Question

The amount of time required for acertain type of car repair at service garage is normally distributed with mean of 45vminuite and srandard deviation of 80 minuite the service manage plan to have work work begin on acustomer car 10 minuite sfter the car is dropped off snd he tells the customer thst the each eill be ready eihin 1 hour total time .What is the required working time allotment such that there is 90% chance that the repair will be completed within that yime

Answer 1

You have a normal distribution with mu = 45 and sigma = 8.0. Let X be the amount of time it takes to complete the repair on a customer's car. To finish in one hour you must have X ≤ 50 so the first question is to find

Pr(X > 50).

Because of the way that the normal table is presented is is easiest to calculate this by

Pr(X > 50) = 1 - Pr(X ≤ 50).

You need to change this to a question about the standard normal variable Z using the transformation

Z = (X - mu)/sigma = (X - 45)/8.0

Thus the first question can be answered by using the normal table to find

Pr(X ≤ 50) = Pr(Z ≤ (50 - 45)/8.0) = Pr(Z ≤ 0.625)

For the second question you want to find the time T so that

Pr(X ≤ T) = 0.75