Question

4.- The number of clients arriving per hour at a certain automotive workshop is supposed to follow a

poisson distribution with mean equal to 7.

A) Calculate the probability that more than 10 clients arrive in a two-hour period.

B) What is the average number of arrivals during a two-hour period?

C) Calculate the probability that no more than 4 clients will arrive between 4 and 5 in the afternoon.

Answer 1

Let X be the random variable that denotes the number of clients arriving per hour at a certain automotive workshop.

Given, X is supposed to follow a poisson distribution with mean equal to 7.

Therefore, X Poisson ( = 7)

Let Y be the random variable that denotes the number of clients that arrive in a 2-hour period.

Therefore, Y Poisson (m = 7 * 2 = 14)

A) Answer :

Y Poisson (m = 14)

The pmf of Y is

P(Y = y) = (e^-14 * 14^y) / y! ; y = 0, 1, 2, .........

              = 0                        ; otherwise

P(Y > 10) = 1 - P(Y 10)

                = 1 - (e^-14 * 14^y) / y!

                = 1 - 0.1757

                = 0.8243

Therefore, the probability that more than 10 clients arrive in a two-hour period is 0.8243

B) Answer :

Y Poisson (m = 14)

E(Y) = m = 14

Therefore, the average number of arrivals during a two-hour period is 14.

C) Answer :

X Poisson ( = 7)

The pmf of X is

P(X = x) = (e^-7 * 7^x) / x! ; x = 0, 1, 2, .........

              = 0                    ; otherwise

P(X 4) = (e^-7 * 7^x) / x!

               = 0.173

Therefore, the probability that no more than 4 clients will arrive between 4 and 5 in the afternoon is 0.173