In: Statistics and Probability
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 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