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?

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:

Substituting values to get:

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

i can you answer this question sweetie!

step-by-step explanation:

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕