Numerical and Statistical Methods

F.Y. B.Sc.(IT) : Sem. II Numerical and Statistical Methods Time : ½ Hrs.] Prelim Question Paper Solution [Marks : 75 Q. Attempt any THREE of the following : [5] Q.(a) What is a mathematical model? With the help of a flowchart, explain the of solving an engineering problem. Ans.: A mathematical model can be broadly defined as a formulation or equation that expresses the essential features of a physical system or process in mathematical terms. In a very general sense, it can be represented as a functional relationship of the form : Dependent variable f indepdent forcing, parameters, variables functions () where the dependent variable is a characteristic that usually reflects the behavior or state of the system; the independent variables are usually dimensions, such as time and space, along which the system s behavior is being determined; the parameters are reflective of the system s properties or composition; and the forcing functions are external influences acting upon the system. The actual mathematical expression of Eq. () can range from a simple algebraic relationship to large complicated sets of differential equations. For example, on the basis of his observations, Newton formulated his second law of motion, which states that the time rate of change of momentum of a body is equal to the resultant force acting on it. The mathematical expression, or model, of the second law is the well-known equation F ma () where F net force acting on the body (N, or kg m/s ), m mass of the object (kg), and a its acceleration (m/s ) Q.(b) Explain the terms : (i) Significant figures, (ii) Accuracy, (iii) Precision, (iv) Truncation error, (v) Round-off error

Vidyalankar : F.Y. B.Sc. (IT) NSM Ans.: (i) Significant figures : The concept of a significant figure or digit has been developed to formally designate the reliability of a numerical value. The significant digits of a number are those that can be used with confidence. (ii) Accuracy : Accuracy refers to how closely a computed or measured value agrees with a true value. (iii) Precision : Precision refers to how closely individual computed or measured values agree with each other. (iv) Truncation error : These errors are generated when only required significant digits are retained in the number and remaining are discarded. (v) Round-off error : Rounding off errors are errors arising from the process of rounding off during computation. Q.(c) Suppose that you have the task of measuring the lengths of a bridge and a rivet and come up with 9999 and 9 cm, respectively. If the true values are 0,000 and 0 cm, respectively, compute (i) the true error and (ii) the true percent relative error for each case. Ans.: (i) The error for measuring the bridge is Error True value Approximate value For bridge 0,000 9,999 cm For rivet 0 9 cm (ii) The percent relative error for the bridge is (Truevalueapprox.value) (E R ) Truevalue (E P ) (0, 0009, 999) 0,000 % error E P 0.000 00 0.0 % Relative Error for rivet 09 (E R ) 0 0 0. % Error (E P ) 0. 00 0% 0,000 0.000 Q.(d) Use zero-through fourth-order Taylor series expansions to approximate the function f(x) 0.x 4 0.5x 3 0.5x 0.5x +. from x i 0 with h. That is, predict the function's value at x i+. Ans.: Because we are dealing with a known function, we can compute values for f(x) between 0 and. The results indicate that the function starts at f(0). and then curves downward to f() 0.. Thus, the true value that we are trying to predict is 0.. The Taylor series approximation with n 0 is f(x i ).

Prelim Question Paper Solution Q.(e) Evaluate and interpret the condition number for sin x f(x) for x.000 cos x sin x uv vu u sin x Ans.: f(x) cosx v v + cos x x.000 f(.000 ) 6366. f(.000 ).05 0 7 f'(x) 0000.97 f(x) Q.(f) Evaluate e 5 using the two formulae : e x 3 4 x x x x +...! 3! 4! and e x x 3 4 e x x x x... 3! 4! and compare with the true value 6.737947 0 3. Use five terms to evaluate each series and compute true and approximate relative errors as terms are added. Ans.: e 5 3 4 (5) 5 5 5 5 55 5 + 5 3.7083! 3! 43 6 4 e 5 5 3 4 e 5 5 5 5 5 65 5 5! 3! 4! 6 4! 65.375 0.0596 Error for first formula True value Approximate value 6.737947 0 3 3.70833 37.059 Error while using second formula : True Value Approximate Value 6.737947 0 3 0.0596 0.0085 Q. Attempt any THREE of the following: [5] Q.(a) Find the roots of the equation x 3.x + 7.45x + 4 0 using Regula-Falsi method correct up to 4 decimal places. Ans.: f() () 3.() + 7.45() + 4 33.() + 7.45() + 4 33 476. + 8.95 + 4.5 f() () 3.() + 7.45() + 4 78 756.8 + 89.4 + 4 0.6 f() f() < 0 a, b af[b] bf[a] (0.6) (.5) x f[b] f[a] 0.6 (.5) 8.655 3.85 383.6 3.85.75 3

Vidyalankar : F.Y. B.Sc. (IT) NSM x.75 f(x ) f(.7) (.7) 3.(.7) + 7.45(.7) + 4 394.0 5.4 + 83. + 4 3.8 f(.7) f() < 0 a.7, b f(a) 3.8, f(b) 0.6 af[b] bf[a] (.7) (0.6) () ( 3.8) x f[b] f[a] 0.6 ( 3.8) 46.438.6 0.63.8 84.3 05.78.958 f(x ) f(.958) (.95) 3.(.95) + 7.45(.95) + 4 403.0 59.0 + 83.40 + 4 0.6 f(.95) f() < 0, a.95, b f(a) 0.6, f(b) 0.6 af[b] bf[a] (.95) (0.6) () ( 0.6) x 3 f[b] f[a] (0.6) ( 0.6) 48.607. x 3.996 03. x.75 x.958 x 3.996 approximate root.889 or.990 Q.(b) Find the roots of the equation x tan x near 4 using Newton Raphson method correct up to 4 decimal places. Ans.: f(x) x. tan x f(x) tan x + x.sec x Given x 0 4 f(x 0 ) f(4) x tan x 4 tan 4 3.63 f(x 0 ) tan x + x.sec x 0.500 f(x n) x n + x n f'(x n) f(x ) 0 3.63 x x 4 0 f'(x ) 0.500 0 x 3.6548 f(x ) f(3.6548) x tan x.0597 f(x ) f(3.6548) tan x + x.sec x f(x ) x x f'(x ) x 3.6548.0597 5.379 x 3.4578 f(x ) 0.33, f(x ) 4.55 x 3 x f(x ) f'(x ) 3.4578 0.33 4.55 x 3 3.456 x 3.6548, x 3.4578, x 3 3.456 approx. root 3.456 4

Prelim Question Paper Solution Q.(c) The following table shows the number of students and range of marks. Find the number of students who have secured less than 45 marks. Marks 30 40 40 50 50 60 60 70 70 80 No. of Students 3 45 3 7 5 Ans.: Marks less than x No. of Students 40 3 50 73 60 4 70 59 80 90 h x x 0 50 40 0 x x 0 + hk k 4540 0 0.5 By Newtons forward difference Interpolation formula. k(k 3 Y(x) y 0 + k y 0 + ) k(k ) (k ) y y 0 0! 3! k(k)(k )(k 3) 4 + y0 4! 0.5(0.5) Y(45) 3 + 0.5 4 + 9 + 0.5 (0.5 ) (0.5 ) ( 5) 5 + 0.5 (0.5 ) (0.5 ) (0.5 3) 37 4! x y y y 3 y 4 y 40 3 4 50 73 5 9 5 60 4 35 6 37 70 59 3 4 80 90 y(45) 47.868 Q.(d) Given : x 3 4 5 6 7 8 f(x) 0.0 0.004 0.0 0. 0.5 0.57 0.35 0.3 find f(7.5) using Newton s backward interpolation formula. Ans.: Newton s backward interpolation formula Y(x) kyn k(k) k(k) (k) 3 x y y... n n n!! 3! x y y y 3 y 4 y 5 y 6 y 7 y 0.0 0.006 0.0 0.06 0.6 0.57.08.087 0.004 0.06 0.084 0.54 0.30 0.564.006 3 0.0 0. 0.07 0.47 0.63 0.44 4 0. 0.03 0.077 0.6 0.75 5 0.5 0.07 0.039 0.059 6 0.57 0.068 0.0 7 0.35 0.088 8 0.37 5

Vidyalankar : F.Y. B.Sc. (IT) NSM ( 0.5)( 0.088) ( 0.5)( 0.5 )(0.0) Y(7.5) 8!! ( 0.5)( 0.5 )( 0.05 )(0.059) 3! ( 0.5)( 0.5 )( 0.05 )( 0.5 3)(0.75) 4! ( 0.5)( 0.5)( 0.05)( 0.53)( 0.54)(0.44) 5! ( 0.5)( 0.5)( 0.5)( 0.5 3)( 0.54)( 0.5 5)(.006) 6! ( 0.5)( 0.5)( 0.05)( 0.53)( 0.54)( 0.55)( 0.56)(.087) 7! 0.98 Q.(e) Determine the real root of f(x) 4x 3 6x + 7x.3 using bisection method correct upto 3 decimal places. Ans.: f(x) 4x 3 6x + 7x.3 f(0) 4(0) 3 6(0) + 7(0).3.3 f() 4() 3 6() + 7().3.7 f(0) f() < 0 Root lies in (0, ) Iterations are x x x i a b f(x i 0.0000.0000 0.5000 0.000 0.0000 0.5000 0.500 0.865 0.500 0.5000 0.3750 0.3078 0.3750 0.5000 0.4375 0.0509 0.4375 0.5000 0.4687 0.0748 0.4375 0.438 0.453 0.00 0.4375 0.453 0.4453 0.094 0.4458 0.453 0.445 0.0036 The solution after successful iteration is 0.445. Q.(f) Given log 0.300, log 3 0.477, log 5 0.6990 and log 7 0.845. Using Lagrange s formula, find log 47. Ans.: x 0 x x x 3 x log log 3 log 5 log 7 Y 0.300 0.477 0.6990 0.845 f(x) by using Lagrange s Interpolation is given by : f(x) (xx ) (xx ) (xx ) 3 f(x ) 0 (x x ) (x x ) (x x ) 0 0 0 3 + + + (xx ) (xx ) (xx ) 0 3 f(x ) (x x ) (x x )(x x ) 0 3 (xx ) (xx ) (xx ) 0 3 f(x ) (x x ) (x x ) (x x ) 0 3 (xx ) (xx ) (xx ) 0 f(x ) 3 (x x ) (x x ) (x x ) 3 0 3 3 6

Prelim Question Paper Solution (3.850.0986) (3.850.6094) (3.850.9459) 0.30 (0.693.0986) (0.693.6094) (0.6939.9459) (3.8500.693) (3.850.6094) (3.850.9450) + 0.477 (.0986 0.693) (.0986.6094) (.0986.9459) + (3.850 0.693) (3.850.0986) (3.850.9459) 0.6990 (.6094 0.693) (.6094.0986) (.6094.9459) + (3.850 0.693) (3.850.0986) (3.850.6094) 0.845 (.94590.693) (.9459.0986) (.9459.6094).670 Q.3 Attempt any THREE of the following: [5] Q.3(a) Solve the following simultaneous equations by Gauss -Jordan elimination method: x + 6x x 3 4, 5x x + x 3 9, x 3 3x 4x 4 Ans.: x + 6x x 3 4 () 5x x + x 3 9 () x 3 3x 4x 4 (3) Operate () 5 () 9 6x + x 3 64 Operate (3) + 3 () 5x x 7 3 The new reduced system of equations are x + 6x x 3 4 (4) 9 6x x 64 (5) 3 5x x 7 (6) 3 Operate (4) + 6 (5) x x 0 3 6 6 Operate (6) + 5 6 (5) 9 x 3 3 3 the new reduced system of equations are x x 0 (7) 3 6 9 6x x 64 (8) 3 9 x3 3 Operate (7) Operate (8) 3 (9) 6 (9) (7) 9 3 x 4 9, x 9 44 9 6x 44 9 x 9 x3 3 89 9 (9) 3 x 3 96 9 (9) 9 7

Vidyalankar : F.Y. B.Sc. (IT) NSM Final solution is x 9 3.860 x 89 9 3.689 x 3 96 9 3.303 Q.3(b) Solve the following simultaneous equations by Gauss-Seidel method : 0x + x + x 3, x + 0x + x 3 3, x + x + 0x 3 4 x x 3 Ans.: x 0 3x x 3 x 0 4x x x 3 0 () () (3) Iteration : x 0, x 3 0 (x ) 0 0 0 0 (x ) 0. Put x., x.06, x 3 0 (x ) 0 3() 0.06 0 (x 3 ) 0 4(.) (.06) 0 (x 3 ) 0 0.948 Iteration : (x ) 0.999 (x ).0053 (x 3 ) 0.999 Iteration 3 : (x ) 0.9995 (x ).000 (x 3 ).0000 After 3 iterations x x x 3 Q.3(c) Evaluate x 0 e x dx using trapezoidal rule and Simpson s 3/8 rule. Ans.: Let n 5 ( ) Hence h 5 0. x 0..4.6.8.0 Y 0 0.63 Y 0.583 Y 0.538 Y 3 0.4988 Y 4 0.4637 Y 5 0.433 Trapezoidal Rule x x0 ydx h [y 0 + (y + y + + y n ) + y n ] 0.5674 By Simpson s 3/8 th rule 0.5075 Q.3(d) Evaluate Ans.: Let n 6 0 x 0 0, x n, h sin x dx 5 4 cos x x x n 0 n using Simpson s 3/8 th rule. 0 6 6 8

Prelim Question Paper Solution Simpson s th 3 rule 8 y 3h 8 [(y 0 + y n ) + (terms of y which are multiple of 3) + 3(remaining terms of y)] a 0, b, n 6 x 0 /6 /3 / /3 5/6 y 0 0.095 0.070 0.999 0.5 0.63 0 y 3 [ (0 + 0) + (terms of y which one multiple of 3) + 8 6 3(remaining terms of y)] y 0.40 Q.3(e) Solve dy log (x + y); y() for x. and x.4 using Euler s modified dx method, taking h 0.. Ans.: f(x, y) log (x + y) y 0 (), x 0, y 0, h 0. f(x 0 y 0 ) log(x 0 + y 0 ) log ( + ) log 3.0986 To find y at x. y(.) y 0 + f(x 0 y 0 )h + log 3 0..97 f(x y ) f(.,.7) log(. +.7).95 y(.4) y + f(x y ) * h.97 +.95 * 0..465 y(.4).465 Q.3(f) Solve dy dx y x, where y(0), to find y(0.) using Runge-Kutta method. y x y x Ans.: f(x, y) y x x 0 0 y 0 h 0. k h.f(x 0, y 0 ) 0. f(0, ) k 0. k hf (x 0 + h, y 0 + k ) k 0. f(0.05, 0.) k 0.06 k (k k ) k 0.08 y y 0 + k y + 0.08 y.08 at x 0. Q.4 Attempt any THREE of the following: [5] Q.4(a) Fit a straight line to the x and y values in the two rows: x 3 4 5 6 7 y 0.5.5.0 4.0 3.5 6.0 5. Ans.: M 7 which is odd Let three required set line for best fit be Y a + bx () 9

Vidyalankar : F.Y. B.Sc. (IT) NSM Normal equations are y ma + b x () xy ax + bx (3) Consider the table no. x y xy x 0.5 0.5.5 5.0 4 3.0 6.0 9 4 4.0 6.0 6 5 3.5 7.5 5 6 6.0 36.0 36 7 5. 36.4 49 x 8 y 3.7 xy 8 x 40 Substituting in equation () and (3) we get 3.7 7a + 8 b 8.4 8 a + 40 b Solving we get a 0.045, b 0.848 Required straight line is y 0.045 + 0.848 x Q.4(b) Fit a second degree parabola for the following : x.5.5 0.5 0 0.5.5 y 4.3 4.83 5.7 5.47 6.6 6.79 7.3 Ans.: Here x 4.5, y 0.7, x 5.5 x 3 3.6, x 4 65.3 xy 6.86 x y 30.0 m 7 The equation of second degree parabola is y a + bx + cx Solve y na + bx + cx xy a x + b x + cx 3 x y ax + bx + cx 4 Substitute the values 0.7 79 4.5b + 5.5c () 6.86 4.5a + 5.5b 3.65c () 30.0 5.5a 3.65 b + 65.35c (3) Solving we get, a 6.0, b 0.76, c 0.003 second degree parabola is, y 6.09 + 0.768x + 0.003 x Q.4(c) Use multiple regression to fit the following data : x x y 0 0 5 0.5 9 3 0 4 6 3 7 7 0

Prelim Question Paper Solution Ans.: x x y x z y x z x y z 0 0 5 0 5 0 0 0 0 4 00 0 0.5 9 6.5 8 5.5 8 3 0 0 3 0 0 4 6 3 6 9 4 8 7 7 49 79 4 89 54 x 6.5, y 4, z 54 x 76.5, z 944, xy 48 xz 43.50 yz 00 y a + bx + cz required regression. y ma + b x + c z y x a x + b x + c z x y z a z + b z x + c z Substitute the values in eq. (), (), (3) 4 6a + 6.5b + 54c () 48 6.5 a + 76.5 b + 43.5 c () 00 54 a + 43.5 b + 944 c (3) by solving equation (), (), (3) simultaneously and removing the variables we get, a.667, b.333, c 0.333 Q.4(d) Maximize 50x + 00y subject to 0x + 5y 500, 4x + 0y 000, x +.5y 450 and x 0; y 0. Ans.: Maximize Z 50x + 00y Subject to 0x + 5y 500 () 4x + 0y 000 () x +.5y 450 (3) and x 0, y 0 convert the given constraints into equations 0x + 5y 500 4x + 0y 000 x +.5y 450 Consider 0x + 5y 500 we get x 0 50 y 500 0 Consider 4x + 0y 000 we get x 0 500 y 00 0 500 y Consider 4x + 0y 000 we get x 0 450 y 300 0 400 300 00 b () () Feasible region is ABCDA A(0, 0), B(0, 00), C(00, 0) D(50, 0) 00 0 a c (3) d 00 00 300 400 500 x

Vidyalankar : F.Y. B.Sc. (IT) NSM Points Z 50x + 00y A 0 B 0000 C 000 D 500 Optimal Value of z is at C X 00 and Y 0 is optimal solution. Q.4(e) A firm makes two types of furniture chairs and tables. The contribution for each product as calculated by the accounting department is Rs. 0 per chair and Rs. 30 per table. Both products are processed on three machines M, M and M 3.The time required in hours by each product and total time available in hours per week on each machine are as follows: MACHINE CHAIR TABLE AVAILABLE TIME M 3 3 36 M 5 50 M 3 6 60 How should the manufacturer schedule his production in order maximize contribution? Ans.: Maximize Z 0 x + 30y Subject to 3x + 3y 36 5x + y 50 x + 6y 60 Non Negative constraints x 0, y 0 3x + 3y 36 5x + y 50 x + 6y 60 x 0 x 0 0 x 0 30 y 0 y 5 0 y 0 0 Vertex Z 0x + 30y O (0, 0) 0 A (0, 0) 300 5 y B (5, 8) 340 0 C (4, 5) 30 D (0, 0) 00 5 0 a b The maximum contribution come upto point b (5, 8) as x 5 and y 8 5 0 a c d 5 0 5 0 5 30 x Q.4(f) An aged person must receive 4000 units of vitamin, 50 units of minerals and 400 calories a day. A dietician advises to thrive on two foods Fand F that cost Rs 4 and Rs respectively per unit of food. It one unit of F contains 00 units of vitamins, unit of mineral and 40 calories and one unit of F contains 00 units of vitamins units of minerals and 40 calories, formulate a linear programming model to minimize the cost of diet.

Prelim Question Paper Solution Ans.: Product Food F Food F Requirements Vitamins 00 00 4000 Minerals 50 Calories 40 40 40 Cost /Units 4 Z 4x + y Subject to 00x + 00 y 4000 x + y 50 40x + 40 400 x, y 0 Q.5 Attempt any THREE of the following: [5] Q.5(a) State and explain the properties of distribution functions. Ans.: Let x be a random variable, the function f defined for all real values x by f(x) P[X x] for all real x is called distribution function. A distribution function is also called as cumulative probability distribution function. Properties of distribution function : ) If f is the distribution function of random variable x if a < b then P[a < x b] f(b) f(a) ) Values of all distribution function lies between 0 and i.e. 0 f(x) for all x 3) All distribution functions are monotonically decreasing i.e. if x < y then f(x) < f(y). 4) f() f() lim f(x) x lim f(x) x 0 5) If x is discrete random variable then f(x) P(x ) xi x 6) If values of discrete random variable x are like x < x < x 3 < x 4 then P(V n + ) f(x n + ) f(x n ) 7) If x is discrete random variable then distribution function is step function. Q.5(b) A random variable x has the following probability distributions : x 0 3 4 5 6 7 p(x) 0 k k k 3k k k 7k + k (i) Find k, (ii) Evaluate P(x < 6), P (x 6) and P(0 < x < 5), (iii) If P(x c) >, find the minimum value of c, and (iv) Determine the distribution function of x. Ans.: i) Find k, Sum of all probabilities is unity x7 p(x) x0 0 + k + k + 3k + k + k + 7k + k k 0 ii) Evaluate P(x < 6) P(x 0) + P(x ) + P(x ) + P(x 3) + P(x 4) + P(x 5) 0 + k + k + k + 3k + k 8 00 3

Vidyalankar : F.Y. B.Sc. (IT) NSM iii) If P(x c) >, find the minimum value of c P(x 0) P(x 0) 0 False not P(x ) P(0) + P(x ) 0 + 0 0 0. not P(x ) P(0) + P() + P() 0 + 0 + 0 0.3 not P(x 3) P(0) + P() + P() + P(3) 0 + 0 + 0 + 0.5 0 P(x 4) P(0) + P() + P() + P(3) +P(4) 0 + 0 + 0 + 0 + 3 0 0.8 condition is true. Minimum value of c 4 not iv) f(0) P[x 0] P(0) 0 f() P[x ] P(0) + P() 0 0 0 f() P[x ] P(0) + P() + P() 0 + 3 0 0 0 f(3) P[x 3] P(0) + P() + P() + P(3) 0 0 0 0 f(4) P[x 4] P(0) + P() + P() + P(3) + P(4) 3 0 8 0 0 0 0 0 f(5) P[x 5] P(0) + P() + P() + P(3) + P(4) + P(5) f(6) 83 00 3 3 0 0 0 0 0 00 00 ; f(7) P(x 7) 8 00 Q.5(c) The price for a litre of whole milk is uniformly distributed between Rs. 45 and Rs. 55 during July in Mumbai. Give the equation and graph the pdf for X, the price per litre of whole milk during July. Also determine the percent of stores that charge more than Rs. 54 per litre. Ans.: The equation of pdf is f(x) 0. 45 < x < 55 55 45 0 0 elsewhere The pdf is shown in the figure below: 4

Prelim Question Paper Solution The probability P(X > 54) is shaded in the figure below: The area of the shaded region is 0.* 0. Hence 0.0* 00 0% of the stores charge more than Rs. 54/ per litre. Q.5(d) What is exponential distribution? Suppose the time till death after infection with Cancer, is exponentially distributed with mean equal to 8 years. If X represents the time till death after infection with Cancer, then find the percentage of people who die within five years after infection with Cancer. Ans.: Exponential distribution : The exponential probability distribution is a continuous probability distribution that is useful in describing the time it takes to complete some task. The pdf for an exponential probability distribution and e.788 five decimal places. f(x) x e for x 0 The graph for the pdf of a typical exponential distribution is shown : The percentage of people who die within five years after infection with cancer is 5 8 P(X 5) e e 0.65 0.535 0.465 i.e. 46.5% Q.5(e) The probability mass function of a random variable X is zero except at the points i 0,,. At these points it has the values p (0) 3c 3, p() 4c 0c, p() 5c for some c > 0. (i) Determine the value of c. (ii) Compute the following probabilities, P (X < ) and P ( < X < ). (iii) Describe the distribution function and draw its graph. (iv) Find the largest x such that F (x) < (v) Find the smallest x such that F (x) 3 Ans.: Given : P(0) 3a 3, P() 4a 0a and P() 5a and 0 for all the other values (i) We know that if p(x) is a PMF, then P(x) P(0) + P() + P() 3a 3 + 4a 0a + 5a 3a 3 0a + 9a 0 (a ) (3a 7a + ) 0 x 5

Vidyalankar : F.Y. B.Sc. (IT) NSM (a ) (3a ) (a ) 0 a,, 3 If a, then P(0) 3 >, which is not possible. Similarly, a 3 P(0) 3 3 9 3 P() 4a 0a 4 0 3 9 9 P() 5a 5 3 3 6 9 The probability distribution function is x 0 P(X x) 9 **Consider a as c 9 (ii) P(X < ) P(X 0) + P(X ) 3 9 9 9 3 P( < X ) P (X ) 6 9 3 (iii) The distribution function is F(x) 0, x < 0,0 x 9 3, x 9, x 6 9 (iv) Since F(x), the largest value of x for which f(x) < is x. 3 Since F(x) 3 for x and F(x) for x, the smallest value of x for which f(x) 3 is x. Q.5(f) The monthly worldwide average number of airplane crashes of commercial airlines is.. What is the probability that there will be (i) more than such accidents in the next month? (ii) more than 4 such accidents in the next months? Ans.: The number of crashes over a period of time is simply a random variable with a Poisson distribution. In this case, the sum of two passion random variables is just a new random variable with the new rate being the sum of the old rates. (a) We have X ~ Poisson (x :.); this is just P(X > ) P(X ). (.) e e.e! 0.377 (b) Let X be the number of accidents that happen in a two month period. By additively of the Poisson, X ~ Poisson (x :. +.). Thus P(X > 4) P(X 4) 0.4488694556849 6