Question

The shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is skewed right. However, records indicate that the mean time is 11.2 minutes, and the standard deviation is 4.2 minutes.

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

A. The sample size needs to be less than or equal to 30.
B. The sample size needs to be greater than or equal to 30.
C. The normal model cannot be used if the shape of the distribution is skewed right.
D. Any sample size could be used.

(b) What is the probability that a random sample of n = 45 oil changes results in a sample mean time less than 10 minutes? The probability is approximately 17. (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 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.

Answer 1

a) The sample size needs to be greater than or equal to 30. b) The probability is approximately 0.170. c) The mean oil-change time is approximately 8.7635 minutes.

Step-by-step explanation:

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

b) To find the probability that a random sample of n = 45 oil changes results in a sample mean time less than 10 minutes, we can use the z-score formula and the standard normal distribution. The z-score for a sample mean of 10 minutes with a population mean of 11.2 minutes and a standard deviation of 4.2 minutes is calculated as z = (10 - 11.2) / (4.2 / sqrt(45)). Using a standard normal table or a calculator, we find that the probability is approximately 0.1706, which rounds to 0.170.

c) To find the value of the mean oil-change time that corresponds to a 10% probability, we can use the z-score formula and the standard normal distribution. The z-score for a probability of 0.10 is -1.2816. Solving for x in the equation -1.2816 = (x - 11.2) / (4.2 / sqrt(45)), we find that the mean oil-change time is approximately 8.7635 minutes.