Question

Fast Auto Service provides oil and lube service to cars.

The time required to service a car is normally distributed.

The mean amount of time to service a car is 30.0 minutes and a

The standard deviation in the amount of time to service a car is 5.5 minutes.

a. Determine the probability that a random car can be serviced in less than 20 minutes.

b. If it takes longer than 45 minutes to service a car, the driver receives a $10 refund. What percentage of cars would receive this refund?

Answer 1

Solution:

1) Ler X be a random variable which the time required to service the cars.

Given that, X~N(30.0, 5.5²)

i.e.  μ = 30.0 minutes and σ = 5.5 minutes

a) We have to obtain P(X < 20 minutes).

We know that, X ~N(μ, σ^{​​​​​​2}) then,

Using "pnorm" function of R we get, P(Z < -1.81818) = 0.0345

Hence, the probability that a random car can be serviced in less than 20 minutes is 0.0345.

b) We have to find P(X > 45 minutes).

We know that, X ~N(μ, σ^{​​​​​​2}) then,

Using "pnorm" function of R we get, P(Z > 2.72727) = 0.0032

0.0032 = 0.32%

Hence, 0.32% of cars would receive the refund.

Please rate the answer. Thank you.