IE 230 Probability & Statistics in Engineering I. Closed book and notes. 60 minutes.

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1 Closed book and notes. 60 minutes. A summary table of some univariate continuous distributions is provided. Four Pages. In this version of the Key, I try to be more complete than necessary to receive full credit for each answer. For example, full credit was given in Problem 4(b) for writing simply Y 1 = 0 implies that X 1 =/ 1. Therefore, for x 1 = 2, 3, 4, 5, 6, the pmf is f X 1 (x Y 1 =0 1 ) = (1 / 6) / (5 / 6) = 1 / 5. And zero elsewhere. BWSch Score Exam #3b, Fall 2000 Cover Sheet, Page 0 of 4 Schmeiser

2 Closed book and notes. 60 minutes. 1. True or false. (for each, 2 points if correct, 1 point if left blank.) (a) T For every normal distribution, the mean, the median, and the mode have the same numerical values. (b) T If σ X is positive, then 2 µ X X µ P( 2 X 6) = P X σ X σ X 6 µ X σ X regardless of the distribution of X. (c) T Of the lifetime distributions, only exponential distributions have the memoryless property. (d) T All exponential distributions have the same shape. (e) T If f Y is a probability mass function for the random variable Y, then f Y (c ) = P(Y = c ) for every real number c. (f) T Consider a random vector (X 1,...,X n ). There are then n marginal cumulative distributions F Xi and one joint cumulative distribution function F X 1,...,X n. (g) T If (X, Y ) is a continuous random vector, then always P(a X<b) = b a f XY (x, y ) dy dx. (h) F If X and Y are dependent random variables, then f X (x ) =/ f (x ) X for Y =y every real number x and every real number y. (i) F For every random variable X, F X (1) =. 2. Tables for the normal distribution invariably assume that the mean is zero and the variance is one. Why? <(a)> Any values for the mean and variance could have been selected. But zero mean and unit variance simplify computation. (b) Zero mean and unit variance are unique, the only values for which the table could be built, because the conversion to the standard normal distribution is closed form: Z =(X µ X ) / σ X. (c) Zero mean and unit variance are unique, the only values for which the table could be built, because these are the values of the standard normal distribution. (d) Any values for the mean and variance could have been selected. In fact, there are books containing tables of normal-distribution values, with one table for each possible value of the mean and of the variance. Exam #3b, Fall 2000 Page 1 of 4 Schmeiser

3 3. Let U 1 and U 2 be random numbers, such as come from MSExcel s function "rand". Assume that both are uniformly distributed on the (0, 1) interval and are independent of each other. Define a third random variable X = U 1 + U 2. Under the assumptions, the density function of X is a symmetric triangle, with lower bound 0 and upper bound 2. (a)are the events U 1 =.5 and U 2 <.5 independent? <Yes> No Not meaningful (b) Are the events U 1 =.5 and X<0.5 independent? Yes <No> Not meaningful (c) Are U 1 and X mutually exclusive? Yes No <Not meaningful> (d) What is the value of f X (1), the density function at the mode? The base of the triangle, (0, 2), has length 2. The triangle s height is therefore 1, because its area must be one (by the def. of pdf). Because the triangle is symmetric, the highest point is at the center, x = 1. That is, f X (1) = 1. (Sketching the pdf is not necessary, but might be helpful.) 4. Toss, independently, a six-sided die three times. Let X i denote the number of dots facing up on the i th toss. Let Y i denote the number of times that i dots face up. (a) Write the joint probability mass function f X 1, X 3. For x i = 1, 2,..., 6 for i = 1, 2, 3, f X 1, X 3 (x 1, x 2, x 3 ) = P(X 1 = x 1 = x 2, X 3 = x 3 ) definition of joint pmf = P(X 1 = x 1 )P(X 2 = x 2 )P(X 3 = x 3 ) tosses are independent = (1 / 6) (1 / 6) (1 / 6) six equally likely sides = 1 / 216 simplifying. The joint pdf is zero elsewhere. (b) Write the joint probability mass function f Y 1, Y 2, Y 3. (Y 1, Y 2, Y 3 ) is multinomial with n = 3 trials and p 1 = p 2 = p 3 = 1 / 6 and p 4 = 3 / 6. Here the fourth random variable is 3 Y 1 Y 2 Y 3, the number of 4, 5, and 6s. More specifically, for positive integers satisfying y 1 + y 2 + y 3 3, 3 f Y 1, Y 2, Y (y 3 1, y 2, y 3 ) = (1 / 6) y 1 (1 / 6) y 2 (1 / 6) y 3 (3 / 6) 3 y y y y 1, y 2, y 3,3 (y 1 y 2 y 3 ) = 3! (1 / 6) y y y y 1! y 2!y 3!(3 y 1 y 2 y 3 )! f Y 1, Y 2,...,Y 6 is zero elsewhere. (c) Write the conditional probability mass function f X 1. Y 1 =0 f X 1 (x Y 1 =0 1 ) = P(X 1 = x 1, Y 1 = 0) / P(Y 1 = 0). If x 1 =/ 1, 2,..., 6, then f X 1 (x Y 1 =0 1 ) = 0. If x 1 = 1,,2,..., 6, then f X 1 (x Y 1 =0 1 ) = P(X 1 = x 1 )P(Y 1 = 0 X 1 = x 1 ) / P(Y 1 = 0). For x 1 = 1, P(Y 1 = 0 X 1 = 1) = 0, which implies that f (1) = X 0. 1 Y 1 =0 Exam #3b, Fall 2000 Page 2 of 4 Schmeiser

4 Otherwise, for x 1 = 2, 3, 4, 5, 6, the pmf is f X 1 (x Y 1 =0 1 ) = (1 / 6) ((5 / 6)2 / (5 / 6) 3 = 1 / 5. Exam #3b, Fall 2000 Page 3 of 4 Schmeiser

5 IE 230 Probability & Statistics in Engineering I 5. Consider the tabled distribution x y f XY (x, y ) 1 2 1/ / / / 8 (a) Find P(X<2.5, Y<3). Because all four pairs (x, y ) satisfy the inequalities P(X<2.5, Y<3) = 1. (b) Find f X (1.5). Because X = 1.5 never occurs, f X (1.5) = P(X = 1.5) = 0. (c) Find F (3). X Y =1 F (3) = P(X X Y =1 3 Y = 1) P(X =.5, Y = 1) f = XY (.5, 1) = = 1/2 = 1. P(Y = 1) f Y (1) 1/2 (d) Find E(X ). E(X ) = Σ Σ xf XY (x, y ) allx ally = ( 1) * 1 / 8) + (.5) (1 / 4) + (.5) (1 / 2) + (1) (1 / 8) = (e) Find E(X Y<4). Because all possible values of Y are less than four, E(X Y<4) = E(X ) =.125 Exam #3b, Fall 2000 Page 4 of 4 Schmeiser

6 6. (Montgomery and Runger, 5-72). Suppose that the log-ons to a computer network follow a Poisson process with an average of three counts per minute. (a) What is the mean time between counts? Let X denote the time between successive log-ons. Then X is exponential with rate λ=3 log-ons per minute. Therefore,E(X ) = 1 / λ=1 / 3 minute. (b) What is the standard deviation of the time between counts? X being exponential implies that σ X = 1 / λ=1 / 3 minute.. (c) Determine the time x such that the probability that at least one count occurs before time x minutes is P(X x ) = e λx = = e λx x = ln(0.05) / λ x (d) Find the probability that each of the next three minutes has exactly one log-on. Let N i denote the number of log-ons in the i th minute. Then N i is Poisson with mean λ=3 log-ons, for i = 1, 2, 3. Therefore P(N i = 1) = e λ λ 1 / 1! = 3e Finally, P(N 1 = 1, N 2 = 1, N 3 = 1) = P(N 1 = 1) P(N 2 = 1) P(N 3 = 1) Exam #3b, Fall 2000 Page 5 of 4 Schmeiser

7 Continuous Distributions: Summary Table random distribution range cumul. probability expected variance variable name dist. func. density func. value X general (, ) P(X x ) df (y ) dy y =x xf (x )dx (x µ) 2 f (x )dx = F (x ) = f (x ) =µ=µ X =σ 2 2 =σ X = F X (x ) = f X (x ) = E(X ) = V(X ) = E(X 2 ) µ 2 X continuous [a, b ] uniform x a b a 1 b a a + b 2 (b a ) 2 12 sum of normal (, ) Table II random (or variables Gaussian) 1 x µ 2 e 2 σ µ σ 2 2πσ time to exponential [0, ) 1 e λx λ e λx 1 / λ 1 / λ 2 Poisson count time to r th Erlang [0, ) Poisson count Σ k =r e λx (λx ) k k! λ r x r 1 e λx r/λ r/λ 2 (r 1)! lifetime gamma [0, ) numerical λ r x r 1 e λx r/λ r/λ 2 Γ(r ) lifetime Weibull [0, ) 1 e βx β 1 e (x/δ)β (x/δ)β δγ(1+ 1 ) δ 2 Γ( ) µ δ β β β Definition. For any r>0, the gamma function is Γ(r ) = 0 x r 1 e x dx. Result. Γ(r ) = (r 1)Γ(r 1). In particular, if r is a positive integer, then Γ(r ) = (r 1)!. Definition. In a multinomial experiment, let X i denote the number of trials that result in outcome i for i = 1, 2,..., k. (Then X 1 + X X k = n.) The random vector (X 1,...,X k ) has a multinomial distribution with joint pmf n! x P(X 1 = x 1 = x 2,...,X k = x k ) = 1 x 2 x p x 1! x 2!... 1 p...pk k 2 x k! Exam #3b, Fall 2000 Schmeiser

8 when each x i is a nonnegative integer and x 1 + x x k = n ; zero elsewhere. Exam #3b, Fall 2000 Schmeiser