IE 230 Seat # Closed book and notes. 60 minutes. Cover page and four pages of exam. No calculators. Score Exam #3a, Spring 2002 Schmeiser
Closed book and notes. 60 minutes. 1. True or false. (for each, 3 points if correct, 2 points if left blank.) (a) T F If X is normally distributed with mean µ X and standard deviation σ X, then P(X >µ X ) = P(X µ X ). (b) T F If X is normally distributed with mean µ X and standard deviation σ X, then 0.65 < P(µ X σ X <X <µ X +σ X ) <.75. (c) T F If Y is exponentially distributed with rate λ, then P(Y <0) = 0. (d) T F If approximating a binomial-distribution probability with a Poissondistribution probability, the continuity correction is useful. (e) T F If is a joint probability mass function, then (x, y ) = P(X x, Y y ) for all real numbers x and y. (f) T F If is a joint probability density function, then (x, y ) = P(X x, Y y ) for all real numbers x and y. (g) T F If X and Y are independent random variables, then P(X 6, Y 8) = f X (6) f Y (8), regardless of whether f X and f Y are pmfs or pdfs. (h) T F If (X 1, X 2,...,X 6 ) is a multinomial random vector, then X 1 + X 2 is a binomial random variable. (i) T F If X i is the time between counts i 1 and i in a Poisson process, the distribution of X 1 + X 2 + X 3 is Erlang. Exam #3a, Spring 2002 Page 1 of 4 Schmeiser
2. (Montgomery and Runger, 6 61) The yield in pounds from a day s production is normally distributed with a mean of 1500 pounds and a variance of 10000 pounds squared. Assume that the yields on different days are independent random variables. Let X i be the yield on day i for i = 1, 2,..., 5. (a) Sketch the pdf of X i. Label and scale both axes. Horizontal and vertical axis. Label with x and f X (x ). Sketch a bell curve, centered at µ X = 1500. The points of inflection should be one standard deviation from µ X. Here one standard deviation is the square root of the variance, σ X = 100. Scale the vertical axis at zero and somewhere else, probably the mode. The pdf integrates to one, so the value at the mode should be about 0.004. (The exact value is 1 / ( 2πσ) 0.004.) (b) What is the probability that the yield exceeds 1400 pounds on a randomly selected day? (Approximate as necessary.) P(X i > 1400) = P(X i > µ X σ X ) 0.84 using the approximation that about 68% of the normal distribution falls within one standard deviation of the mean. (Half of 68% is 34% to the left of the mean. The right half is 50%, for a total of 84%.) (c) Let p denote your answer to Part (b). What is the probability that production exceeds 1400 pounds all five days of a randomly selected week? (That is, determine P(X 1 > 1400, X 2 > 1400,..., X 5 > 1400).) P(X 1 > 1400, X 2 > 1400,..., X 5 > 1400) = 5 Π i =1 P(X i > 1400) = p 5 Exam #3a, Spring 2002 Page 2 of 4 Schmeiser
3. Result: E(XY ) = E(X ) E(Y ). The following proof assumes that X and Y are continuous. Choose exactly one reason for each step. (i) Axiom 1 (ii) Axiom 2 (iii) Axiom 3 (iv) Definition of expected value (v) Density functions integrate to one. (vi) Independence (vii) Total Probability (viii) Definition of pdf (ix) Calculus (No probability result needed) E(XY ) = xy f XY (x, y ) dx dy < (iv) > = xy f X (x ) f Y (y ) dx dy < (vi) > = xfx (x ) dx yfy (y ) dy < (ix) > = E(X )E(Y ) < (iv) > 4. The Weibull cdf is F (x ) = 1 exp[ (x /δ) β ]ifx>0 and zero elsewhere. Write P(2 X 5) in terms of F. The Weibull distribution is continuous, so P(2 X 5) = P(2 <X 5) = P(X 5) P(X 2) = F (5) F (2) Exam #3a, Spring 2002 Page 3 of 4 Schmeiser
5. Recall: The exponential cdf is F (x ) = 1 e λx if x>0 and zero elsewhere. The pdf is f (x ) =λe λx if x>0 and zero otherwise. The associated mean value is 1 / λ. (From Montgomery and Runger, Problem 5 114) The CPU of a personal computer has a lifetime that is exponentially distributed with a mean lifetime of six years. (a) What is the value of λ, including its units. λ=1 / 6 (failure per year) (b) Suppose that you have owned this CPU for three years. What is the probability that the CPU fails sometime during the next four years? P(X <7 X>3) = P(X<4) memoryless property = P(X 4) X is continuous = F (4) definition of cdf = 1 e (1 / 6)(4) X is exponential = 1 e 2 / 3 simplify 6. Consider the joint pdf (a, b ) = ab for 0 <a <c and 0 <b <c and zero elsewhere. Determine the value of the constant c. Every pdf must integrate to one, which gives us one equation with the unknown c : Let s rewrite the pdf using the more-common dummy variables x and y : (x, y ) = xy for 0 <x <c and 0 <y <c and zero elsewhere. f XY (x, y ) dx dy = 1 0 c 0 c xy dx dy = 1 c 4 / 4 = 1 c = 2 Exam #3a, Spring 2002 Page 4 of 4 Schmeiser