# Hw9Sol - SIEO 3600(IEOR Majors Introduction to Probability...

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SIEO 3600 (IEOR Majors) Solution to Assignment #9 Introduction to Probability and Statistics April 11, 2010 Solution to Assignment #9 ( Uniform, Exponential and Normal random variables ) 1. If X is a normal random variable with parameters μ = 10, σ 2 = 26, compute (a) P ( X > 5); (b) P (4 < X < 16); (c) P ( X < 8); (d) P ( X < 20); (e) P ( X > 16). Hint: express these probabilities in terms of the standard normal CDF function Φ and use the tables at the back of the textbook. Solution: Let Z be a standard normal random variable, i.e., Z N (0 , 1). (a) P ( X > 5) = P ± X - μ σ > 5 - μ σ ² = P ± Z > 5 - 10 26 ² = 1 - Φ ± 5 - 10 26 ² = 0 . 8366. (b) P (4 < X < 16) = P ± 4 - μ σ < X - μ σ < 16 - μ σ ² = P ± 4 - μ σ < Z < 16 - μ σ ² = Φ ± 16 - 10 26 ² - Φ ± 4 - 10 26 ² = 0 . 7607. (c) P ( X < 8) = P ± X - μ σ < 8 - μ σ ² = P ± Z < 8 - 10 26 ² = Φ ± 8 - 10 26 ² = 0 . 3474. (d) P ( X < 20) = P ± X - μ σ < 20 - μ σ ² = P ± Z < 20 - 10 26 ² = Φ ± 20 - 10 26 ² = 0 . 9751. (e) P ( X > 16) = P ± X - μ σ > 16 - μ σ ² = P ± Z > 16 - 10 26 ² = 1 - Φ ± 16 - 10 26 ² = 0 . 1197. 2. The annual rainfall (in inches) in a certain region is normally distributed with μ = 40, σ = 4. What is the probability that in 2 of the next 4 years the rainfall will exceed 50 inches? Assume that the rainfalls in diﬀerent years are independent. Solution: Let Z be a standard normal, i.e., Z N (0 , 1). We have that the probability of rainfall exceeding 50 in one year is P ( X > 50) = P ³ X - 40 4 > 50 - 40 4 ´ = P ( Z > 2 . 5) = 1 - Φ(2 . 5) = 1 - 0 . 9938 = 0 . 0062 p. Let N be the number of years in which the rainfall exceed 50 inches in n = 4 years. Then it is clear that N is a binomial random variable with parameter n = 4 and p = 0 . 0062. Hence, the desired probability is P ( N = 2) = ³ 4 2 ´ p 2 (1 - p ) 2 . 3. A random variable X is said to have a lognormal distribution if Z log X is normally distributed. If X is lognormal with E [ Z ] = μ and var( Z ) = σ 2 , then (a) determine the probability distribution function (PDF) of X . (b) What is P ( X x )?

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2 SIEO 3600, Solution to Assignment #9 Solution: Since Z = log X , we must have X = e Z , where Z N ( μ,σ 2 ). We ﬁrst solve (b). (b) For x > 0, we compute the CDF of X as F X ( x ) = P ( X x ) = P ( e Z x ) = P ( Z log x ) = Z log x -∞ 1 2 πσ 2 e - ( z - μ ) 2 2 σ 2 dz. (a) To compute the PDF of X , we diﬀerentiate its CDF obtained in (b), i.e., by the chain rule of calculus, we have f X ( x ) = d dx F X ( x ) = 1 2 πσ 2 e - (log x - μ ) 2 2 σ 2 · (log x ) 0 = 1 2 πσ 2 e - (log x - μ ) 2 2 σ 2 · 1 x . 4. In the midterm of a probability course, the scores that students get follow a uniform distri- bution from 20 to 100. Due to popular grieving, the professor decides to rescale the results. The formula he uses is that the curved score
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Hw9Sol - SIEO 3600(IEOR Majors Introduction to Probability...

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