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Unformatted text preview: (b) [7 points] What is the probability that the the first three lightbulbs last at least 5 years (all together)? Solution: The number of failures in the first five years is a Poisson random variable with rate 5 × 1 / 2. So the probability that we have less than three is P (none) + P (one) + P (two) = e 5 / 2 1 + 5 / 2 + (5 / 2) 2 2 ≈ 54 . 38% . 4. John and James want to meet for lunch. John arrives at a time normally distributed with a mean of 12:02PM and standard deviation of 3 minutes. James arrives at a time normally distributed with a mean of 12:05PM and standard deviation of 4 minutes. Assume that the arrival times are independent. (a) [6 points] What is the probability that James arrives before John? Solution: Let X be John’s arrival time in minutes after 12:00 noon, and let Y be the same for James. We want to find P ( Y < X ) = P ( Y X < 0). Note that Y X has normal distribution and that E ( Y X ) = E ( Y ) E ( X ) = 5 2 = 3 , V ar ( Y X ) = V ar ( Y ) + ( 1) 2 V ar ( X ) = 3 2 + 4 2 = 25, and SD ( Y X ) = p V ar ( Y X ) = 5 . Therefore, P ( Y < X ) = P Y X 3 5 < 3 5 = Φ 3 5 = 1 Φ 3 5 ≈ 25 . 25% . (b) [6 points] What is the probability that the two men arrive within 3 minutes of each other? Page 3 Math 431: Exam 2 Solution: Using what we found above P (  Y X  ≤ 3) = P ( 3 ≤ Y X ≤ 3) = P 3 3 5 ≤ Y X 3 5 ≤ 3 3 5 = P 6 5 ≤ Y X 3 5 ≤ = Φ(0) Φ 6 5 ≈ . 5 . 115 = 38 . 5% . 5. Let X and Y be independent uniform (0 , 1) distributions. Let Z = X 2 + Y . (a) [6 points] Find E ( Z ). Solution: E ( Z ) = E ( X 2 + Y ) = Z 1 Z 1 x 2 + y dx dy = 5 6 . (b) [6 points] Find P ( Z ≤ 1). Solution: The Z ≤ 1 if and only if Y ≤ 1 X 2 , which describes a region of the unit square. Its area (divided by the area of the unit square) gives us the probability: P ( Z ≤ 1) = Z 1 1 x 2 dx = 2 3 . (c) [6 points] Find P ( X > 1 / 2  Z ≤ 1)....
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 Spring '12
 Miller
 Normal Distribution, Probability, Probability theory, Exponential distribution

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