Question

The number of messages sent to a computer website is a Poisson random variable with a mean of 5 messages per hour. a. What is the probability that 5 messages are received in 1 hours? b. What is the probability that fewer than 2 messages are received in 0.5 hour? c. Let Y be the random variable defined as the time between messages arriving to the computer bulletin board. What is the distribution of Y? What is the mean of Y? d. What is the probability that the time between messages exceeds 15 minutes?

Answer 1

a) Let X be the number of messages in I hour X follows Poisson distribution with param eter (rate) λ-5 XPoisson (A- 5) e_λλκ e-5(5)5 (0.006 738) × (3125) 120 P(X = 5) = -5! Use excel function POISSON.DIST( X, mean, cumulative) to calculate Px (5): POISSON.DIST( 5, 5, FALSE ) 0.175 467 369 768 0.175467 P(X=5)= 0.1755 b) Let X be the number of messages in 0.5 hour The rate of 5 messages in 1 hour is same as 5/1 0.5 2.5 in 0.5 hour X follows Poisson distribution with param eter (rate) λ = 2.5 XPoisson (A 2.5) e_λ X e-2.5(25)0 = 0.082085 _ e_λλ! e-25(25)-= 0.205 212 _ Answer 0.082 0850.205 212-0.287 297

Use excel function POISSON.DIST X , mean, cumulative) to calculate P(X < 1)-FX (1): POISSON.DIST( 1, 2.5. TRUE ) 0.287 297 495 184 P(X < 2) = 0.2873 c) 5 messages per hour is same as 1 message per 12 minutes. Y follows exponential distribution with B- 12 minutes Y~Exp B 12) Expected value: E(Y) = μ=β= 12 (given mean) mean: 12 d) Y~Exp B 12) Pdf is given by: f(y)--e-ws,0 < y Pdf: f() ev/12, 0 < y, and zero otherwise P(Y〈 y) = 1 _ e(-y/12) for y > 0 → P(Y > y) = e-y/12 P(Y > 15) = exp(_ 15/β) = e-125 0.286505 0.2865 (given mean) 12 P(Y > 15) = 0.2865