Question

The shape of the distribution of the time required to get an oil change at a 20 minute oil change facility. However, records indicate the mean time is 21.3 minutes and the standard deviation is 3.9 minutes.
Suppose the manager agrees to pay each employee a​ $50 bonus if they meet a certain goal. On a typical​ Saturday, the​ oil-change facility will perform 40 oil changes between 10 A.M. and 12 P.M. Treating this as a random​ sample, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below? This will be the goal established by the manager.

Answer 1

Answer:

For a mean oil change time of 20.51 minutes there would be a​ 10% chance of being at or​ below

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 21.3, \sigma = 3.9, n = 40, s = \frac{3.9}{\sqrt{40}} = 0.6166$

Treating this as a random​ sample, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below?

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$-1.28 = \frac{X - 21.3}{0.6166}$

$X - 21.3 = -1.28*0.6166$

$X = 20.51$

For a mean oil change time of 20.51 minutes there would be a​ 10% chance of being at or​ below