Solution the number of failures in the first five

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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 < 0 - 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

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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 0 = Φ(0) - Φ - 6 5 0 . 5 - 0 . 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 0 Z 1 0 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 0 1 - x 2 dx = 2 3 . (c) [6 points] Find P ( X > 1 / 2 | Z 1). Solution: P ( X > 1 / 2 | Z 1) = P ( X > 1 / 2 and Z 1) P ( Z 1) .
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