Question

The shape of the distribution of the time required to get an oil change at a 10 minute oil change facility is unknown. However, records indicate that the mean time is 11.2 minutes and the standard deviation is 4.8 minutes
(a) to compute probabilities regarding the sample mean using the normal model, what size sample would be required?
(b) What is the probability that a random sample of n=45 oil changes results in a sample mean time less than 10 minutes?

Answer 1

Answer:

The answer is below

Step-by-step explanation:

a) For a normal model the sample size has to be equal or greater than 30 so that it can be a normal distribution.

b) Given that:

μ = 11.2 minutes, σ = 4.8 minutes, n = 45

The z score determines how many standard deviations the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma} \\For\ a\ sample\ size(n)\\z=\frac{x-\mu}{\sigma/\sqrt{n} }$

For x < 10 minutes

$z=\frac{x-\mu}{\sigma/\sqrt{n} }\\\\ z=\frac{10-11.2}{4.8/\sqrt{45} }= -1.68$

Therefore from the normal distribution table, P(x < 10) = P(z < -1.68) =  0.0465