Problem 126E

Refer to Exercise 3.122. Assume that arrivals occur according to a Poisson process with an average of seven per hour. What is the probability that exactly two customers arrive in the two-hour period of time between

a 2:00 P.M. and 4:00 P.M. (one continuous two-hour period)?

b 1:00 P.M. and 2:00 P.M. or between 3:00 P.M. and 4:00 P.M. (two separate one-hour periods that total two hours)?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. During a given hour, what are the probabilities that

a no more than three customers arrive?

b at least two customers arrive?

c exactly five customers arrive?

Solution:

Step 1 of 2:

For a departmental store arrival of the customers is 7 per hour.

- The claim is to find the probability that exactly two customers arrive in the two-hour period of time, 2:00 P.M and 4:00 P.M. ( one continuous two-hour period)

The mean of the distribution is = 7 per hour and 14 per two hours

Let X follows the Poisson distribution with probability density function

P(X) =

P( x = 2) =

= 0.0000815

Hence, the probability that exactly two customers arrive in the two-hour period of time, 2:00 P.M and 4:00 P.M. ( one continuous two-hour period) is 0.0000815

Step 1 of 2:

b)

The claim is to find the probability that exactly two customers arrive in the two-hour period of time, 1:00 P.M and 2:00 P.M or between 3:00 P.M and 4:00 P.M ( two separate one-hour periods that total two hours.

P( x = 2) =

= 0.0000815

Hence, The claim is to find the probability that exactly two customers arrive in the two-hour period of time, 1:00 P.M and 2:00 P.M or between 3:00 P.M and 4:00 P.M ( two separate one-hour periods that total two hours is 0.0000815