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 11.4 minutes and the standard deviation is 4.5 minutes.

a) To compute the 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 = 35 oil changes results in a sample mean time of less than 10 minutes?
1) Choose the required sample size:
i) Sample size need to be greater than 30
ii) The normal model cannot be used if the shape of the distribution is unknown
iii) Any sample size could be used.
iv) Sample size needs to be less than 30
2) The probability is approximately ___________. (round to 4 decimal places as needed)

Answer 1

a) Sample size need to be greater than 30

b) The probability is approximately 0.0571

Explanation:

a) For a normal distribution, the sample size has to be greater than 30. A sample size greater than 30 makes it to be an approximate normal distribution.

b) Given that:

μ = 11.2 minutes, σ = 4.5 minutes, n = 35

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

`z=(x-\mu)/(\sigma) \\For\ a\ sample\ size(n)\\z=(x-\mu)/(\sigma/√(n) )`

For x < 10 minutes

`z=(x-\mu)/(\sigma/√(n) )\\\\ z=(10-11.2)/(4.5/√(35) )= -1.58`

Therefore from the normal distribution table, P(x < 10) = P(z < -1.58) = 0.0571

The probability is approximately 0.0571