp(−x

2)

dx by Gauss integration formula with n = 3.

0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕𝟒𝟔𝟖 2

C

1

Evaluate ∫ −1

dx by using Gaussian quadrature formula with one point, 1 + x2

two points & three points. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐, 𝟏. 𝟓, 𝟏. 𝟓𝟖𝟑𝟑𝟑 H

3

1

Evaluate I = ∫ 0

dx by Gaussian formula with one point, two − point 1+x

and three − points. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟔𝟔𝟔𝟔𝟕, 𝟎. 𝟔𝟗𝟐𝟑𝟏, 𝟎. 𝟔𝟗𝟑𝟏𝟐 C

4

3

Evaluate ∫ sinx dx using Gauss Quadrature of five points. Compare the 1

result with analytic value. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟓𝟑𝟎𝟑𝟏, 𝟏. 𝟓𝟑𝟎𝟐𝟗

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

W-19 (3)

UNIT-3 » Numerical Integration

[36]

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-4 » Numerical Solution of ODE

[37]

UNIT- 4 » NUMERICAL METHODS FOR O.D.E. ❖ PICARD METHOD: dy = f(x, y) and y(x0 ) = y0 , then dx

If

x

yn = y0 + ∫ f(x, yn−1 ) dx ; n = 1,2,3, … x0

Note: (1). We stop the process, When yn+1 = yn , up to the desired decimal places. (2). This method is applicable only to a limited class of equations in which successive integrations can be performed easily.

METHOD – 1: PICARD METHOD 1

C

Using Picard’s method solve

dy − 1 = xy with initial condition y(0) = 1, dx

compute y(0.1) correct to three decimal places. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟎𝟓 H

2

Solve

dy = 3 + 2xy , Where y(0) = 1, for x = 0.1 by Picard’s method. dx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟑𝟏𝟎𝟎 T

3

Obtain Picard’s second approximation solution of the initial value problem dy = x 2 + y 2 for x = 0.4 correct places, given that y(0) = 0. dx 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟒) = 𝟎. 𝟎𝟐𝟏𝟒

C

4

Obtain Picard’s second approximate solution of the initial value problem dy x2 = , y(0) = 0 dx y 2 + 1 𝟏 𝟑 𝐱𝟗 𝐱 𝟏𝟓 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝟐 = 𝐱 − + − 𝟑 𝟖𝟏 𝟏𝟐𝟏𝟓

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UNIT-4 » Numerical Solution of ODE H

5

Using Picard’s method solve

[38]

dy = x + y 2 , y(0) = 1 up to 2nd dx

approximation. 𝟑 𝟐 𝐱𝟒 𝐱𝟓 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝟐 = 𝟏 + 𝐱 + 𝐱 𝟐 + 𝐱 𝟑 + + 𝟐 𝟑 𝟒 𝟐𝟎

❖ TAYLOR SERIES: If

dy = f(x, y) and y(x0 ) = y0 , then dx

y(xn ) = yn = yn−1 +

h ′ h2 ′′ h3 ′′′ yn−1 + yn−1 + y + ⋯ ; n = 1,2,3 … 1! 2! 3! n−1

✓ Where, xn = x0 + nh ; n = 1,2,3, … ✓ Here, h is step size and n is the number of steps.

METHOD – 2: TAYLOR’S SERIES METHOD H

1

Find by Taylor’s series method the value of y at x = 0.1 and x = 0.2 to four places of decimal, for

dy = x 2 − 1; y(0) = 1. dx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟎. 𝟗𝟎𝟎𝟑, 𝐲(𝟎. 𝟐) = 𝟎. 𝟖𝟎𝟐𝟔 H

2

Evaluate y(0.1)correct to four decimal places using Taylor’s series method if

dy = y 2 + x , y(0) = 1. dx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟔𝟓 C

3

Use Taylor’s series method to solve the equation

dy = 3x + y 2 to dx

approximate y when x = 0.1 given that y(0) = 1. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟒𝟓 H

4

Using Taylor series method, find y(0.1) and y(0.2) correct to four decimal places, if y(x) satisfies

dy = x − y 2 , y(0) = 1. dx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟎. 𝟗𝟏𝟑𝟕, 𝐲(𝟎. 𝟐) = 𝟎. 𝟖𝟓𝟏𝟏

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UNIT-4 » Numerical Solution of ODE 5

C

[39]

Solve the Ricatti’s equation y ′ = x 2 + y 2 using the Taylor’s series method for the initial condition y(0) = 0.Where 0 ≤ x ≤ 0.2.

W-19 (3)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟐) = 𝟎. 𝟎𝟒 6

H

Using Taylor’s series method, find y(1.1)correct to four decimal places, given that

1 dy = x y 3 , y(1) = 1. dx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟏. 𝟏) = 𝟏. 𝟏𝟎𝟔𝟖 7

H

Solve

dy = 1 + y 2 , y(0) = 0, using Taylor’s series method and find the dx

value of y for x = 0.2, taking h = 0.2. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟐) = 𝟎. 𝟐𝟎𝟐𝟕

❖ EULER’S METHOD (RK 1ST ORDER METHOD): If

dy dx

= f(x, y) and y(x0 ) = y0 , then

𝐲𝐧+𝟏 = 𝐲𝐧 + 𝐡 [𝐟(𝐱 𝐧 , 𝐲𝐧 )] ; 𝐧 = 𝟎, 𝟏, 𝟐, … Where, xn = x0 + nh ; h is step size ; n = 1,2,3, …

METHOD – 3: EULER’S METHOD H

1

Using Euler’s method solve for y at x = 0.1 from

dy = x + y + xy, dx

y(0) = 1, in five steps. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟐𝟒 H

2

Given

dy = x + y , with y(0) = 1 using Euler’s method to find y(0.1) & dx

y(0.2). 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟎𝟓𝟎, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟑𝟏𝟎 H

3

Use Euler method to find y(0.2) given that

dy 2x = y − , y(0) = 1; h = 0.1 dx y

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟐) = 𝟏. 𝟏𝟗𝟏𝟖

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W-19 (3)

UNIT-4 » Numerical Solution of ODE T

4

Use Euler’s method to find y(2)from the differential equation

[40] dy = x + 2y dx

, y(1) = 1, Taking h = 0.1. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟐) = 𝟗. 𝟓𝟖𝟓𝟓 C

5

Solve initial value problem

dy = x√y , y(1) = 1 and hence find y(1.5) by dx

taking h = 0.1 using Euler’s method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟏. 𝟓) = 𝟏. 𝟔𝟕 H

6

Using Euler’s method, find an approximate value of corresponding to x = 0.3, given that

dy y − x = ; y(0) = 1. Take h = 0.1. dx y + x

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟑) = 𝟏. 𝟐𝟓𝟒𝟒 T

7

Use Euler’s method to find an approximation value of y at x = 0.1 for the initial value problem

dy = x − y 2 ; y(0) = 1. dx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟎. 𝟗𝟏𝟏𝟑 H

8

Use Euler’s method to find the approximate value of y at x = 1.3 taking h = 0.1. Given

dy y − x = and y(1) = 2. dx √xy

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐. 𝟏𝟗𝟑𝟒

❖ IMPROVED EULER’S METHOD: ✓ It is also known as Heun’s Method OR Modified Euler’s Method OR R-K 2nd Order Method. If

dy dx

= f(x, y) and y(x0 ) = y0 , then RK 4th Order Formula

1 yn+1 = yn + (K1 + K 2 ); n = 0,1,2, … 2 ✓ Where, xn = x0 + nh ; n = 1,2,3, … ✓ Here, h is step size and n is the number of steps.



K1 = h ⋅ f(xn , yn )



K 2 = h ⋅ f(xn + h, yn + K1 )

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UNIT-4 » Numerical Solution of ODE

[41]

METHOD – 4: IMPROVED EULER’S METHOD 1

C

Use 2nd order Range Kutta method to find y(1.1)provided

dy = x2 + y2 dx

with y(1) = 1. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟏. 𝟏) = 𝟏. 𝟐𝟑𝟑𝟒 2

T

Use Runge Kutta second order method to find the approximate value of y(0.2) given that

dy = x − y 2 and y(0) = 1 and h = 0.1. dx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟐) = 𝟎. 𝟖𝟓𝟐𝟑 3

C

Given that y = 1.3 when x = 1 and

dy = 3x + y use second order RK dx

method to approximate y , when x = 1.2 use step size 0.1. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟏. 𝟐) = 𝟐. 𝟑𝟏𝟑𝟓 4

H

Given that

dy = x + y 2 and y = 1 at x = 0. Find an approximate value of y dx

at x = 0.3 by Modified Euler method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟑) = 𝟏. 𝟒𝟖𝟎𝟏

❖ RUNGE KUTTA 4TH If

dy dx

ORDER METHOD:

= f(x, y) and y(x0 ) = y0 , then RK 4th Order Formula

1 yn+1 = yn + (K1 + 2K 2 + 2K 3 + K 4 ); n = 0,1,2, … 6 Where, xn = x0 + nh ; h is step size; n = 1,2,3, …



K1 = h ⋅ f(xn , yn )



K 2 = h ⋅ f (xn + 2 , yn +



K 3 = h ⋅ f (xn + , yn +



K 4 = h ⋅ f(xn + h, yn + K 3 )

h

K1 2

h

K2

2

2

) )

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-4 » Numerical Solution of ODE

[42]

METHOD – 5: RK 4TH ORDER METHOD C

1

State the formula for Runge Kutta method of fourth order and use it to calculate y(0.2) given that

dy = x + y , y(0) = 1 taking h = 0.1. dx

W-19 (4)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟒𝟐𝟗 H

2

Describe y(0.1) and y(0.2). Correct to four decimal places from dy = 2x + y , y(0) = 1 use fourth order RK method. dx 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟓𝟓, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟔𝟒𝟐

H

3

Using Runge Kutta fourth order method solve for y(0.1)and y(0.2) given that

dy = xy + y 2 , dx

y(0) = 1.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟔𝟗, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟕𝟕𝟒 C

4

dy y 2 − x 2 Using Runge Kutta method to fourth order, solve = with dx y 2 + x 2 y(0) = 1 at x = 0.2 with a Step − size of 0.1. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟐) = 𝟏. 𝟏𝟗𝟔𝟎

H

5

Use fourth order Runge Kutta method to find the value of y at x = 1 y−x given that y ′ = such that y(0) = 1. (Take h = 0.5) y+x 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟏) = 𝟏. 𝟒𝟗𝟗𝟐

H

6

Apply Runge Kutta fourth order method, to find an approximate value of y when x = 0.2 in steps of 0.1, if

dy = x + y 2 , given that y = 1 when dx

x = 0. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟔𝟓, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟕𝟑𝟔 C

7

Using the Range Kutta method of fourth order, find y at x = 0.1 given differential equation

dy = 3ex + 2y, y(0) = 0 by taking h = 0.1 and also dx

compare the solution with exact solution. 𝐀𝐧𝐬𝐰𝐞𝐫: By RK 4th Method: 𝐲(𝟎. 𝟏) = 𝟎. 𝟑𝟒𝟖𝟕 ; By Exact Method: 𝐲(𝟎. 𝟏) = 𝟎. 𝟑𝟒𝟖𝟎

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UNIT-4 » Numerical Solution of ODE H

8

[43]

Apply Runge Kutta fourth order method to calculate y(0.2) and y(0.4) given

dy 2x = y − , y(0) = 1. dx y

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟐) = 𝟏. 𝟏𝟖𝟑𝟐, 𝐲(𝟎. 𝟒) = 𝟏. 𝟑𝟒𝟏𝟕 H

9

Solve initial value problem

dy = −2xy 2 ; y(0) = 1 with h = 0.2 for y(0.2) dx

using Runge − Kutta fourth order method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟎. 𝟐) = 𝟎. 𝟗𝟔𝟏𝟓 T

10

Use the Runge Kutta 4th order method with h = 0.1 to find the approx. solution for y(1.1) for the initial value problem

dy = 2xy ; y(1) = 1. dx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝟏. 𝟏) = 𝟏. 𝟐𝟑𝟑𝟕

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

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UNIT-4 » Numerical Solution of ODE

[44]

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UNIT-5 » Curve Fitting

[45]

UNIT-5 » CURVE FITTING ❖ CURVE FITTING: ✓ Curve fitting is the process of finding the ‘best-fit’ curve for a given set of data. It is the representation of the relationship between two variables by means of an algebraic equation. ❖ THE METHOD OF LEAST SQUARE: ✓ The method of least squares assumes that the best-fit curve of a given type is the curve that has the minimum sum of the square of the deviation (least square error) from a given set of data. ✓ Suppose that the data points are (x1 , y1 ), (x2 , y2 ), … , (xn , yn ), where x is independent and y is dependent variable. Let the fitting curve f(x) has the following deviations (or errors or residuals) from each data points. 𝐝𝟏 = 𝐲𝟏 − 𝐟(𝐱 𝟏 ), 𝐝𝟐 = 𝐲𝟐 − 𝐟(𝐱 𝟐 ), … , 𝐝𝐧 = 𝐲𝐧 − 𝐟(𝐱 𝐧 ). ✓ Clearly, some of the deviations will be positive and others negative. Thus, to give equal weightage to each error, we square each of these and form their sum; that is, 𝐃 = 𝐝𝟏 𝟐 + 𝐝𝟐 𝟐 + ⋯ + 𝐝𝐧 𝟐 ✓ Now, according to the method of least squares, the best fitting curve has the property that 𝐧 𝟐

𝟐

𝟐

𝐧 𝟐

𝐃 = 𝐝𝟏 + 𝐝𝟐 + ⋯ + 𝐝𝐧 = ∑ 𝐝𝐢 = ∑[𝐲𝐢 − 𝐟(𝐱 𝐢 )]𝟐 = 𝐚 𝐦𝐢𝐧𝐢𝐦𝐮𝐦. 𝐢=𝟏

𝐢=𝟏

❖ FITTING A STRAIGHT LINE 𝐲 = 𝐚 + 𝐛𝐱 (LINEAR APPROXIMATION): ✓ Suppose the equation of a straight line of the form y = a + bx is to be fitted to the n-data points (x1 , y1 ), (x2 , y2 ), … , (xn , yn ), n ≥ 2, ✓ Where a is y-intercept and b is its slope. ✓ For the general point (xi , yi ), the vertical distance of this point from the line y = a + bx is the deviation di , then di = yi − f(xi ) = yi − a − bxi ✓ Applying method of least squares, the values of a and b are so determined as to minimize

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UNIT-5 » Curve Fitting

[46]

n

D = ∑(yi − a − bxi )2 i=1

✓ This will be so if n

∂D = 0 ⇒ −2 ∑(yi − a − bxi ) = 0 ∂a i=1 n

∂D = 0 ⇒ −2 ∑ xi (yi − a − bxi ) = 0 ∂b i=1

✓ Simplifying and expanding the above equations, we have n

n

n

∑ yi = a ∑ 1 + b ∑ xi , i=1 n

i=1

i=1

n

n

∑ xi yi = a ∑ xi + b ∑ xi 2 i=1

i=1

i=1

✓ Which implies n

n

∑ yi = an + b ∑ xi i=1 n

… (1)

i=1 n

n

∑ xi yi = a ∑ xi + b ∑ xi 2 i=1

i=1

… (2)

i=1

✓ According to the principle of least squares, these partial derivatives must be equal to zero. Hence, equating zero and simplifying. We obtain following normal equations for the best fitting straight line y = a + bx. ∑y = n a + b∑x

∑ xy = a ∑ x + b ∑ x 2 }

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UNIT-5 » Curve Fitting

[47]

METHOD – 1: FITTING A STRAIGHT LINE 𝐲 = 𝐚 + 𝐛 𝐱 1

C

x

−1

0

1

2

y

1

0

1

4

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟏 + 𝐱 H

2

x

1

2

3

4

5

Y

1

3

5

6

5

W-19 (4)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟎. 𝟕 + 𝟏. 𝟏 𝐱 3

C

x

1

2

3

4

5

y

14

27

40

55

68

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟏𝟑. 𝟔𝐱 4

T

x

1

3

4

6

8

9

11

14

1

y

1

2

4

4

5

7

8

9

1

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟎. 𝟒𝟖𝟕𝟖 + 𝟎. 𝟔𝟒𝟐𝟑𝐱 H

5

Find y when x = 1.8. x

2

5

6

9

11

y

2

4

6

9

10

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = −𝟎. 𝟎𝟐𝟒𝟒 + 𝟎. 𝟗𝟒𝟑𝟏𝐱, 𝐘(𝟏. 𝟖) = 𝟏. 𝟔𝟕𝟑𝟐 6

C

x

20.5

32.7

51.0

73.2

95.7

y

765

826

873

942

1032

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟕𝟎𝟐. 𝟏𝟕𝟐𝟏 + 𝟑. 𝟑𝟗𝟒𝟖𝐱 H

7

Find y when x = 1.8. x

1

1.2

1.4

1.6

y

0

0.182

0.336

0.47

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = −𝟎. 𝟕𝟔𝟗𝟔 + 𝟎. 𝟕𝟖𝟐𝐱, 𝐘(𝟏. 𝟖) = 𝟎. 𝟔𝟑𝟖

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UNIT-5 » Curve Fitting H

8

[48]

Fit a straight line to the following data, and find y when x = 2.4. x

1

2

3

4

5

6

7

y

0.5

2.5

2.0

4.0

3.5

6.0

5.5

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟎. 𝟎𝟕𝟏𝟒𝟑 + 𝟎. 𝟖𝟑𝟗𝟑𝐱; 𝐲(𝟐. 𝟒) = 𝟐. 𝟎𝟖𝟓𝟖 T

9

If P is the pull required to lift a load W by means of a pulley block, find a linear approximation of the form P = mW+c connecting P and W, using the following data: P

12

15

21

25

W

50

70

100

120

Where P and W are taken in kg. wt. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏 = 𝟐. 𝟐𝟕𝟓𝟗 + 𝟎. 𝟏𝟖𝟕𝟗𝐖 C

10

A simply supported beam carries a concentrated load P(lb) at its midpoint. Corresponding to various values of P, the maximum deflection Y(in) is measured. The data are given below P

100

120

140

160

180

200

Y

0.45

0.55

0.60

0.70

0.80

0.85

Find a law of the form Y = a + bP using least square method. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟎. 𝟎𝟒𝟕𝟔 + 𝟎. 𝟎𝟎𝟒𝟏𝐏

❖ FITTING A PARABOLA 𝐲 = 𝐚 + 𝐛𝐱 + 𝐜𝐱 𝟐 BY LEAST SQUARE APPROXIMATION: ✓ Consider a set of n pairs of the given values (x, y) for fitting the curve y = a + bx + cx 2 . The residual R = y − (y = a + bx + cx 2 ) is the difference between the observed and estimated values of y. We have to find a, b, c such that the sum of the squares of the residuals is minimum (least). n

Let S = ∑[y − (a + bx + cx 2 )]2 … … (1) 1

✓ Differentiating S with respect to a, b, c and Equating zero. ✓ We obtain following normal Equations for the best fitting y = a + bx + cx 2 curve (parabola) of second degree.

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UNIT-5 » Curve Fitting

[49]

∑ y = n a + b ∑ x + c ∑ x2 ∑ x y = a ∑ x + b ∑ x2 + c ∑ x3 ∑ x2 y = a ∑ x2 + b ∑ x3 + c ∑ x4

METHOD – 2: FITTING A PARABOLA 𝐲 = 𝐚 + 𝐛𝐱 + 𝐜𝐱 𝟐 H

1

W-19 (4)

x

1

2

3

4

y

6

11

18

27

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟑 + 𝟐𝐱 + 𝐱 𝟐 2

C

Find the least squares approximations of second degree for the following data: x

−1

0

1

2

3

y

5

6

21

50

93

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟔 + 𝟖𝐱 + 𝟕 𝐱 𝟐 3

C

x

1

2

3

4

5

y

5

12

26

60

97

Find f(6). 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟏𝟎. 𝟒 − 𝟏𝟏. 𝟎𝟖𝟓𝟕𝐱 + 𝟓. 𝟕𝟏𝟒𝟑𝐱 𝟐 ; 𝐟(𝟔) = 𝟏𝟒𝟗. 𝟔𝟎𝟎𝟔 4

C

Fit second degree polynomial to the data, (1,3), (2,7), (3,13), (4,21), (5,31). 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟏 + 𝐱 + 𝐱 𝟐

H

5

x

0

1

2

3

4

y

1

1.8

1.3

2.5

2.3

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟏. 𝟎𝟕𝟕𝟏 + 𝟎. 𝟒𝟏𝟓𝟕𝐱 − 𝟎. 𝟎𝟐𝟏𝟒𝐱 𝟐 T

6

x

0

1

2

3

4

5

6

y

14

18

23

29

36

40

46

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟏𝟑. 𝟒𝟓𝟐𝟑 + 𝟒. 𝟗𝟔𝟒𝟐𝐱 + 𝟎. 𝟎𝟖𝟑𝐱 𝟐

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UNIT-5 » Curve Fitting T

7

[50]

x

2

4

6

8

10

y

3.07

12.85

31.47

57.38

91.29

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟎. 𝟗𝟗𝟏𝟗𝐱 𝟐 − 𝟎. 𝟖𝟓𝟓𝟎𝐱 + 𝟎. 𝟔𝟗𝟔 T

8

x

−1

0

1

2

y

2

1

0

−2

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟏. 𝟏𝟓 − 𝟏. 𝟎𝟓𝐱 − 𝟎. 𝟐𝟓𝐱 𝟐 H

9

𝐱

−𝟐

−𝟏

𝟎

𝟏

𝟐

𝐲

−𝟏

𝟎

𝟎

𝟏

𝟓

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟏. 𝟑𝐱 + 𝟎. 𝟓𝐱 𝟐 H 10

Fit a second-degree parabola y = ax 2 + bx + c in least square sense for the following data: x

1

2

3

4

5

y

10

12

13

16

19

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟎. 𝟐𝟖𝟓𝟕𝐱 𝟐 + 𝟎. 𝟒𝟖𝟓𝟕𝐱 + 𝟗. 𝟒

❖ NON-POLYNOMIAL APPROXIMATION OR NON-LINEAR REGRESSION: ✓ 𝐲 = 𝐚𝐞𝐛𝐱 yields (1). Taking Logarithm on both sides log y = loga + bx (2). Denoting logy = Y and loga = A , the above equation becomes Y = A + bx Which is a straight line. (3). From above equation A, b can be found & consequently a=Antilog A can be calculated. ✓ 𝐲 = 𝐚𝐱 𝐛 yields (1). Taking Logarithm on both sides log y = log a + blog x (2). Denoting log y = Y , loga = A and log x = X , we obtain Y = A + bX, which is a straight line.

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UNIT-5 » Curve Fitting

[51]

(3). From above equation A, b can be found & consequently a=Antilog A can be calculated. ✓ 𝐲 = 𝐚𝐛𝐱 yields (1). Taking Logarithm on both sides log y = log a + x log b (2). Denoting logy = Y , log a = A , and log b = B , we obtain Y = A + Bx , which is a straight line. (3). From above equation A, B can be found & consequently a=Antilog A and b=Antilog B can be calculated.

METHOD – 3: NON-LINEAR REGRESSION H

1

Determine 𝐚 and 𝐛 so that 𝐲 = 𝐚𝐞𝐛𝐱 fits the data: x

1

2

3

4

y

7

11

17

27

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟒. 𝟒𝟔𝟖𝟎, 𝐛 = 𝟎. 𝟒𝟒𝟖𝟓 2

C

Determine a and b so that y = aebx fits the data : X

1

5

7

9

12

y

10

15

12

15

21

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟗. 𝟒𝟕𝟓𝟒, 𝐛 = 𝟎. 𝟎𝟓𝟗𝟎 3

T

Find the exponential curve y = aebx to the following data. X

0

2

4

6

8

y

150

63

28

12

5.6

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟖. 𝟕𝟏𝟔𝟐, 𝐛 = −𝟎. 𝟏𝟕𝟖𝟖 H

4

Fit a curve of the form y = ax b to the following data. X

20

16

10

11

14

y

22

41

120

89

56

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟐𝟕𝟏𝟖𝟏. 𝟒𝟖𝟏𝟏, 𝐛 = −𝟐. 𝟑𝟔𝟎𝟖

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UNIT-5 » Curve Fitting C

5

[52]

Fit a curve of the form y = ax b to the following data. X

1

2

3

4

5

y

0.5

2

4.5

8

12.5

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟎. 𝟓, 𝐛 = 𝟐 T

6

By the method of least square fit a curve of the form y = ax b to the following data: x

2

3

4

5

y

27.8

62.1

110

161

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟕. 𝟑𝟖𝟖𝟑, 𝐛 = 𝟏. 𝟗𝟐𝟗𝟑 C

7

By the method of least square fit a curve of the form y = ax b to the following data: x

2

3

4

5

6

y

8.3

15.4

33.1

65.2

126.4

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟐. 𝟎𝟒𝟗𝟔, 𝐛 = 𝟏. 𝟗𝟗𝟏𝟔 H

8

Fit a curve of the form y = abx for the data and hence find the estimation for y when x = 8. X

1

2

3

4

5

6

7

y

87

97

113

129

202

195

193

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟕𝟑. 𝟕𝟒𝟏𝟔, 𝐛 = 𝟎. 𝟏𝟓𝟔𝟎, 𝐲(𝟖) = 𝟐𝟓𝟔. 𝟖𝟑𝟏𝟓

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

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UNIT-6 » Basic Probability

[53]

UNIT 6 – BASIC PROBABILITY ❖ INTRODUCTION: ✓ The words PROBABLE and POSSIBLE CHANCES are quite familiar to us. ✓ We use these words when we are sure of the result of certain events. These words convey the sense of uncertainty of occurrence of events. ✓ PROBABILITY is the word we use to calculate the degree of the certainty of events.

W’19

✓ There are two types of approaches in the theory of PROBABILITY. (1). Classical Approach – By Blaise Pascal (2). Axiomatic Approach – By A. Kolmogorov ✓ An experiment, in which we know all the possible outcomes in advance but which of them will occur is known only after the experiment is performed, is called a Random Variable. ❖ EXAMPLES: Event

Outcomes

Toss Coin

Head or Tail

Roll Die

1, 2, 3, 4, 5, 6

Pick a Card

Ace,2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King (Clubs, Diamonds, Hearts, Spades)

Throw a stone

It comes back surely. (Why do we think about possibility)

❖ RANDOM EXPERIMENTS: ✓ An experiment is called random experiment if it satisfies conditions: ✓ It has more than one possible outcome. ✓ It is not possible to predict the outcome in advance. ❖ SAMPLE SPACE: ✓ The set of outcomes is called the sample space of the experiment. ✓ It is denoted by “U”. ✓ Each element of the sample space is called a SAMPLE POINT.

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UNIT-6 » Basic Probability

[54]

✓ If a sample space is in one-one correspondence with a finite set, {𝐱 ∈ ℕ | 𝟏 ≤ 𝐱 ≤ 𝐧 , 𝐧 ∈ ℕ}, then it is called a finite sample space. Otherwise it is knowing as an infinite sample space. ❖ EXAMPLES: ✓ Finite Sample Space: An experiment of tossing a coin twice. U = {H, T} × {H, T} = {HH, HT, TH, TT} ✓ Infinite Sample Space: An experiment of tossing a coin until a HEAD comes up for the first time. U = {H, TH, TTH, TTTH, TTTTH, TTTTTH, … } ❖ EVENT: ✓ A subset of sample space is known as Event. Each member is called Sample Point. ✓ Example: Experiment U : Tossing a coin twice. 𝐔 = {𝐇𝐇, 𝐇𝐓, 𝐓𝐇, 𝐓𝐓} Event A : getting TAIL both times. 𝐀 = {𝐓𝐓} Event B : getting TAIL exactly once. 𝐀 = {𝐇𝐓, 𝐓𝐇} ❖ DEFINITIONS: ✓ The subset ∅ of a sample space is called “Impossible Events”. ✓ The subset 𝐔(𝐢𝐭𝐬𝐞𝐥𝐟) of a sample space is called “Sure/Certain Events”. ✓ If Subset contains only one element, it is called “Elementary/Sample Events”. ✓ If Subset contains more than one element, it is called “Compound/Decomposable Events”. W-19(1) ✓ The set contains all elements other than event A is known as “Complementary Event” of A. It is known as A’. ✓ A Union of Events A and B is Union of sets A and B (As per set theory). ✓ An Intersection of Events A and B is Intersection of sets A and B (As per set theory). ✓ If 𝐀 ∩ 𝐁 = 𝛟, Events are called Mutually Exclusive Events (Disjoint set).

W-19(1)

✓ Set Notation: 𝐀 ∩ 𝐁 = { 𝐱 | 𝐱 ∈ 𝐀 𝐀𝐍𝐃 𝐱 ∈ 𝐁 }

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UNIT-6 » Basic Probability

[55]

✓ If 𝐀 ∪ 𝐁 = 𝐔, Events are called Mutually Exhaustive Events. Set Notation: 𝐀 ∪ 𝐁 = { 𝐱 | 𝐱 ∈ 𝐀 𝐎𝐑 𝐱 ∈ 𝐁 } ✓ If 𝐀 ∩ 𝐁 = 𝛟 and 𝐀 ∪ 𝐁 = 𝐔, Events are called Mutually Exclusive & Exhaustive Events. ✓ Suppose 𝐀 𝟏 , 𝐀 𝟐 , 𝐀 𝟑 ,…, 𝐀 𝐧 are Events. ✓ If, (1). 𝐀 𝐢 ∩ 𝐀 𝐣 = 𝛟 for 𝐢 = 𝟏, 𝟐, … , 𝐧 ; 𝐣 = 𝟏, 𝟐, … , 𝐧 ; 𝐢 ≠ 𝐣. (2). ⋃𝐧𝐢=𝟏 𝐀 𝐢 = 𝐔 ✓ Events 𝐀 𝟏 , 𝐀 𝟐 , 𝐀 𝟑 ,…, 𝐀 𝐧 are said to be Mutually Exclusive and Exhaustive Events. ✓ Difference Events A − B = A − (A ∩ B) = {x |x ∈ U, x ∈ A AND x ∉ B } B − A = B − (A ∩ B) = {x |x ∈ U, x ∈ B AND x ∉ A } ✓ Symmetric Difference Set A Δ B = (A − B) ∪ (B − A)

METHOD – 1: BASIC EXAMPLES ON PROBABILITY C

1

Define Mutually Exclusive and EXHAUSTIVE events with a suitable example.

C

2

Describe the sample space for the indicated random experiments. (1). A coin is tossed 3 times. (2). A coin and die is tossed simultaneously. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐒 = {𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓𝐇, 𝐓𝐓𝐓} 𝐒 = {𝐇𝟏, 𝐇𝟐, 𝐇𝟑, 𝐇𝟒, 𝐇𝟓, 𝐇𝟔, 𝐓𝟏, 𝐓𝟐, 𝐓𝟑, 𝐓𝟒, 𝐓𝟓, 𝐓𝟔}

C

3

Four cards are labeled with A, B, C and D. We select and two cards at random without replacement. Describe the sample space for the experiments. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐒 = {𝐀𝐁, 𝐀𝐂, 𝐀𝐃, 𝐁𝐂, 𝐁𝐃, 𝐂𝐃}

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UNIT-6 » Basic Probability H

4

[56]

A balanced coin is tossed thrice. If three tails are obtained, a balance die is rolled. Otherwise the experiment is terminated. Write down the elements of the sample space. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐒 = { 𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓𝐇, 𝐓𝐓𝐓𝟏, 𝐓𝐓𝐓𝟐, 𝐓𝐓𝐓𝟑, 𝐓𝐓𝐓𝟒, 𝐓𝐓𝐓𝟓, 𝐓𝐓𝐓𝟔}

C

5

Let a coin be tossed. If it shows head, we draw a ball from a box containing 3 identical red and 4 identical green balls and if it shows a tail, we throw a die. What is the sample space of experiments? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐒 = {𝐇𝐑 𝟏 , 𝐇𝐑 𝟐 , 𝐇𝐑 𝟑 , 𝐇𝐆𝟏 , 𝐇𝐆𝟐 , 𝐇𝐆𝟑 , 𝐇𝐆𝟒 , 𝐓𝟏, 𝐓𝟐, 𝐓𝟑, 𝐓𝟒, 𝐓𝟓, 𝐓𝟔}

H

6

An experiment consists of rolling a die and then tossing a coin once if the number on the die is odd. If the number on the die is even the coin is tossed twice. Write sample space. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐒 = 𝐒𝐎 ∪ 𝐒𝐄 Where, 𝐒𝐎 = {𝟏𝐇, 𝟑𝐇, 𝟓𝐇, 𝟏𝐓, 𝟑𝐓, 𝟓𝐓} 𝐒𝐄 = {𝟐𝐇𝐇, 𝟐𝐇𝐓, 𝟐𝐓𝐇, 𝟐𝐓𝐓, 𝟒𝐇𝐇, 𝟒𝐇𝐓, 𝟒𝐓𝐇, 𝟒𝐓𝐓, 𝟔𝐇𝐇, 𝟔𝐇𝐓, 𝟔𝐓𝐇, 𝟔𝐓𝐓}

C

7

A coin is tossed 3 times. Give the elements of the following events: Event A: Getting at least two heads Event B: Getting exactly two tails Event C: Getting at most one tail Event D: Getting at least one tail Find A ∩ B , C ∩ D′ , A ∪ C , B ∩ C , A′ ∪ C′. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐒 = {𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓H, 𝐓𝐓𝐓} EVENT 𝐀 = {𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐓𝐇𝐇} EVENT 𝐁 = {𝐇𝐓𝐓, 𝐓𝐇𝐓, 𝐓𝐓𝐇} EVENT 𝐂 = {𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐓𝐇𝐇} EVENT 𝐃 = {𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓𝐇, 𝐓𝐓𝐓}

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UNIT-6 » Basic Probability H

8

[57]

Describe the sample space associated with the experiment of selecting a child at random from three families each with a boy and a girl. Also write the elements of the following events: A: There is at most one boy in the selection. B: The selection consists of only girl. C: The selection has exactly two girls. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐒 = {𝐆𝐆𝐆, 𝐆𝐆𝐁, 𝐆𝐁𝐆, 𝐆𝐁𝐁, 𝐁𝐆𝐆, 𝐁𝐆𝐁, 𝐁𝐁𝐆, 𝐁𝐁𝐁} EVENT 𝐀 = {𝐆𝐆𝐆, 𝐁𝐆𝐆, 𝐆𝐁𝐆, 𝐆𝐆𝐁} EVENT 𝐁 = {𝐆𝐁𝐁, 𝐁𝐆𝐁} EVENT 𝐂 = {𝐆𝐆𝐁, 𝐆𝐁𝐆, 𝐁𝐆𝐆}

C

9

An integer from 1 to 50 is selected at random. Write the elements of the following events. Event A: Randomly selected integer is a multiple of 2. Event B: Randomly selected integer is a multiple of 10. Event C: Randomly selected integer is a multiple of 4. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐒 = {𝟏, 𝟐, 𝟑, 𝟒, … , 𝟓𝟎} EVENT 𝐀 = {𝟐, 𝟒, 𝟔, … , 𝟓𝟎} EVENT 𝐁 = {𝟏𝟎, 𝟐𝟎, 𝟑𝟎, 𝟒𝟎, 𝟓𝟎} EVENT 𝐂 = {𝟒, 𝟖, 𝟏𝟐, 𝟏𝟔, … , 𝟒𝟖}

C

10

In how many different ways can the director of a research laboratory choose 2 chemists from 7 applicants and 3 physicists from 9 applicants? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏𝟕𝟔𝟒

❖ DESCRIPTION OF DIFFERENT EVENTS IN WORDS AND SET NOTATION ARE GIVEN IN THE FOLLOWING TABLE.

No.

Description of Events in Words

Set Notation of Event

1.

A is an event.

A

2.

Event does not occur.

A’

3.

Event B is surely occurring when A occurs.

B⊂A

4.

Impossible Events.

ϕ

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UNIT-6 » Basic Probability

[58]

5.

Sure Events.

U (sample space)

6.

Events A and B are Mutually exclusive

A∩B=ϕ

7.

Events A and B are Exhaustive

A∪B= U

8.

Among A and B events, Only B occurs

9.

Among A and B events, Only one event occurs

10.

Events A and B occur together

A∩B

11.

At least one of the events A and B occurs

A∪B

12.

Among the events A,B,C Only A occurs

A ∩ B′ ∩ C′

13.

Among the events A,B,C Only A and B occurs

A ∩ B ∩ C′

B − A or B ∩ A′ (A − B) ∪ (B − A)

❖ DEFINITION: PROBABILITY FUNCTION ✓ Let 𝐔 be a finite sample space and 𝐒 be its power set. Suppose that a set function 𝐏: 𝐒 ⟶ 𝐑 satisfies following axioms. ✓ Axiom 1: For every A ∈ S, P(A) ≥ 0 ✓ Axiom 2: P(U) = 1 ✓ Axiom 3: ∀A1 ∈ S, A2 ∈ S, ifA1 ∩ A2 = ϕ, then P(A1 ∪ A2 ) = P(A1 ) + P(A2 ). ✓ Then P is called a Probability function. For event A ∈ S, P(A) is called the probability of the event A . The triplet (U, S, P) is called probability space. ❖ MEANING OF DEFINITION: ✓ Axiom 1: Probability of any event is non-negative real number. ✓ Axiom 2: Probability of certain event U is 1. At most value of probability of any event is

1.

✓ Axiom 3: If A1 , A2 , A3 ,…, An ∈ S are mutually exclusive (n ≥ 2), P(A1 ∪ A2 ∪ A3 ∪ … ∪ An ) = P(A1 ) + P(A2 ) + P(A3 ) + ⋯ + P(An ) ❖ DEFINITION: EQUIPROBABLE EVENTS ✓ Let U = {x1 , x2 , . . , xn } be a finite sample space. If P({x1 }) = P({x2 }) = P({x3 }) = ⋯ = P({xn }), then the elementary events {x1 }, {x2 }, {x3 }, … , {xn } are called Equiprobable Events.

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UNIT-6 » Basic Probability

[59]

❖ DEFINITION: PROBABILITY OF AN EVENT ✓ If a finite sample space associated with a random experiment has "𝐧" equally likely (Equiprobable) outcomes (elements) and of these "𝐫“ (𝟎 ≤ 𝐫 ≤ 𝐧) outcomes are favorable for the occurrence of an event 𝐀, then probability of 𝐀 is defined as follow. 𝐏(𝐀) =

𝐟𝐚𝐯𝐨𝐫𝐚𝐛𝐥𝐞 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 𝐫 = 𝐭𝐨𝐭𝐚𝐥 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 𝐧

❖ THEOREMS ✓ Theorem 1. For the Impossible Event, P(ϕ) = 0. ✓ Theorem 2. Complementation Rule: For every Event A, P(A′ ) = 1 − P(A). ✓ Theorem 3. If A ⊂ B, then P(B − A) = P(B) − P(A) and P(A) ≤ P(B) ✓ Corollary 1. For every event A, 0 ≤ P(A) ≤ 1. ✓ Corollary 2. For any events A and B, ✓ Theorem 4. Addition Rule for Arbitrary Events: ✓ Let S be sample space and A, B and C be any events in S, then



P(A ∪ B) = P(A) + P(B) − P(A ∩ B)



P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(C ∩ A)



+P(A ∩ B ∩ C)



P(A ∩ B′) = P(A) − P(A ∩ B)



P(A′ ∩ B′ ) = P(A ∪ B)′ = 1 − P(A ∪ B) (De Morgan’s Rule)



P(A′ ∪ B′ ) = P(A ∩ B)′ = 1 − P(A ∩ B) (De Morgan’s Rule)

METHOD – 2: PROBABILITY OF EVENT H

1

9

If probability of an event A is 10, what is the probability of the event “not A”. Answer: 𝟎. 𝟏

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UNIT-6 » Basic Probability C

2

[60]

If A and B are two mutually exclusive events with P(A) = 0.30 , P(B) = 0.45. Find the probability of A′ , A ∩ B , A ∪ B , A′ ∩ B′ . Answer: 𝟎. 𝟕 , 𝟎 , 𝟎. 𝟕𝟓, 𝟎. 𝟐𝟓

C

3

A fair coin is tossed twice. Find the probability of (1) getting H exactly once. (2) getting T at least once. 𝐀𝐧𝐬𝐰𝐞𝐫:

H

4

𝟏 𝟑 , 𝟐 𝟒

One card is drawn at random from a well shuffled pack of 52 cards. Calculate the probability that the card will be (1) An Ace (2) A card of black colour (3) A diamond (4) Not an ace 𝐀𝐧𝐬𝐰𝐞𝐫:

C

5

𝟏 𝟏 𝟏 𝟏𝟐 , , , 𝟏𝟑 𝟐 𝟒 𝟏𝟑

Four cards are drawn from the pack of cards. Find the probability that (i) all are diamonds (ii) there is one card of each suit (iii) there are two spades and two hearts. [ Hint: Spades ♠ , Club

, Diamond , Heart

]

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟎𝟎𝟐𝟔, 𝟎. 𝟏𝟎𝟓𝟓, 𝟎. 𝟎𝟐𝟐𝟓 H

6

A class consist of 6 girls and 10 boys. If a committee of 3 is chosen at random from the class, find the probability that (1) Three boys are selected (2) Exactly 2 girls are selected 𝐀𝐧𝐬𝐰𝐞𝐫: (𝟏) 𝟎. 𝟐𝟏𝟒𝟑 (𝟐) 𝟎. 𝟐𝟔𝟕𝟗

H

7

A box contains 5 red, 6 white and 2 black balls. The balls are identical in all respect other than colour. 1. One ball is drawn at random from the box. Find the probability that the selected ball is black. 2. Two balls are drawn at random from the box. Find the probability that one ball is white and one is red. 𝐀𝐧𝐬𝐰𝐞𝐫:

𝟐 𝟓 , 𝟏𝟑 𝟏𝟑

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UNIT-6 » Basic Probability C

8

[61]

A room has three lamp sockets. From a collection of 10 light bulbs of which only 6 are good. A person selects 3 at random and puts them in the socket. What is the probability that the room will have light? 𝐀𝐧𝐬𝐰𝐞𝐫:

𝟐𝟗 𝟑𝟎

❖ CONDITIONAL PROBABILITY: ✓ Let S be a sample space and A and B be any two events in S. Then the probability of the occurrence of event A when it is given that B has already occurred is expressed by the symbol P(A⁄B) and is defined as 𝐏(𝐀⁄𝐁) =

𝐏(𝐀 ∩ 𝐁) ; 𝐏(𝐁) > 𝟎, 𝐏(𝐁)

✓ Which is known as conditional probability of the event A relative to event B. ✓ Similarly, the conditional probability of the event B relative to event A is 𝐏(𝐁⁄𝐀) =

𝐏(𝐁 ∩ 𝐀) ; 𝐏(𝐀) > 𝟎. 𝐏(𝐀)

❖ PROPERTIES OF CONDITIONAL PROBABILITY: ✓ Let A1 , A2 and B be any three events of a sample space 𝐒,then 𝐏(𝐀 𝟏 ∪ 𝐀 𝟐 ⁄𝐁) = 𝐏(𝐀 𝟏 ⁄𝐁) + 𝐏(𝐀 𝟐 ⁄𝐁) − 𝐏(𝐀 𝟏 ∩ 𝐀 𝟐 ⁄𝐁); 𝐏(𝐁) > 𝟎. ✓ Let A and B be any two events of a sample space 𝐒, then 𝐏(𝐀′ ⁄𝐁) = 𝟏 − 𝐏(𝐀⁄𝐁); 𝐏(𝐁) > 𝟎. ❖ THEOREM (MULTIPLICATION RULE): ✓ Let S be a sample space and A and B be any two events in 𝐒, then P(A ∩ B) = P(A) ⋅ P(B⁄A); P(A) > 0. ✓ Also, P(A ∩ B) = P(B) ⋅ P(A⁄B); P(B) > 0. ✓ Corollary: Let S be a sample space and A, B and C be three events in S,then P(A ∩ B ∩ C) = P(A) ⋅ P(B⁄A) ⋅ P(C⁄A ∩ B).

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UNIT-6 » Basic Probability

[62]

❖ INDEPENDENT EVENTS: ✓ Let A and B be any two events of a sample space S, then A and B are called independent events if (A ∩ B) = P(A) ⋅ P(B) . ✓ It also means that, P(A⁄B) = P(A) and P(B⁄A) = P(B). ✓ This means that the probability of A does not depend on the occurrence or nonoccurrence of B, and conversely. ✓ Remarks: ✓ Let A, B and C are said to be Mutually independent, if P(A ∩ B) = P(A) ⋅ P(B) ,

P(B ∩ C) = P(B) ⋅ P(C)

P(C ∩ A) = P(C) ⋅ P(A)

P(A ∩ B ∩ C) = P(A) ⋅ P(B) ⋅ P(C)

,

✓ Let A, B and C are said to be pairwise independent, if P(A ∩ B) = P(A) ⋅ P(B)

P(B ∩ C) = P(B) ⋅ P(C)

P(C ∩ A) = P(C) ⋅ P(A)

METHOD – 3: CONDITIONAL PROBABILITY C

1

For two independent events A and B if P(A) = 0.3 and P(A ∪ B) = 0.6, find P(B). Answer: 𝟎. 𝟒𝟐𝟗

H

2

1

2

If A and B are independent events, where P(A) = 4 , P(B) = 3. Find P(A ∪ B). 𝐀𝐧𝐬𝐰𝐞𝐫:

T

3

1

4

3

11

If P(A) = 3 , P(B) = 4 and P(A ∪ B) = 12 . Find P(A⁄B). 𝐀𝐧𝐬𝐰𝐞𝐫:

H

𝟑 𝟒

𝟐 𝟗 3

7

If A and B are independent events, with P(A) = 8 , P(B) = 8. Find P(A ∪ B) Find P(A⁄B)and P(B⁄A). 𝐀𝐧𝐬𝐰𝐞𝐫:

𝟓𝟗 𝟓𝟗 𝟓𝟗 , , 𝟔𝟒 𝟓𝟔 𝟐𝟒

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UNIT-6 » Basic Probability 5

C

1

1

3

4

If P(A) = , P(B′ ) =

and P(A ∩ B) =

[63] 1 6

, then find P(A ∪ B), P(A′ ∩ B′ )

and P(A′ ⁄B′ ). 𝐀𝐧𝐬𝐰𝐞𝐫: H

6

𝟏𝟏 𝟏 𝟏 , , 𝟏𝟐 𝟏𝟐 𝟑

A person is known to hit the target in 3 out of 4 shots, whereas another person is known to hit the target in 2 out of 3 shots. Find the probability of

W-19 (3)

the target being hit at all when they both try. 𝐀𝐧𝐬𝐰𝐞𝐫: H

7

𝟏𝟏 𝟏𝟐

In a box, 100 bulbs are supplied out of which 10 bulbs have defects of type A, 5 bulbs have defects of type B and 2 bulbs have defects of both the type. Find the probability that 1. a bulb to be drawn at random has a B type defect under the condition that it has an A type defect, and 2. a bulb to be drawn at random has no B type defect under the condition that it has no A type defect. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟐, 𝟎. 𝟗𝟔𝟔𝟕

C

8

In producing screws, let A mean “screw too slim” and B “screw too short”. Let p(A) = 0.1 and let the conditional probability that a slim screw is also too small be P(B/A) = 0.2. What is the probability that the screw that we pick randomly from a lot produced will be both too slim and too short? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟎𝟐

C

9

Two students X and Y work independently on a problem. The probability 3

2

that X will solve it is 4 and probability that Y will solve it is 3. What is the probability that problem will be solved? 𝐀𝐧𝐬𝐰𝐞𝐫:

𝟏𝟏 𝟏𝟐

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W-19 (3)

UNIT-6 » Basic Probability C

10

[64]

In a group of 200 students 40 are taking English,50 are taking mathematics, and 12 are taking both (a) If a student is selected at random, what is the probability that the student is taking English? (b) A student is selected at random from those taking mathematics. What is the probability that the student is taking English? (c) A student is selected at random from those taking English, what is the probability that the student is taking mathematics? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟐𝟎 , 𝟎. 𝟐𝟒 , 𝟎. 𝟑

H 11

A person is known to hit the target in 3 out of 4 shots, whereas another person is known to hit the target in 2 out of 3 shots. Find the probability of the target being hit at all when they both try. (What is the probability that the target will be hit?) 𝐀𝐧𝐬𝐰𝐞𝐫:

𝟏𝟏 𝟏𝟐

❖ TWO-DIMENSIONAL RANDOM VARIABLE: ✓ Let S be the sample space associated with a random experiment E. Let X = X(s) and Y = Y(s) be two functions each assigning a real number to each outcome. Then (X, Y) is called a two-dimensional random variable. ❖ TWO-DIMENSIONAL DISCRETE RANDOM VARIABLE ✓ If the possible values of (X, Y) are finite or countable infinite, (X, Y) is called a twodimensional discrete random variable. ✓ Example: Consider the experiment of tossing a coin twice. The sample space S = {HH, HT, TH, TT}. Let X denotes the number of head obtained in first toss and Y denotes the number of head obtained in second toss. Then S

HH HT TH TT

X(S)

1

1

0

0

Y(S)

1

0

1

0

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UNIT-6 » Basic Probability

[65]

✓ Here, (X, Y) is a two-dimensional random variable and the range space of (X, Y) is {(1, 1), (1, 0), (0, 1), (0, 0)} which is finite & so (X, Y) is a two-dimensional discrete random variables. Further, Y=0 Y=1 Y=2 X=0

0.25

0.25



X=1

0.25

0.25



X=2







❖ TWO- DIMENSIONAL CONTINUOUS RANDOM VARIABLE ✓ If (X, Y) can assume all values in a specified region R in the xy-plane, (X, Y) is called a twodimensional continuous random variable. ❖ JOINT PROBABILITY MASS FUNCTION (DISCRETE CASE) ✓ If (X, Y) is a two-dimensional Discrete Random Variable such that P(X = xi , Y = yj ) = pij , then pij is called the joint probability mass function of (X , Y) provided Pij ≥ 0 For all i & j and ∑i ∑j Pij = 1. ❖ THE MARGINAL PROBABILITY FUNCTION (DISCRETE CASE) ✓ The marginal probability function is defined as ✓ 𝐏𝐗 (𝐱) = ∑𝐲 𝐏(𝐗 = 𝐱, 𝐘 = 𝐲) , 𝐏𝐘 (𝐲) = ∑𝐱(𝐗 = 𝐱, 𝐘 = 𝐲) ✓ Example: The joint probability mass function (PMF) of X and Y is Y=0 Y=1 Y=2 X=0

0.1

0.04

0.02

X=1

0.08

0.2

0.06

X=2

0.06

0.14

0.3

✓ The marginal Probability Mass Function of X is



PX ( X = 0) = 0.1 + 0.04 + 0.02 = 0.16



PX ( X = 1) = 0.08 + 0.2 + 0.06 = 0.34



PX ( X = 2) = 0.06 + 0.14 + 0.3 = 0.5

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UNIT-6 » Basic Probability

[66]

✓ The marginal Probability Mass Function of Y is



PY ( Y = 0) = 0.1 + 0.08 + 0.06 = 0.24



PY ( Y = 1) = 0.04 + 0.2 + 0.14 = 0.38



PY ( Y = 2) = 0.02 + 0.06 + 0.3 = 0.38

❖ JOINT PROBABILITY DENSITY FUNCTION (CONTINUOUS CASE) ✓ If (X, Y) is a two-dimensional continuous Random Variable such that P (x −

dx 2

≤X≤x+

dx 2

, y−

dy 2

≤Y≤y+

dy 2

) = f(x, y)

✓ It is called the joint probability density function of (X, Y), provided f(x, y) ≥ 0, for all (x, y) ϵ D; Where D is range of space and ∬ f(x, y) dx dy = 1 D d

b

In particular, P(a ≤ X ≤ b, c ≤ Y ≤ D) = ∫c ∫a f(x, y) dx dy ❖ THE MARGINAL PROBABILITY FUNCTION (CONTINUOUS CASE) ✓ The marginal probability function is defined as ∞

FX (x) = ∫ f(x , y)dy ,



FY (y) = ∫ f(x, y)dx

−∞

−∞

✓ Example: Joint probability density function of two random variables X & Y is given by x 2 − xy ; 0 < x < 2 and − x < y < x f(x, y) = { 8 0 ; otherwise ✓ The marginal probability density function of X is x

x

−x

−x

x

x 2 − xy 1 2 xy 2 x3 (x) FX = ∫ f(x, y) dy = ∫ dy = (x y − ) = ;0 < x < 2 8 8 2 −x 4 ✓ The marginal probability density function of Y is 2

2

2

x 2 − xy 1 x3 x2y 1 y FY (y) = ∫ f(x, y) dx = ∫ dx = ( − ) = − ; −x < y < x 8 8 3 2 0 3 4 0

0

✓ Remark: The marginal distribution function of (X, Y) is x



F1 (x) = ∫ ∫ f(x, y) dxdy , −∞ −∞



y

F2 (y) = ∫ ∫ f(x, y) dxdy −∞ −∞

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UNIT-6 » Basic Probability

[67]

❖ INDEPENDENT RANDOM VARIABLES: ✓ Two random variables X and Y are defined to be independent if P(X = x, Y = y) = PX (x) . PY (y) If X and Y are discrete f(x, y) = FX (x) . FY (y) If X and Y are continuous ✓ Example: The joint probability mass function (PMF) of X and Y is

Y=0 Y=1 Y=2 X=0

0.1

0.04

0.02

X=1

0.08

0.2

0.06

X=2

0.06

0.14

0.3

✓ The marginal Probability Mass Function of X=0 is PX ( X = 0) = 0.1 + 0.04 + 0.02 = 0.16 ✓ The marginal Probability Mass Function of Y=0 is PY ( Y = 0) = 0.1 + 0.08 + 0.06 = 0.24 PX (0) PY (0) = 0.16 × 0.24 = 0.0384 But P(X = 0, Y = 0) = 0.1 ∴ P(X = 0, Y = 0) ≠ PX (0) . PY (0) ∴ X and Y are not independent random variables.

❖ EXPECTED VALUE OF TWO-DIMENSIONAL RANDOM VARIABLE: ✓ Discrete case: E(X) = ∑ xi PX (xi ) and E(Y) = ∑ yi PY (yi ) ✓ Continuous case: E(X) = ∬ x f(x, y)dx dy and E(Y) = ∬ y f(x, y)dy dx (; where R is given region) R

R

METHOD-6: EXAMPLES ON TWO-DIMENSIONAL RANDOM VARIABLE C

1

The joint probability mass function of (X, Y) is given by P(x, y) = K(2x + 3y) Where x = 0,1,2 and y = 1,2,3. Find the marginal probability of X. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐊 =

𝟏 𝟏𝟖 𝟐𝟒 𝟑𝟎 & , , 𝟕𝟐 𝟕𝟐 𝟕𝟐 𝟕𝟐

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UNIT-6 » Basic Probability H

2

[68]

X, Y are two random variables having joint mass function P(x, y) = 1 27

(2x + y) where x = 0, 1, 2 and y = 0, 1, 2. Find the marginal

probabilities. 𝐀𝐧𝐬𝐰𝐞𝐫: C

3

𝟑 𝟗 𝟏𝟓 𝟔 𝟗 𝟏𝟐 , , , , , 𝟐𝟕 𝟐𝟕 𝟐𝟕 𝟐𝟕 𝟐𝟕 𝟐𝟕 1

1

1

Let P(X = 0, Y = 1) = 3 , P(X = 1, Y = −1) = 3 , P(X = 1, Y = 1) = 3. Is it the joint probability mass function of X and Y? if yes, Find the marginal probability function of X and Y. 𝟏

𝟐

𝟏

𝟐

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏𝐗 (𝟎) = 𝟑 , 𝐏𝐗 (𝟏) = 𝟑 & 𝐏𝐘 (−𝟏) = 𝟑 , 𝐏𝐘 (𝟏) = 𝟑 H

4

A two-dimensional random variable (X, Y) have a bivariate distribution given by P(X = x, Y = y) =

x2 +y 32

for x = 0,1,2,3 & y = 0,1. Find the

marginal distributions of X and Y. Also check the independence of X & Y. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐗: H

5

𝟏 𝟑 𝟗 𝟏𝟗 𝟏𝟒 𝟏𝟖 , , , 𝐚𝐧𝐝 𝐘: , , 𝐍𝐎 𝟑𝟐 𝟑𝟐 𝟑𝟐 𝟑𝟐 𝟑𝟐 𝟑𝟐

For given joint probability distribution of X and Y, find p(X ≤ 1, Y = 2), p(X ≤ 1), P(Y ≤ 3), P(X < 3, Y ≤ 4). Also check the independence of X & Y. Y=1 Y=2 Y=3 Y=4 Y=5 Y=6 X=0

0

0

1/32

2/32

2/32

3/32

X=1

1/16

1/16

1/8

1/8

1/8

1/8

X=2

1/32

1/32

1/64

1/64

0

2/64

𝐀𝐧𝐬𝐰𝐞𝐫:

𝟏 𝟕 𝟐𝟑 𝟗 , , , , 𝐍𝐎 𝟏𝟔 𝟖 𝟔𝟒 𝟏𝟔

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UNIT-6 » Basic Probability C

6

[69]

The following table represents the joint probability distribution of discrete random variable (X, Y), find P(X ≤ 2, Y = 3), P(Y ≤ 2) & P(X + Y < 4). Y=1 Y=2 Y=3 Y=4 Y=5 Y=6 X=0

0

0

1/32

2/32

2/32

3/32

X=1

1/16

1/16

1/8

1/8

1/8

1/8

X=2

1/32

1/32

1/64

1/64

0

2/64

𝐀𝐧𝐬𝐰𝐞𝐫: C

7

𝟏𝟏 𝟑 𝟏𝟑 , , 𝟔𝟒 𝟏𝟔 𝟑𝟐

Out of five cars two have tyre problems and one has break problem and two are in good running condition. Two cars are required for the

W-19 (4)

journey. If two cars are selected among five at random and if X denotes the number with tyre problem, Y denotes with break problem then find the marginal probability function of X and Y. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩(𝟎, 𝟎) = 𝟎. 𝟏, 𝐩(𝟏, 𝟎) = 𝟎. 𝟒, 𝐩(𝟐, 𝟎) = 𝟎. 𝟏, 𝐩(𝟎, 𝟏) = 𝟎. 𝟐, 𝐩(𝟏, 𝟏) = 𝟎. 𝟐 C

8

Suppose that 2 batteries are randomly chosen without replacement from the group of 12 batteries which contains 3 new batteries, 4 used batteries and 5 defective batteries. Let X denote the number of new batteries chosen and Y denote the number of used batteries chosen then, find the joint probability distribution. 𝟏𝟎 𝟏𝟓 𝟑 𝟐𝟎 , 𝐩(𝟏, 𝟎) = , 𝐩(𝟐, 𝟎) = , 𝐩(𝟎, 𝟏) = , 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟏𝟐 𝟔 𝐩(𝟏, 𝟏) = , 𝐩(𝟎, 𝟐) = 𝟔𝟔 𝟔𝟔

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩(𝟎, 𝟎) =

C

9

Check the following functions are PDF or not: 3

(a)f(x, y) = {4 (b)f(x, y) = {

+ xy ; 0 ≤ x < 1 and 0 ≤ y < 1 0

;

otherwise

3(x 2 + y 2 ) ; 0 ≤ x < 1 and 0 ≤ y < x 0 ; otherwise

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐘𝐄𝐒

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UNIT-6 » Basic Probability C

10

[70]

Cxy ; 0 < x < 4 and 1 < y < 5 Let f(x, y) = { is the joint density 0 ; otherwise function of two random variables X & Y, then find the value if C. 𝐀𝐧𝐬𝐰𝐞𝐫:

C

11

𝟏 𝟗𝟔

If X and Y are two random variables having joint density function: 1

f(x, y) = {8

(6 − x − y) ; 0 ≤ x < 2 and 2 ≤ y < 4 0 ;

otherwise

Find P(X < 1 ∩ Y < 3), P(X + Y < 3). 𝟑 𝟏 𝐀𝐧𝐬𝐰𝐞𝐫: , 𝟖 𝟐𝟒 H 12

Suppose, two-dimension continuous random variable (X, Y) has Pdf given by f(x, y) = {

6x 2 y ; 0 < x < 1 and 0 < y < 1 . 0 ; elsewhere 1

1

(a) Verify: ∫ ∫ f(x, y) dx dy = 1 0

0 3

(b)Find P (0 < X < , 4

1 3

< Y < 2) & P(X + Y < 1)

𝟑 𝟏 𝐀𝐧𝐬𝐰𝐞𝐫: , 𝟖 𝟏𝟎 C

13

The joint pdf of a two-dimensional random variable (X, Y) is given by f(x, y) = {

2 ; 0 < x < 1, 0 < y < x , 0; elsewhere

Find the marginal density function of X and Y. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐅𝐗 (𝐱) = 𝟐𝐱, 𝟎 < 𝐱 < 𝟏, H 14

𝐅𝐘 (𝐲) = 𝟐(𝟏 − 𝐲), 𝟎 < 𝐲 < 𝟏

The joint probability function of two random variables X and Y is given by f(x, y) = {

W-19 (4)

k(x + 2y); 0 < x < 1,0 < y < 2 0 ; elsewhere

Find the marginal density function of X and Y. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐅𝐗 (𝐱) = 𝐤[𝟐𝐱 + 𝟒], 𝟎 < 𝐱 < 𝟏 𝟏 𝐅𝐘 (𝐲) = 𝐤 [ + 𝟐𝐲] , 𝟎 < 𝐲 < 𝟐 𝟐

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UNIT-6 » Basic Probability C

15

[71]

Check the independence of X and Y for the following: 1

(1 + xy) ; −1 < x < 1, −1 < y < 1 f(x, y) = {4 0 ; elsewhere 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐍𝐎 C

16

The random variables X and Y have the following joint probability distribution. What is the expected value of X and Y? Y=0 Y=1 Y=2 X=0

0.2

0.1

0.2

X=1

0

0.2

0.1

X=2

0.1

0

0.1

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕, 𝟏. 𝟏 C

17

Consider the joint density function for X and Y, f(x, y) = x 2 y 3 ; 0 < x < 1 & 0 < y < x, find the expected value of X. 𝐀𝐧𝐬𝐰𝐞𝐫:

𝟏 𝟑𝟐

❖ THEOREM (TOTAL PROBABILITY): ✓ If B1 and B2 are two mutually exclusive and exhaustive events and P(B1 ), P(B2 ) ≠ 0 , then for any event A. 𝐏(𝐀) = 𝐏(𝐁𝟏 ) ⋅ 𝐏(𝐀⁄𝐁𝟏 ) + 𝐏(𝐁𝟐 ) ⋅ 𝐏(𝐀⁄𝐁𝟐 ) ✓ Corollary: If B1 , B2 and B3 are mutually exclusive and exhaustive events and P(B1 ), P(B2 ), P(B3 ) ≠ 0 , then for any event A. 𝐏(𝐀) = 𝐏(𝐁𝟏 ) ⋅ 𝐏(𝐀⁄𝐁𝟏 ) + 𝐏(𝐁𝟐 ) ⋅ 𝐏(𝐀⁄𝐁𝟐 ) + +𝐏(𝐁𝟑 ) ⋅ 𝐏(𝐀⁄𝐁𝟑 ) ❖ BAYES’ THEOREM: ✓ Let B1 , B2 , B3 … , Bn be an n-mutually exclusive and exhaustive events of a sample space S and let A be any event such that P(A) ≠ 0, then 𝐏(𝐁𝐢 ⁄𝐀) =

𝐏(𝐁𝐢 ) ⋅ 𝐏(𝐀⁄𝐁𝐢 ) 𝐏(𝐁𝟏 ) ⋅ 𝐏(𝐀⁄𝐁𝟏 ) + 𝐏(𝐁𝟐 ) ⋅ 𝐏(𝐀⁄𝐁𝟐 ) + 𝐏(𝐁𝟑 ) ⋅ 𝐏(𝐀⁄𝐁𝟑 ) + ⋯ + 𝐏(𝐁𝐧 ) ⋅ 𝐏(𝐀⁄B𝐧 )

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UNIT-6 » Basic Probability

[72]

METHOD – 4: TOTAL PROBABILITY AND BAYE’S THEOREM C

1

Three boxes contain 10%, 20%, and 30% of defective finger joints. A finger joint is selected at random which is defective. Determine probability that it comes from i)1st box ii) 2nd box iii) 3rd box. 𝐀𝐧𝐬𝐰𝐞𝐫:

H

2

𝟏 1 𝟏 , , 𝟔 𝟑 𝟐

In a certain assembly plant, three machines, B1, B2 and B3, make 30%, 45% and 25%, respectively, of the products. It is known form the past experience that 2%, 3% and 2% of the products made by each machine respectively, are defective. Now, suppose that a finished product is randomly selected. What is the probability that it is defective? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟎𝟐𝟒𝟓

C

3

State Bayes’ theorem. A microchip company has two machines that produce the chips. Machine-I produces 65% of the chips, but 5% of its chips are defective. Machine-II produces 35% of the chips, but 15% of its chips are defective. A chip is selected at random and found to be defective. What is

the

probability

that

it

came

from

Machine-I?

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟑𝟖𝟐𝟒 H

4

State Bayes’ theorem. In a bolt factory, three machines A, B and C manufacture 25%, 35% and 40% of the total product respectively. Out Of

W-19 (4)

these outputs 5%, 4% and 2% respectively, are defective bolts. A bolt is picked up at random and found to be defective. What are the Probabilities that it was manufactured by machine A, B and C? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟑𝟔𝟐𝟑, 𝟎. 𝟒𝟎𝟓𝟕, 𝟎. 𝟐𝟑𝟏𝟖 C

5

A company has two plants to manufacture hydraulic machine. Plant I manufacture 70% of the hydraulic machines and plant II manufactures 30%. At plant I, 80% of hydraulic machines are rated standard quality and at plant II, 90% of hydraulic machine are rated standard quality. A machine is picked up at random and is found to be of standard quality. What is the chance that it has come from plant I? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟔𝟕𝟒𝟕

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UNIT-6 » Basic Probability H

6

[73]

There are two boxes A and B containing 4 white, 3 red and 3 white,7 red balls respectively. A box is chosen at random and a ball is drawn from it, if the ball is white, find the probability that it is from box A. 𝐀𝐧𝐬𝐰𝐞𝐫:

𝟒𝟎 𝟔𝟏

❖ RANDOM VARIABLE:

W-19(1)

✓ An experiment, in which we know all the possible outcomes in advance but which of them will occur is known only after the experiment is performed, is called a Random Variable. ✓ Types of Random Variables (1). Discrete random variable (2). Continuous random variable ❖ DISCRETE RANDOM VARIABLE: ✓ A random variable, which can take only finite, countable, or isolated values in a given interval, is called discrete random variable. ✓ i.e. A random variable is one, which can assume any of a set of possible values which can be counted or listed. ✓ For example, the numbers of heads in tossing coins, the number of auto passengers can take on only the values 1, 2, 3 and so on. ✓ Note: Discrete random variables can be measured exactly. ❖ CONTINUOUS RANDOM VARIABLE: ✓ A random variable, which can take all possible values that are infinite in a given interval, is called Continuous random variable. ✓ i.e. a continuous random variable is one, which can assume any of infinite spectrum of different values across an interval which cannot be counted or listed ✓ For example, measuring the height of a student selected at random, finding the average life of a brand X tyre etc. ✓ Note: Continuous random variables cannot be measured exactly.

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UNIT-6 » Basic Probability

[74]

❖ PROBABILITY FUNCTION: ✓ If for random variable X, the real valued function f(x) is such that P(X = x) = f(x), then f(x) is called Probability function of random variable X. ✓ Probability function f(x) gives the measures of probability for different values of X say x1 , x2 , … . , xn . ✓ Sometimes Probability density function is denoted by p(x). ❖ PROBABILITY DENSITY FUNCTION:

W-19(1)

✓ If X is a continuous random variable then its probability function f(x) is called continuous probability function OR probability density function. It is defined as below b

P(a < x < b) = ∫ f(x)dx a

✓ Note: (1). f(x) ≥ 0 ∞

(2). ∫−∞ f(x)dx = 1 ❖ PROBABILITY MASS FUNCTION: ✓ If X is a discrete random variable then its probability function f(x) is discrete probability function. It is also called probability mass function, then P(X = xi ) = f(xi ). ❖ PROPERTIES OF PROBABILITY FUNCTION: ✓ If f(x) is a probability function of random variable X then it possess the following Properties: (1). f(x) is positive for any value of x i.e. f(x) ≥ 0 for all x. (2). Total of all values of f(x) for different x is always one. i.e. ∑ f(x) = 1, summation is taken for all values of x. (3). f(x) measures the probability for any given value of x. ❖ MATHEMATICAL EXPECTATION: ✓ If X is a discrete random variable having various possible values x1 , x2 , … . , xn and if f(x) is the probability function, the mathematical Expectation or simply expectation of X is defined and denoted by E(X).

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UNIT-6 » Basic Probability

[75]

n

n

n

E(X) = ∑ xi ⋅ f(xi ) = ∑ xi ⋅ p(xi ) = ∑ xi ⋅ pi i=1

i=1

i=1

✓ Where, ∑ni=1 pi = p1 + p2 + ⋯ pn = 1 ✓ If X is, a continuous random variable having probability density function f(x), expectation ∞

of X is defined as E(X) = ∫−∞ x f(x) dx ✓ Note: E(X) is also called the mean value of the probability distribution of x and is denoted by μ. ❖ PROPERTIES OF MATHEMATICAL EXPECTATION: (1). Expected value of constant term is constant. i.e. E(c) = c (2). If c is constant, then E(cX) = c ∙ E(X) (3). E(X 2 ) = ∑ni=1 xi2 ⋅ pi (4). If a and b are constants, then E(aX ± b) = aE(X) ± b aX+b

(5). If a, b and c are constants, then E (

c

1

) = c [aE(X) + b]

(6). If X and Y are two random variables, then E(X + Y) = E(X) + E(Y) (7). If X and Y are two independent random variables, then E(X ∙ Y) = E(X) ∙ E(Y) (8). If g(x) is any function of random variable X and f(x) is probability density function, then E{g(x)} = ∑ g(x) ∙ f(x) ❖ VARIANCE OF A RANDOM VARIABLE: ✓ Variance is a characteristic of random variable X and it is used to measure dispersion (or variation) of X. ✓ If X is a discrete random variable with probability density function f(x), then expected value of [X − E(X)]2 is called the variance of X and it is denoted by V(X). ✓ i.e. V(X) = E[X − E(X)]2 = E(X 2 ) − [E(X)]2 ❖ PROPERTIES OF VARIANCE: (1). V(c) = 0, Where c is a constant. (2). V(cX) = c 2 V(X) , where c is a constant.

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UNIT-6 » Basic Probability

[76]

(3). V(X + c) = V(X), Where c is a constant. (4). If a and b are constants, then V(aX + b) = a2 V(X). (5). If X and Y are the independent random variables, then V(X + Y) = V(X) + V(Y) V(X) = E(X 2 ) − μ2 ❖ STANDARD DEVIATION OF RANDOM VARIABLE: ✓ The positive square root of V(X) (Variance of X) is called standard deviation of random variable X and is denoted by σ. ✓ i.e. S.D. σ = √V(X). ✓ Note: σ2 OR s2 is called variance of V(X).

METHOD – 5: PROBABILITY FUNCTION, EXPECTATION, VARIANCE C

1

Which of the following functions are probability function? 1 x 1 1−x

(1). f(x) = (2) (2)

; x = 0,1

1 x

(2). f(x) = (− 2) , x = 0,1,2 (3).

(4).

x

−1

0

1

f(x)

0.5

0.8

−3

x

1

2

3

f(x)

0.3

0.5

0.2

Answer: YES, NO, NO, YES C

2

For the probability function f(x) = 𝐀𝐧𝐬𝐰𝐞𝐫:

k , −∞ < x < ∞, find k . 1 + x2

𝟏 𝛑

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UNIT-6 » Basic Probability T

3

[77]

The probability function of a random variable X is p(x) =

2x + 1 , x = 1,2,3,4,5,6. 48

(a)Verify whether p(x) is probability function?

(b)Also find E(X).

Answer: YES, 4.23 C

4

Find the expected value of a random variable X having the following probability distribution. X

−5

−1

0

1

5

8

P(X = x)

0.12

0.16

0.28

0.22

0.12

0.1

Answer: 0.8600 C

5

A random variable X has the following function. X

0

1

2

3

4

5

6

7

P(X = x)

0

2k

3k

k

2k

k2

7k 2

2k 2 + k

Find the value of k and then evaluate (i)P(X < 6)(ii)P(X ≥ 6)(iii)P(0 < x < 5). [Hint: (i)P(X < 6) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) (ii)P(X ≥ 6) = 1 − P(X < 6) (iii) P(0 < X < 5) = P(1) + P(2) + P(3) + P(4) ] 𝐀𝐧𝐬𝐰𝐞𝐫: (𝐢)𝟎. 𝟖𝟏𝟎𝟎, (𝐢𝐢)𝟎. 𝟏𝟗𝟎𝟎, (𝐢𝐢𝐢)𝟎. 𝟖𝟎𝟎𝟎 C

6

An urn contains 10 white and 3 black balls, while another urn contains 3 white and 5 black balls. Two balls are drawn from the first urn and put into the second urn and then a ball is drawn from the later. What is the probability that it is a white ball? 𝐀𝐧𝐬𝐰𝐞𝐫:

C

7

𝟓𝟗 𝟏3𝟎

A machine produces on average of 500 items during the first week of the month and average of 400 items during the last week of the month. The probability for these being 0.68 and 0.32. Determine the expected value of the production. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝟔𝟖

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W-19 (4)

UNIT-6 » Basic Probability T

8

[78]

In a business, the probability that a trader can get profit of Rs. 5000 is 0.4 and probability for loss of Rs. 2000 is 0.6. Find his expected gain or loss. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟖𝟎𝟎

C

9

Three coins are tossed together and let random variable x be the number of heads in each outcome. Then find (a) Probability Distribution, (b) Mean (c) Standard deviation. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏(0) =

H 10

𝟏 𝟑 𝟑 𝟏 , 𝐏(𝟏) = , 𝐏(𝟐) = , 𝐏(𝟑) = , 𝐄(𝐗) = 𝟏. 𝟓, 𝛔(𝐗) = 𝟎. 𝟖𝟔𝟔 𝟖 𝟖 𝟖 𝟖

Three balanced coins are tossed, then find the mathematical expectation of tails. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟓

H 11

There are 8 apples in a box, of which 2 are rotten. A person selects 3 Apples at random from it. Find the expected value of the rotten Apples. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎. 𝟕𝟓

C

12

There are 3 red and 2 white balls in a box and 2 balls are taken at random from it. A person gets Rs. 20 for each red ball and Rs. 10 for each white ball. Find his expected gain. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝟐

H 13

There are 10 bulbs in a box, out of which 4 are defectives. If 3 bulbs are taken at random, find the expected number of defective bulbs. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏. 𝟏𝟗

T

14

(a)A contestant tosses a coin and receives $5 if heads appears and $1 if tail appears. What is the expected value of a trial? (b)A contestant receives $4.00 if a coin turns up heads and pays $3.00 if it turns tails. What is the expected value of a trail? 1

[ Hint: P(H) = P(T) = 2]. 𝐀𝐧𝐬𝐰𝐞𝐫: (𝐚)$𝟑. 𝟎𝟎 , (𝐛)$𝟎. 𝟓𝟎

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UNIT-6 » Basic Probability H 15

[79]

The following table gives the probabilities that a certain computer will malfunction 0,1,2,3,4,5, or 6 times on any one day: NO. of malfunction x:

0

1

2

3

4

5

6

Probability f(x)

0.17

0.29

0.27

0.16

0.07

0.03

0.01

Find mean and standard deviation of this probability distribution. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐦𝐞𝐚𝐧 = 𝐄(𝐗) = 𝐕(𝐗) = 𝟏. 𝟖 C

16

The probability distribution of a random variable X is given below. X

−2

−1

0

1

2

P(x)

1 12

1 3

p

1 4

1 6

Find (i) p

(ii)E(X) (iii) E(2X + 3) (iv)E(X 2 + 2) (v) V(X) 𝟏 𝟏 𝟏𝟗 𝟒𝟑 , 𝐄(𝐗) = , 𝐄(𝟐𝐗 + 𝟑) = , 𝐄(𝐗 𝟐 + 𝟐) = , 𝟔 𝟏𝟐 𝟔 𝟏𝟐 𝟐𝟐𝟕 𝐕(𝐗) = 𝟏𝟒𝟒

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩 =

H 17

The probability distribution of a random variable x is as follows. Find p, E(x). X

0

1

2

3

4

5

p(x)

P

1 5

1 10

P

1 20

1 20

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩 = 𝟎. 𝟑𝟎, 𝐄(𝐱) = 𝟏. 𝟕𝟓𝟎𝟎 H 18

For a random variable X, if E(2X − 10) = 20 and E(X 2 ) = 400. Find the standard deviation of X. 𝐀𝐧𝐬𝐰𝐞𝐫: √𝟏𝟕𝟓

C

19 C If mean and standard deviation of a random variable x are 5 and 5 respectively. Find E(X 2 ) and E(2X + 5)2 . 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐄(𝐗 𝟐 ) = 𝟓𝟎, 𝐄(𝟐𝐗 + 𝟓)𝟐 = 𝟑𝟐𝟓

❖ BINOMIAL PROBABILITY DISTRIBUTION: ✓ This distribution is associated with repeated trials of an experiment.

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UNIT-6 » Basic Probability

[80]

❖ BERNOULLI TRIALS: ✓ Suppose a random experiment has two possible outcomes, which are complementary, say success (S) and failure (F). If the probability p(0 < p < 1) of getting success at each of the n trials of this experiment is constant, then the trials are called Bernoulli trials. ❖ BINOMIAL DISTRIBUTION: ✓ P(X = x) = p(x) = nCx px qn−x ; x = 0,1,2 … , n. Where, nCx =

n! x! (n − x)!

✓ q = probability of failure in each trial=1 − p ❖ EXAMPLES OF BINOMIAL DISTRIBUTION: (1). Number of defective bolts in a box containing n bolts. (2). Number of post-graduates in a group of n people. (3). Number of oil wells yielding natural gas in a group of n wells test drilled. (4). Number of machines lying idle in a factory having n machines. ❖ PROPERTIES OF BINOMIAL DISTRIBUTION: ✓ The Binomial distribution holds under the following conditions. (1). The number of trials n is finite. (2). There are only two possible outcomes, success or failure. (3). The trials are independent of each other. (4). The probability of success p is constant for each trial. ❖ NOTE: ✓ The mean and variance of binomial distribution with parameters n and p are defined as follow, ✓ 𝐌𝐞𝐚𝐧 μ = E(X) = np ✓ 𝐕𝐚𝐫𝐢𝐚𝐧𝐜𝐞 V(X) = σ2 = npq ✓ 𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 = σ = √npq ✓

nCx

= nCn−x

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UNIT-6 » Basic Probability ✓

nC0

[81]

= nCn = 1

✓ We use math expression " ≥ " for at least and " ≤ " for at most.

METHOD – 6: BINOMIAL DISTRIBUTION C

1

Write assumption of Binomial Distribution.

C

2

Find the binomial distribution for n = 4 and p = 0.3. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏(𝐗 =. 𝟎) = 𝟎. 𝟐𝟒𝟎𝟏, 𝐏(𝐗 = 𝟏) = 𝟎. 𝟒𝟏𝟏𝟔, 𝐏(𝐗 = 𝟐) = 𝟎. 𝟐𝟔𝟒𝟔, 𝐏(𝐗 = 𝟑) = 𝟎. 𝟎𝟕𝟓𝟔, 𝐏(𝐗 = 𝟒) = 𝟎. 𝟎𝟎𝟖𝟏

3

C

Obtain the binomial distribution for which mean is 10 and variance is 5. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏(𝐗 = 𝐱) =

H

4

20Cx (𝟎. 𝟓)

𝐱 (𝟎.

𝟓)𝟐𝟎−𝐱

Determine binomial distribution whose mean is 4 and variance is 3 and hence evaluate P(X ≥ 2). 𝟏 𝐱 𝟑 𝟏𝟔−𝐱 ( 𝐂 ; 𝐗 = 𝟎, 𝟏, … 𝟏𝟔 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏(𝐗 = 𝐱) = { 𝟏𝟔 𝐱 𝟒) (𝟒) 𝟎 ; 𝐨𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞 𝐏(𝐗 ≥ 𝟐) = 𝟎. 𝟗𝟑𝟔𝟓

H

5

For the binomial distribution with n = 20, p = 0.35. Find Mean, Variance and Standard deviation. 𝐀𝐧𝐬𝐰𝐞𝐫: (𝐚)𝟕. 𝟎𝟎𝟎𝟎 (𝐛)𝟒. 𝟓𝟓, (𝐜)𝟐. 𝟏𝟑𝟑𝟏

C

6

If the probability of a defective bolt is 0.1 Find mean and standard deviation of the distribution of defective bolts in a total of 400. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝛍 = 𝟒𝟎, 𝛔 = 𝟔

C

7

Write formula for binomial distribution when n and p is known. If n=4 and p=0.2, then prepare binomial distribution for the random variable x. Also, find E(X), V(X), E(3x + 7), V(3x + 7). 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏(𝐗) = 𝟒𝐂𝐱 (𝟎. 𝟐)𝐱 (𝟎. 𝟖)𝟒−𝐱 , 𝐄(𝐗) = 𝟎. 𝟖, 𝐕(𝐗) = 𝟎. 𝟔𝟒, 𝐄(𝟑𝐱 + 7) = 𝟗. 𝟒, 𝐕(𝟑𝐱 + 𝟕) = 𝟓. 𝟕𝟔

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UNIT-6 » Basic Probability C

8

[82]

20% of the bulbs produced are defective. Find probability that at most 2 bulbs out of 4 bulbs are defective. Answer: 𝟎. 𝟗𝟕𝟐𝟖

H

9

12% of the tablets produced by a tablet machine are defective. What is the probability that out of a random sample of 20 tablets produced by the machine, 5 are defective? Answer: 𝟎. 𝟎𝟓𝟔𝟕

H 10

If 3 of 12 car drivers do not carry driving license, what is the probability that a traffic inspector who randomly checks 3 car drivers, will catch 1 for not carrying driving license. (Use binomial dist.) 𝐀𝐧𝐬𝐰𝐞𝐫:

H 11

𝟐𝟕 𝟔𝟒

The probability that India wins a cricket test match against Australia is 1

given to be 3. If India and Australia play 3 tests matches, what is the probability that (1). India will lose all the three test matches. (2). India will win at least one test match. Answer: (𝐢) 𝟎. 𝟐𝟗𝟔𝟑, (𝐢𝐢) 𝟎. 𝟕𝟎𝟑𝟕 C

12

What are the properties of Binomial Distribution? The average percentage of failure in a certain examination is 40. What is the probability that out of a group of 6 candidates, at least 4 passed in examination? Answer: 𝟎. 𝟓𝟒𝟒𝟑

T

13

The probability that in a university, a student will be a post-graduate is 0.6. Determine the probability that out of 8 students (i) None (ii) Two (iii) At least two will be post-graduate Answer: (𝐢)𝟎. 𝟎𝟎𝟎𝟕, (𝐢𝐢)𝟎. 𝟎𝟒𝟏𝟑, (𝐢𝐢𝐢)𝟎. 𝟗𝟗𝟏𝟒

C

14

The probability that a person catch swine flu virus is 0.001. Find the probability that out of 3000 persons (i) exactly 3, (ii) more than 2 person will catch the virus. (F (2; λ)=0.42) 𝐀𝐧𝐬𝐰𝐞𝐫: (𝐢) 𝟎. 𝟐𝟐𝟒𝟐 (𝐢𝐢) 𝟎. 𝟓𝟕𝟔𝟗

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UNIT-6 » Basic Probability T

15

[83]

The probability that an infection is cured by a particular antibiotic drug within 5 days is 0.75. Suppose 4 patients are treated by this antibiotic drug. What is the probability that (a) no patient is cured (b) exactly two patient are cured (c) At least two patients are cured. Answer: (𝐚) 𝟎. 𝟎𝟎𝟑𝟗, (𝐛)𝟎. 𝟐𝟏𝟎𝟗, (𝐜)𝟎. 𝟗𝟒𝟗𝟐

C

16

Assume that on the average one telephone number out of fifteen called between 1 p.m. and 2 p.m. on weekdays is busy. What is the probability that if 6 randomly selected telephone numbers were called (i) not more than three, (ii) at least three of them would be busy? Answer: 𝟎. 𝟗𝟗𝟗𝟕, 𝟎. 𝟎𝟎𝟓𝟏

H 17

An unbiased coin is tossed 6 times. Find the probability of getting (i) exactly 4 heads, (ii) at least 4 heads. 𝐀𝐧𝐬𝐰𝐞𝐫: (𝐢) 𝟎. 𝟐𝟑𝟒𝟒, (𝐢𝐢) 𝟎. 𝟑𝟒𝟑𝟖

C

18

A multiple-choice test consists of 8 questions with 3 answer to each question (of which only one is correct). A student answers each question by rolling a balanced dice and checking the first answer if he gets 1 or 2, the second answer if he gets 3 or 4 and the third answer if he gets 5 or 6. To get a distinction, the student must secure at least 75% correct answers. If there is no negative marking, what is the probability that the student secures a distinction? 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐏(𝐗 ≥ 𝟔) = 𝟎. 𝟎𝟏𝟗𝟕

❖ CHEBYSHEV’S INEQUALITY: ✓ The Chebyshev inequality, in general, gives us bounds on probability that how far a random variable X is deviated when both mean μ and variance σ2 of the distribution are known. ✓ Moreover, the bounds given by the inequality is universal; that is, it is the same for all random variable X with a given μ and σ2 , with the drawbacks that the bounds are not sharp in general. The only restriction with the inequality is that X should have finite σ2 .

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UNIT-6 » Basic Probability

[84]

METHOD – 7: CHEBYSHEV’S INEQUALITY C

1

The number of customers who visit a car dealer’s showroom on Sunday morning is a random variable with mean 18 and standard deviation 2.5. What is the probability that on Sunday morning the customer’s will be between 8 and 28? 𝟏𝟓

Answer: 𝐏(𝟖 < 𝐗 < 𝟐𝟖) ≥ 𝟏𝟔 C

2

Determine the smallest value of k in the Chebyshev’s inequality for which the probability is at least 0.95. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐤 𝟐 = √𝟐𝟎

H

3

Suppose X is a random variable such that E(X) = 3 and E(X 2 ) = 13. Calculate a lower bound for the inequality that X lies between -2 and 8 using Chebyshev inequality. Answer: 0.84

H

4

The number of items cleared by an assembly line during a week is a random variable with mean 50 and variance 25. What can be said about the probability that this week’s clearance will be between 40 and 60? Answer: 0.75

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

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UNIT-7 » Basic Statistic

[85]

UNIT-7 » BASIC STATISTICS ❖ INTRODUCTION: ✓ Statistics is the branch of science where we plan, gather and analyze information about a particular collection of objects under investigation. Statistics techniques are used in every other field of science, engineering and humanity, ranging from computer science to industrial engineering to sociology and psychology. ✓ For any statistical problem the initial information collection from the sample may look messy, and hence confusing. This initial information needs to be organized first before we make any sense out of it. ✓ Quantitative data in a mass exhibit certain general characteristic or they differ from each other in the following ways:



They show a tendency to concentrate values, usually somewhere in the center of the distribution. Measures of this tendency are called measures of Central Tendency or averages.



The data vary about a measure of Central tendency and these measures of deviation are called measures of variation or Dispersion.



The data in a frequency distribution may fall into symmetrical or asymmetrical patterns. The measure of the direction and degree of asymmetry are called measures of Skewness.



Polygons of frequency distribution exhibit flatness or peakedness of the frequency curves. The measures of peakedness of the frequency curves are called measures of Kurtosis.

❖ DATA ✓ Data is a set of values of qualitative or quantitative variables. ✓ Types of data (On the basis of Numbers):



Univariate Data: If one variable is required for observation, than it is called Univariate data. e.g. {𝐱 𝟏 , 𝐱 𝟐 , 𝐱 𝟑 , … , 𝐱 𝐧 }.



Bivariate Data: If two variables is required for observation, than it is called Univariate data. e.g. {(𝐱𝟏 , 𝐲𝟏 ) , (𝐱 𝟐 , 𝐲𝟐 ) , (𝐱 𝟑 , 𝐲𝟑 ) , … , (𝐱 𝐧 , 𝐲𝐧 )}.

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UNIT-7 » Basic Statistic



[86]

Multivariate Data: If more than one variable is required to describe the data, than it is called Multivariate data.

❖ UNIVARIATE ANALYSIS: ✓ Univariate analysis involves the examination across cases of one variable at a time. There are three major characteristics of a single variable that we tend to look at:



Distribution (Data)



Central Tendency



Dispersion

❖ DISTRIBUTION:

✓ Distribution of a statistical data set (or a population) is a listing or function showing all the possible values (or intervals) of the data and how often they occur. ✓ Type of distribution (Data):



Distribution of ungrouped data: E.g. Marks of AMEE of 𝟏𝟎 students are 𝟏𝟎, 𝟐𝟓, 𝟐𝟔, 𝟑𝟓, 𝟎𝟑, 𝟎𝟖, 𝟏𝟗, 𝟐𝟗, 𝟑𝟎, 𝟏𝟖.



Distribution of grouped data: (1).

Discrete Frequency Distribution:

E.g. Data of Students using Library during exam time. No. reading hours(xi )

1 2

3

4

No. of hostel students(fi ) 4 7 10 8

(2).

Continuous Frequency Distribution:

E.g. Data of Students using Library during exam time. No. reading hours (xi ) No. of hostel students (fi )

0 − 2 3 − 5 6 − 8 9 − 11 11

7

8

0

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UNIT-7 » Basic Statistic

[87]

❖ SOME DEFINATION: ✓ Exclusive Class: If classes of frequency distributions are 0 − 2, 2 − 4, 4 − 6, … such classes are called Exclusive Classes. ✓ Inclusive Class: If classes of frequency distributions are 0 − 2, 3 − 5, 6 − 8, … such classes are called Inclusive Classes. ✓ Lower Boundary & Upper Boundary: In Class xi – xi+1, Lower Boundary is 𝐱 𝐢 and Upper boundary is 𝐱 𝐢+𝟏 . ✓ Mid-Point of class: It is defined as Lower Boundary + Upper Boundary 2 ❖ CENTRAL TENDENCY: ✓ The central tendency of a distribution is an estimate of the "center" of a distribution of values. There are three major types of estimates of central tendency:



Mean



Median



Mode

❖ MEAN(𝐱̅): ✓ The Mean or Average is probably the most commonly used method of describing central tendency. To compute the mean, add up all the values and divide by the number of values. ✓ Mean for Ungrouped data:



W-19(1)

If data is 𝐱 𝟏 , 𝐱 𝟐 , 𝐱 𝟑 , … , 𝐱 𝐧 . Then Mean 𝐱̅ is defined as below. 𝐧

𝐱𝟏 + 𝐱𝟐 + 𝐱𝟑 + ⋯ + 𝐱𝐧 𝟏 𝐱̅ = = ⋅ ∑ 𝐱𝐢 𝐧 𝐧 𝐢=𝟏

✓ Mean for Grouped data:



Direct Method: •

Suppose, given data (OR Mid Values of Class) is 𝐱 𝟏 , 𝐱 𝟐 , 𝐱 𝟑 , … , 𝐱 𝐧 . Mean is defined as below. 𝐧

𝐱 𝟏 𝐟𝟏 + 𝐱 𝟐 𝐟𝟐 + 𝐱 𝟑 𝐟𝟑 + ⋯ + 𝐱 𝐧 𝐟𝐧 𝟏 𝐱̅ = = ⋅ ∑ 𝐱 𝐢 𝐟𝐢 𝐧 𝐧 𝐢=𝟏

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UNIT-7 » Basic Statistic



[88]

Assumed Mean Method: •

Suppose, given data (OR Mid Values of Class) is 𝐱 𝟏 , 𝐱 𝟐 , 𝐱 𝟑 , … , 𝐱 𝐧 . Let 𝐀 = 𝐱 𝐢 . Mean is defined as below. 𝐧

𝟏 𝐱̅ = 𝐀 + ⋅ ∑ 𝐝𝐢 𝐟𝐢 𝐧

; 𝐖𝐡𝐞𝐫𝐞 𝐝𝐢 = 𝐱 𝐢 − 𝐀 𝐚𝐧𝐝 𝐀 𝐜𝐚𝐧 𝐛𝐞 𝐚𝐧𝐲 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐱 𝐢

𝐢=𝟏



Step Deviation Method: •

Suppose, given data (OR Mid Values of Class) is 𝐱 𝟏 , 𝐱 𝟐 , 𝐱 𝟑 , … , 𝐱 𝐧 with class length 𝐂. Let 𝐀 = 𝐱 𝐢 . Mean is defined as below. 𝐧

𝐜 𝐱̅ = 𝐀 + ⋅ ∑ 𝐮𝐢 𝐟𝐢 𝐧

; 𝐖𝐡𝐞𝐫𝐞 𝐮𝐢 =

𝐢=𝟏



𝐱𝐢 − 𝐀 𝐜

Step deviation method is used for only continuous frequency distribution.

❖ MODE ( Z ): ✓ The Mode is the most frequently occurring value in the set. To determine the mode, you might again order the observations in numerical order and then count each one. The most frequently occurring value is the mode. ✓ Mode of Ungrouped data: •

W-19(1)

Most repeated observation among given data is called Mode of Ungrouped data.

✓ Mode of Grouped data:



Discrete Frequency Distribution: •



The value of variable corresponding to maximum frequency. Continuous Frequency Distribution: l = Lower boundary of Modal Class

𝐟𝟏 − 𝐟𝟎 )×𝐜 𝐙=𝐥+( 𝟐𝐟𝟏 − 𝐟𝟎 − 𝐟𝟐

c = class interval OR class length f1 = Frequency of the modal class. f0 = Frequency of the class preceding the modal class. f2 = Frequency of the class succeeding the modal class.

Where, the modal class is the class with highest frequency

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UNIT-7 » Basic Statistic

[89]

❖ MEDIAN ( M ): ✓ The Median is the value found at the exact middle of the set of values. To compute the median is to list all observations in numerical order and then locate the value in the center of the sample. ✓ Median of Ungrouped Data:

W-19(1)



Let the total number of observations be 𝐧.



If n is odd number, than M = (

n+1 th



If n is even number, than M =

2

)

observation.

n th 2

n 2

th

( ) observation + ( +1) observation 2

.

✓ Median of Grouped Data:



Median of Discrete Grouped Data: •



n+1 th

In case of discrete group data the position of median i.e. ( 2 ) item can be located through cumulative frequency. The corresponding value at this position is value of median. Median of Continuous Grouped Data:

l = lower boundary point of the Median class 𝐧 −𝐅 )×𝐜 𝐌=𝐥+(𝟐 𝐟

n = total number of observation (sum of the frequencies) F = cumulative frequency of the class preceding the median class. f = the frequency of the median class c = class interval 𝐎𝐑 class length 𝐧

Where, Median class = Class whose cumulative frequency with property 𝐦𝐢𝐧 {𝐜𝐟 | 𝐜𝐟 ≥ 𝟐}

❖ MEASURE OF DISPERSION: ✓ Measure of dispersion can be classified into two different categories: ✓ The measure which express the spread of observations in terms of distance between the values of selected observations. It is also known as Quartile deviation.

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UNIT-7 » Basic Statistic

[90]

✓ Dispersion refers to the spread of the values around the central tendency. There are two common measures of dispersion, the range and the standard deviation. ✓ Range: It is simply the highest value minus the lowest value. In our example distribution, the high value is 𝟑𝟔 and the low is 𝟏𝟓, so the range is 𝟑𝟔 − 𝟏𝟓 = 𝟐𝟏. 1

✓ Quartile Deviation: It is defined as Q = 2 (Q3 − Q1 ), Where Q1 and Q3 are the first and third quartiles of the distribution respectively. ✓ Standard Deviation(𝛔): It is a measure that is used to quantify the amount of variation or dispersion of a set of data values. Note: ✓ Range is the simplest measure of dispersion. Since it is based on two extreme observations which may fluctuate. So, it is not reliable measure of dispersion. ✓ Quartile deviation is better than range as it uses 50% of the data. But it ignores 50% of the data, so it is not reliable. ✓ For finding Quartile deviation first we have to arrange the data in ascending order. N+1 th

✓ 𝐅𝐢𝐫𝐬𝐭 𝐐𝐮𝐚𝐫𝐭𝐢𝐥𝐞 (𝐐𝟏 ) = (

)

4

observation

N+1 th

✓ 𝐬𝐞𝐜𝐨𝐧𝐝 𝐐𝐮𝐚𝐫𝐭𝐢𝐥𝐞 (𝐐𝟐 ) = ( ✓ 𝐓𝐡𝐢𝐫𝐝 𝐐𝐮𝐚𝐫𝐭𝐢𝐥𝐞 (𝐐𝟑 ) = (

2

)

3(N+1) th

)

4

observation observation

❖ FORMULA TO FIND STANDARD DEVIATION: Method

Ungrouped Data

Direct Method σ=

∑N √ j=1(xj − x̅)

Actual Mean Method

Grouped Data

2

n

σ=√

∑N ̅) j=1 fj (xj − x n

σ=√

∑ni=1 xi2 − x̅ 2 n

σ=√

∑ni=1 fi xi2 − x̅ 2 ∑ni=1 fi

σ=√

∑ni=1 d2i − x̅ 2 n

σ=√

∑ni=1 fi d2i − x̅ 2 ∑ni=1 fi

Assumed Mean Method

2

❖ VARIANCE ( 𝐕(𝐱) ): ✓ Variance is expectation of the squared deviation. It informally measures how far a set of (random) numbers are spread out from their mean.

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UNIT-7 » Basic Statistic

[91]

✓ 𝐕𝐚𝐫𝐢𝐚𝐧𝐜𝐞 = (𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐃𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 )𝟐 ❖ SKEWNESS: ✓ It is a measure of the asymmetry of the probability distribution of a real-valued random variable about its mean. 𝐒𝐤𝐞𝐰𝐧𝐞𝐬𝐬 =

𝐌𝐞𝐚𝐧 − 𝐌𝐨𝐝𝐞 𝐱̅ − 𝐙 = 𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 𝛔

✓ The Skewness value can be positive or negative, or even undefined. ✓ Negative skew: The left tail is longer;

the

mass

of

the

distribution is concentrated on the right. ✓ Positive skew: The right tail is longer; the mass of the distribution is concentrated on the left. ❖ COEFFICIENT OF QUARTILE DEVIATION: 𝐅𝐨𝐫𝐦𝐮𝐥𝐚: C. D. =

(Q3 − Q1 )/2 (Q3 − Q1 ) = (Q3 + Q1 )/2 (Q3 + Q1 )

❖ COEFFICIENT OF VARIATION 𝛔

✓ Formula: C. V. = 𝐱̅ × 𝟏𝟎𝟎 ✓ The Coefficient of variation is lesser is said to be less variable or more consistent.

METHOD – 1: EXAMPLES ON CENTRAL TENDENCY, DISPERSION AND SKEWNESS 1

C

Define Mode and also give the relationship between Mean, Median & Mode. ̅ 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐙 = 𝟑𝐌 − 𝟐𝐗

2

C

Find the mean, mode and median by the data 2, 8, 4, 6, 10, 12, 4, 8, 14, 16. ̅ = 𝟖. 𝟒, 𝐙 = 𝟖 & 𝟒, 𝐌 = 𝟖 𝐀𝐧𝐬𝐰𝐞𝐫: 𝐗

H

3

Find

mean,

mode

and

median

of

following

observation.

10, 9, 21, 16, 14, 18, 20, 18, 14, 18, 23, 16, 18,4 . 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏𝟓. 𝟔𝟒𝟐𝟗, 𝟏𝟖, 𝟏𝟕

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-7 » Basic Statistic C

4

[92]

Find mean, mode and Median of following data. Number of students

6

4

16

7

8

2

Marks obtained

20

9

25

50

40

80

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝟐. 𝟐𝟑𝟐𝟔, 𝟐𝟓, 𝟐𝟓 H

5

Find mean, mode and Median of following data Marks obtained

18

22

30

35

39

42

45

47

Number of students

4

5

8

8

16

4

2

3

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝟒. 𝟓, 𝟑𝟗, 𝟑𝟕 C

6

The following data represents the no. of foreign visitors in a multinational company in every 10 days during last 2 months. Use the data to the mean, mode and median. X

0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60

No. of visitors

12

18

27

20

17

06

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝟖, 𝟐𝟓. 𝟔𝟐𝟓, 𝟑𝟎. 𝟕𝟒𝟎𝟕 T

7

Calculate Mean, Median and Mode for the following data. Class 50

53

56

59

62

65

68

71

74

− 53

− 56

− 59

− 62

− 65

− 68

− 71

− 74

− 77

3

8

14

30

36

28

16

10

5

F

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟔𝟑. 𝟖𝟐, 𝟔𝟑. 𝟔𝟔, 𝟔𝟑. 𝟐𝟖𝟓𝟕 H

8

A Survey regarding the weights (in kg) of 45 students of class X of a school was conducted and the following data was obtained: Weight

20-25

25-30

30-35

35-40

40-45

45-50

50-55

2

5

8

10

7

10

3

(in kg) No.

of

students Find the mean, mode and median weight. 𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝟖. 𝟖𝟑𝟑𝟑, 𝟑𝟕, 𝟑𝟖. 𝟕𝟓

DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AMEE - 3130908

UNIT-7 » Basic Statistic H

9

[93]

Find the missing frequency when median is 24. Marks Students

0 − 10

10 − 20

20 − 30

30 − 40

40 − 50

15

20

X

14

16

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝟓 H 10 Obtain the mean, mode and median for the following distribution: Mid value

15

20

25

30

35

40

45

50

55

Frequency

2

22

19

14

3

4

6

1

1

Cumulative

2

24

43

57

60

64

70

71

72

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝟓. 𝟖𝟒𝟕𝟐, 𝟐𝟏. 𝟖𝟒𝟕𝟖, 𝟐𝟓. 𝟔𝟓𝟕𝟗 H 11 Obtain the mean, mode and median for the following information: Marks

0