Question

The shape of the distribution of the time required to get an oil change at a 20​-minute ​oil-change facility is unknown.​ However, records indicate that the mean time is 21.4 minutes​, and the standard deviation is 4.2 minutes.

Suppose the manager agrees to pay each employee a​ $50 bonus if they meet a certain goal. On a typical​ Saturday, the​ oil-change facility will perform 45 oil changes between 10 A.M. and 12 P.M. Treating this as a random​ sample, there would be a​ 10% chance of the mean​ oil-change time being at or below what​ value. This will be the goal established by the manager.

There is a​ 10% chance of being at or below a mean​ oil-change time of (answer here) minutes? ​(Round to one decimal place as​ needed.)

Answer 1

For the sample of size 45, we can apply Central Limit Theorem

So, the sampling distribution of mean will be normally distributed

Mean time = 14x60 /45 = 18.67 minutes

standard deviation of mean time(standard error) =

=

= 0.626

Let the time below which there is 10% chance of oil-change time be A

P(X < A) = P(Z < (A - mean)/standard error)

P(X < A) = 0.10

P(Z < (A - 18.67)/0.626) = 0.10

From standard normal distribution table, take Z score corresponding to 0.1

(A - 18.67)/0.626 = -1.28

A = 17.9 minutes

There is 10% chance of being at or below a mean oil change time of 17.9 minutes