Question

the shape of the distribution of the time required to get an oil change in a 15 minute oil change facility is on now. However, records indicate them me Tom is 16.8 minutes, and the center deviation is 4.4 minutes.

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

b.) what is the probability that a random sample of n equals 40 oil changes results in a sample mean time less than 15 minutes?

c.) suppose the manager agreed to pay each employee $50 bonus if they made a certain go on a typical Saturday the oil change facility will perform 40 oil changes between 10 a. m. and 12 p. m. During this is a random sample, what main oil change time will there be a 10% chance of being at or below? This will be the goal established by the manager

Answer 1

1) Even if the underlying distribution is non-normal, the Central Limit Theorem tells us that the distribution of sample means will be at least approximately normal if the sample sizes are more than 30. The number 30 came as a result of simple sampling simulations from different parent populations (Uniform, Normal, Exponential, Triangular) and by the time the sample sizes reached 30-32, the distribution of the means started looking normal. That is the reason for the rule-of-thumb. So the sample size needs to be greater than 30.

2) Sample size n = 40; Population mean min; standard deviation min

Sample mean

we need to find the  probability that the sample mean oil change time is less than 15 minutes

Look up the normal distribution for z < -2.012, we get the probability of 2.21% that a random sample change results in a sample mean time of less than 15 min.

c) We need to find the mean oil change time for which probability is 0.1 or we can write

First, we find the z score associated with the probability of 0.1,

Zc = -1.28

The cut-off time would then be

15.909 minutes mean oil-change time would there be a 10% chance of being at or below for 40 oil changes.