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Name: ID: STT 441 Quiz 4 Apr. 1, 2015 1. The time X (in hours) required to repair a machine is an exponentially distributed random variable with parameter λ = 1/3. The pdf of an exponential distribution is f (x) = 1 exp(−x/λ) λ for x > 0. What is (a) the probability that a repair time exceeds 2 hours? [5pts] Let X denote the time (in hours) required to repair a machine. Then X ∼ Exponential( 13 ). P(X > 2) = Z ∞ 3e −3x 2 dx = −e ∞ −6 = e = 0.0025. −3x 2 (b) the conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours? [5pts] By the memoryless property of exponential random variables, we have P (X ≥ 10|X > 9) = P (X ≥ 9 + 1|X > 9) = P (X ≥ 1) Z ∞ ∞ 3e−3x dx = −e−3x = 0.0497. = 1 1 2. Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters µ = 71 and σ 2 = 6.25. Using the attached standard normal table to compute the following. (a) What percentage of 25-year-old men are over 6 feet and 2 inches (=74 inches) tall ? [5pts] Let X be the height of a a 25-year-old man. Then X − 71 74 − 71 √ > √ 6.25 6.25 74 − 71 = 1 − Φ( √ ) = 1 − Φ(1.2) = 0.1151. 6.25 P (X > 74) = P where Φ(·) is the CDF function of the standard normal distribution. (b) What percentage of men in the 6-footer club are over 6 feet and 5 inches (=77 inches)? (Here you need to compute the conditional probability) [5pts] Let X be the height of a a 25-year-old man. Then P (X > 77|X > 72) = P ({X > 77} ∩ {X > 72})/P (X > 72) = P (X > 77)/P (X > 72). Because, X − 71 77 − 71 √ > √ 6.25 6.25 77 − 71 = 1 − Φ( √ ) = 1 − Φ(2.4) = 0.0082 6.25 P (X > 77) = P and X − 71 72 − 71 √ > √ 6.25 6.25 72 − 71 ) = 1 − Φ(0.4) = 0.3446 = 1 − Φ( √ 6.25 P (X > 72) = P where Φ(·) is the CDF function of the standard normal distribution, we have P (X > 77|X > 72) = P (X > 77)/P (X > 72) = 0.0082/0.3446 = 0.0238. 2 3. If X is an exponential random variable with parameter λ = 1 (The expectation of X is 1). (a) Find the CDF of Y = 3 log X. [5pts] The distribution function of Y is FY (y) = P (Y ≤ y) = P (3 log X ≤ y) = P (X ≤ ey/3 ) Z ey/3 ey/3 e−x dx = −e−x = 1 − exp(−ey/3 ), = 0 0 for all −∞ < y < ∞. (b) Find the probability density function for Y defined in part (a). The pdf of Y is fY (y) = d 1 FY (y) = exp(−ey/3 )ey/3 dy 3 3 − ∞ < y < ∞. [5pts]