Question

Tri-Cities Bank has a single drive-in teller window. On Friday mornings, customers arrive at the drive-in window randomly, following a Poisson distribution at an average rate of 30 per hour.a. How many customers arrive per minute, on average?b. How many customers would you expect to arrive in a 10-minute interval?c. Use equation 13.1 to determine the probability of exactly 0, 1, 2, and 3 arrivals in a 10-minute interval. (You can verify your answers using the POISSON( ) function in Excel.)d. What is the probability of more than three arrivals occurring in a 10-minute interval?

Answer 1

Answer:

a) 0.5 per minutes

b) 5 arrivals expected in 10 minutes

c) P ( x = 0 ) = 0.00673 , P ( x = 1 ) =  0.03368 , P ( x = 2 ) = 0.08422 ,P ( x = 3 ) = 0.14037                  

d)  P ( X >= 4 ) = 0.735                

Step-by-step explanation:

Given:

- The number of customer arriving at window is modeled by Poisson distribution. The distribution is given by:

                         P(x) = ( λ^x ) (e^-λ) / x!            x = 0 , 1 , 2 , 3 , ......

- Average rate λ = 30 / hr

Find:

a. How many customers arrive per minute, on average?

b. How many customers would you expect to arrive in a 10-minute interval?c. Use equation 13.1 to determine the probability of exactly 0, 1, 2, and 3 arrivals in a 10-minute interval.

d. What is the probability of more than three arrivals occurring in a 10-minute interval?

Solution:

- The average rate λ in number of customers that arrive in a minute is given by:

                              λ1 = 30 / 60 = 0.5 arrival per minutes

- The average number of customer that are expected to arrive in 10-minutes window is:

                              λ2 = 10*λ1 = 10*0.5 = 5 arrivals expected in 10 minutes

- The probability of exactly 0,1 , 2 , and 3 arrivals in 10 minute windows:

                   P ( x = 0 ) = ( 5^0 ) (e^-5) / 0! = 0.00673  

                   P ( x = 1 ) = ( 5^1 ) (e^-5) / 1! = 0.03368

                   P ( x = 2 ) = ( 5^2 ) (e^-5) / 2! = 0.08422

                   P ( x = 3 ) = ( 5^3 ) (e^-5) / 3! = 0.14037

- The probability of more than three arrivals occuring in 10-minute interval is:

                  P ( X >= 4 ) = 1 - P ( X =< 3 )

                  P ( X >= 4 ) = 1 - [ P ( x = 0 ) + P ( x = 1 ) +  P ( x = 2 ) + P ( x = 3 ) ]

                  P ( X >= 4 ) = 1 - [ 0.00673 + 0.03368 + 0.08422 + 0.14037 ]

                  P ( X >= 4 ) = 1 - [ 0.265 ]

                  P ( X >= 4 ) = 0.735