Question

​​​​​​​Please post the work so I can understand the process :) Thanks!

1. Telephone calls arrive at the Global Airline reservation office in Louisville according to a Poisson distribution with a mean of 1.6 calls per minute:

a. What is the probability of receiving exactly one call during a one-minute interval?

b. What is the probability of receiving at most 2 calls during a one-minute interval?

c. What is the probability of receiving at least two calls during a one-minute interval?

d. What is the probability of receiving exactly 4 calls during a five-minute interval?

Answer 1

solution:

average calls per minute = = 6

using poisson probability process we know

so,

a) probability of receiving one call during one minute interval = P(X=1)

b) probability that at most two calls during one minute interval =

= P(X=0) + P(X=1) + P(X=2)

= 0.00248 + 0.01487 + 0.04462 = 0.06197

c) probability of receiving atleast two call = 1 - P(X<2)

P(X<2) = P(X=0) + P(X=1) = 0.00248 + 0.01487 = 0.01735

probability of receiving atleast two call = 1 - 0.01735 = 0.98265

d) probability of receiving 4 calls during 5 minute interval

so, average calls in 5 minute = =5*6 =30

P(4;30) =

so, probability of receiving 4 calls during 5 minute interval is approximately 0