Question

The shape of the distribution of the time required to get an oil change at a...

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.)

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