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 21.1 minutes​, and the standard deviation is 4.9 minutes. Complete parts ​(a) through (c).

​(a) To compute probabilities regarding the sample mean using the normal​ model, what size sample would be​ required?

A. The normal model cannot be used if the shape of the distribution is unknown.

B. The sample size needs to be greater than or equal to 30.

C. Any sample size could be used.

D. The sample size needs to be less than or equal to 30.

​(b) What is the probability that a random sample of n=40 oil changes results in a sample mean time less than 20 minutes?

The probability is approximately ______

(Round to four decimal places as​ needed.)

​(c) 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 40 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 _____ minutes.

Answer 1

Given data :

mean time is 21.1 minutes​, and the standard deviation is 4.9 minutes.

​(a) To compute probabilities regarding the sample mean using the normal​ model, what size sample would be​ required?

Ans :B. The sample size needs to be greater than or equal to 30.

​(b) What is the probability that a random sample of n=40 oil changes results in a sample mean time less than 20 minutes?

​(c) 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 40 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

n=40

InvNorm(0.10)= - 1.28

Ans: