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 for an oil change is 11.4 minutes, and the standard deviation for oil- change time is 3.2 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 of less than 10 minutes?

Answer 1

Answer:

a) sample size ≥ 30

b) 0.0017

Step-by-step explanation:

a) The central limit theorem states that for population with mean μ and standard deviation σ and take sufficiently large random samples from the population. This will hold provided the sample size is sufficiently large (usually n > 30) for a normal population.

Therefore the sample size should be greater or equal to 30 i.e n ≥ 30.

b) Given that n = 45, μ = 11.4 minutes and σ = 3.2 minutes, the z score (z) is given by the equation:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }$

Substituting values to get:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }= \frac{10-11.4}{\frac{3.2}{\sqrt{45} } }=-2.93$

Using the z table:

P(x < 10) = P(z < -2.93) = 0.0017

the probability that a random sample of n=45 oil changes results in a sample mean time of less than 10 minutes is 0.0017