Agni College of Technology,Thalambur Coaching Class – Question Paper Statistics and Numerical methods (common to Mechanical,Mechatronics)

Part-A 1. State the basic principles of design of experiments. There are three basic principles of experimental design. They are (1).Randomization (2).Replication (3). Local control (error control) 2. What is the need of Newton’s and Lagrange’s interpolation formulae? To develop interpolation formula for unequally spaced values of x. 3. Bring out the merits and demerits of Taylor series method. The method gives a straight forward adaptation of classic calculus to develop the solution as an infinite series. It is a powerful single method if we are able to find the successive derivatives easily. If f(x,y) involves some complicated algebraic structures then the calculation of higher derivatives becomes tedious and the method fails. This is the major drawback of this method. 4. Find y(0.1) by Euler’s method, If

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

Solution: Given y ' =+ x 2 y 2 , x0 = 0, y0 = 1, h = 0.1

Formula

y1 = y0 +

h h2 h3 y0 '+ y0 ''+ y0 '''+ ..... 1! 2! 3!

y ' =x 2 + y 2

y0 ' =x0 + y0 =(0) 2 + (1) 2 =1

y '' = 2 x + 2 yy '

y0 '' =+ 2 x0 2 y0 y0 ' = 2(0) + 2(1)(1) = 2

2

2

2 + 2[ yy ''+ y ' y '] y ''' = 2 + 2[ y0 y0 ''+ y0 ' y0 '] = 2 + 2[(1)(2) + (1)(1)] = 8 y ''' = (0.1) 2 (0.1)3 (2) + (8) 2 6 y1 =+ 1 0.1 + 0.01 + 0.0013333 1 + (0.1)(1) + y1 = y1 = 1.11133

5. Write the central difference approximations for

dy d 2 y , dx dx 2

Solution:

1 ( yi +1, yi −1 ) 2h 1 y='' 2 ( yi +1 − 2 yi + yi −1 ) h y '=

6. Form the divided difference table for the following data x y

5 7

15 36

22 160

7. State the assumptions in using ANOVA? Solution: The assumptions are (i) The samples are drawn from normal populations. (ii) The samples are drawn independently from these populations. (iii) All populations have the same variance.

1

8. Using Trapezoidal rule, evaluate

dx

∫ 1+ x

2

with h = 0.2 Hence evaluate an

0

approximate value of π

9. State the formula to find the second order derivative using the forward difference

10. Write down the milne’s predictor-corrector formula for solving initial value problem in first order differential equations. Solution: Milne’s predictor formula is yn +1 = yn −3 +

4h (2 y 'n − 2 − y 'n −1 + 2 yn ) 3

h 3

Milne’s corrector formula is yn +1 = yn −1 + ( y 'n −1 + 4 yn + y 'n +1 ) 11. State Lagrange’s interpolation formula. Solution: Let y = f(x) be a function which takes the values y0 , y1 , y2 ...... yn corresponding to xo , x1 , x2 ......xn Then Lagrange’s interpolation formula is

y f= = ( x)

( x − x1 )( x − x2 ) ... ( x − xn ) y ( x0 − x1 )( x0 − x2 ) ... ( x0 − xn ) 0 ( x − xo )( x − x2 ) ... ( x − xn ) y + ( x1 − xo )( x1 − x2 ) ... ( x1 − xn ) 1 + ............................................ + ............................................ +

( x − xo )( x − x1 ) ... ( x − xn−1 ) y ( xn − x0 )( xn − x1 ) ... ( xn − xn−1 ) n

12. Write the merits and demerits of Completely Randomized Design (CRD). Solution: Merits: 1. CRD has a simple lay out. 2. There is complete flexibility as the number of replication is not fixed. 3. The analysis of the design is simple as it results in a one-way classification analysis of variance. 4. Analysis can be performed even if some observation are missing. Demerits: The experimental error is large as compared to the other designs because homogeneity of the units is not taken into consideration. 13. What you understand by Design of an experiment? Solution: Design of experiment is a sequence of steps taken to ensure a scientific analysis leading to valid inference about the hypothesis. The prime objective of the design of experiments is to control the extraneous variables so that the results could be attributed only to the experimental variables. 14. State Inverse interpolation? Solution: Inverse interpolation is the process of finding the value of the argument (x) corresponding to a given value of a function (y) lying between two tabulated functional values. 15.State any two properties of divided differences. Solution: The properties of divided differences are (1) they are symmetric functions of their arguments.

(2) the nth difference is a ratio of two determinants of order (n+1). 16. Define 22 factorial design. Solution: when there are two factors A,B and two levels ‘high’ and ‘low’ for each factor we a 22 factorial design. In spite of its simplicity, the 22 design is a powerful tool to improve products and process. 17. Obtain the finite scheme for the differential equation 2

d2y +y= 5 dx 2

Solution: 1 5 y ( x) = 2 2 y − 2 yi + yi +1 y '' = i −1 h2 yi −1 − 2 yi + yi +1 1 5 − yi = 2 2 2 h 1 5 2 yi −1 − 2 yi + yi +1 − h 2 yi = h 2 2 1 5 yi −1 − [2 + h 2 ] yi + yi +1 =h 2 2 2 2 2 yi −1 −  4 + h  yi + 2 yi +1 = 5h 2 y ''( x) −

18. Define Mean sum of squares. Solution: Mean square =

sum of the squares Degree of freedom

19.Find the parabola of the form y = ax 2 + bx + c passing through the points (0 , 0), (1, 1) and(2, 20). Solution: We use Lagrange’s interpolation formula y=

( x − 1)( x − 2 ) 0 + ( x − 0 )( x − 2 ) 1 + ( x − 0 )( x − 1) 20 ( 0 − 1)( 0 − 2 ) (1 − 0 )(1 − 2 ) ( 2 − 0 )( 2 − 1) = 0 − x ( x − 1) + 10 x ( x − 1) = y 9 x2 − 8x

20. Show that

1 1  = − abcd a

Solution: 1 1 , f (a) = x a 1 1 − 1 f ( a, b) = b a = − b−a ab

= If f ( x)

1 1 − + f (b, c) − f (a, b) 1 c−a 1 bc ab = = = = f ( a , b, c )   c−a c−a abc  c − a  abc f (b, c, d ) − f (a, b, c) f ( a , b, c , d ) = d −a 1 1 − 1 a−d  1 = bcd abc =  = − d −a abcd  d − a  abcd

Therefore

1 1  = − abcd a

21. Compare one-way classification model with two-way classification model. Solution: One way classification 1 2

One factor is involved One set of Hypothesis

Two way classification Two factors involved Two set of Hypothesis

22. Compute y at x = 0.25 by modified Euler’s method given y ' 2= xy y (0) 1 = Solution: Given f(x, y) = 2xy, x0 = 0, y0 = 1, h = 0.25, x1 = 0.25 By modified Euler method, 1  h h  f  xn + , yn + f ( xn , yn )  h  2 2  0.25 0.25  y1 = y0 + hf  0 + ,1 + ( 2 x0 , y0 )  2 2   yn +1 = yn +

= 1 + (0.25)[ f ( 0.125,1 + (0.25)(0)(1) )]

= 1 + 0.25[(2)(0.125)(1)] = 1 + 0.00625 = 1.0625

23. Write down the formula for fourth order Runge-Kutta method Solution: k1 = hf ( xn , yn ) h k   k2 = hf  xn + , yn + 1  2 2  h k   k3 = hf  xn + , yn + 2  2 2  k4 = hf ( xn + h, yn + k3 ) 1 ( k1 + 2k2 + 2k3 + k4 ) 6 yn += yn + ∆y 1

∆= y

24. State Newton’s divided difference formula for interpolation. Solution:

f ( x= ) f ( x0 ) + ( x − x0 ) f ( x0 , x1 ) + ( x − x0 )( x − x1 ) f ( x0 , x1 , x2 ) + ..... + ( x − x0 )( x − x1 ) ..... ( x − xn −1 ) f ( x0 , x1 ,....xn )

PART-B 1(i)

1(ii)

2)

3.(i)

3(ii)

4)

5.(i)

5(ii)

6)

7)(i)

7)(ii)