Question

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 5.5 minutes. (a) Find the probability that a customer has to wait more than nine minutes. (Round your answer to three decimal places.)

(b) Find the probability that a customer is served within the first five minutes. (Round your answer to three decimal places.)

(c) The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she doesn't want to give away free hamburgers to more than 2% of her customers. What should the advertisement say? (Give your answer to the nearest integer that satisfies the conditions.) "If you aren't served within ???? minutes, you get a free hamburger."

Answer 1

Answer to the question)

Since the waiting time follows poisson distribution

given mean time = 5.5

Lambda = 5.5

.

Part a)

P(x > 9) = 1 - P(x < =9)

P(x < = 9) = P(x=0) + P(x=1) + P(x=2) + ..... + P(x=9)

The formula to be used for manual working is :

P(X=x) = e^-(λ) * (λ)^x / x!

we got λ = 5.5

using technology , we can use the excel function as follows to find P(x < = 9)

=POISSON(9,5.5,1)

We get P(x < =9) = 0.9462

P(x>9) = 1 - 0.9462

P(x>9) =0.0538

.

Part b)
P(x<= 5) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5)

Using excel function =poisson(5,5.5,1)

P(x<=5) = 0.5289

.

Part c)

P(X > x) = 2% ~0.02

P(X < =x) = 1 - P(x > x)

P(X < =x) = 1 - 0.02 = 0.98

we know that for more than 9 minutes the probability is 0.05

Thus x has to be larger than 9

by hit and trial method we can plug in x = 10

we get P( x> 10) = 0.025 ~2.5%

[use excel function =1-POISSON(10,5.5,1)]

Next plug in x = 11

we get P(x > 11) = 0.011 ~1.1%

[use excel function =1-POISSON(11,5.5,1)]

Thus we conclude that the time must be 10 minutes

it must state that if you aren't served within 10 minutes you get a free hamburger