5/19/2018 Homework4-Statistics-slidepdf.com http://slidepdf.com/reader/full/homework-4-statistics 1/8 Homework 4 Kien Vu 1 1. (9 pts) The waiting time for service at a hospital emergency department (in hours) follows a distribution with probability density function f(x)=0.5exp(-0.5x) for 0<x. Determine the following: a. (3 pts) P(X<0.5). b. (3 pts) P(X>2). c. (3 pts) Value (in hours) exceeded with probability 0.05. Solution: a.P(X<0.5) = ∫ () = 1 - exp(-0.25) = 0.2212 b.P(X>2) = ∫ () = exp(-1) = 0.3679c.∫ () = 0.05, so exp(-0.5x) – 1 = -0.05, then exp(-0.5x) = 0.95 x = 0.1026 2. (9 pts) The distribution of X is approximated with a triangular probability density function f(x) = 0.0025x-0.075 for 30<x<50 and f(x) = -0.0025x+0.175 for 50<x<70. Determine the cumulative distribution function of X. Use the cumulative distribution function to determine the probability that the random variable is less than 55. Solution: The probability density function of X: f(x) = { The cumulative distribution function of X: F(x) = P(X<55) = ∫ () = ∫ () + ∫ () F(x) = (0.0025*50 2 /2-0.075*50)-(0.0025*30 2 /2-0.075*30)+(0.175*55-0.0025*55 2 /2)-( 0.175*50- 0.0025*50 2 /2) F(x) = 0.7188 3. (12 pts) A show is scheduled to start at 9:00 AM, 9:30 AM, and 10:00 AM. Once the show starts, the gate will be closed. A visitor will arrive at the gate a time uniformly distributed between 8:30 AM and 10:00 AM. Determine the following: a. (3 pts) Cumulative distribution function of the time (in minutes) between arrival and 10:00 AM. b. (3 pts) Mean and variance of the distribution in part (a). c. (3 pts) Probability that a visitor waits less than 10 minutes for a show. d. (3 pts) Probability that a visitor waits more than 20 minutes for a show. Solution: Let x denote the time that a visitor will arrive at the gate So x is a continuous uniform distribution a.The p.d.f: f(x) = 1/[(10-8.5)*60] = 1/90
1. (9 pts) The waiting time for service at a hospital emergency
department (in hours) follows a distribution with probability
density function f(x)=0.5exp(-0.5x) for 0
