Question

the shape of the distribution of the time required to get an oil change at a 20 - minute oil - change facility is skewed right. however, records indicate that the mean time is 21.3 minutes, and the standard deviation is 3.2 minutes. complete parts (a) through (c) below.
(a) to compute probabilities regarding the sample mean using the normal model, what size sample would be required?
choose the required sample size below
a. the normal model cannot be used if the shape of the distribution is skewed right.
b. the sample size needs to be greater than 30.
c. the sample size needs to be less than 30.
d. any sample size could be used.
(b) what is the probability that a random sample of n = 40 oil changes results in a sample mean time less than 20 minutes?
the probability is approximately 0.0051 (round to four decimal places as needed.)
(c) 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.
there would be a 10% chance of being at or below \\(\square\\) minutes. (round to one decimal place as needed.)

Explanation:

Step1: Recall the Z - score formula for sample means

The formula for the Z - score of a sample mean \(\bar{x}\) is \(Z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\), where \(\mu_{\bar{x}}=\mu\) (the population mean) and \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\) (the standard error of the mean). Here, \(\mu = 21.3\), \(\sigma=3.2\), and \(n = 40\). We want to find the value of \(\bar{x}\) such that \(P(\bar{X}\leq\bar{x})=0.10\).

First, find the Z - score corresponding to a cumulative probability of \(0.10\). From the standard normal table, the Z - score \(z\) such that \(P(Z\leq z)=0.10\) is approximately \(z=- 1.28\) (since the standard normal distribution is symmetric, and the Z - score for the 10th percentile is - 1.28).

Step2: Substitute into the Z - score formula and solve for \(\bar{x}\)

We know that \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Rearranging the formula to solve for \(\bar{x}\), we get \(\bar{x}=\mu+z\times\frac{\sigma}{\sqrt{n}}\).

Substitute \(\mu = 21.3\), \(z=-1.28\), \(\sigma = 3.2\), and \(n = 40\) into the formula:

First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{3.2}{\sqrt{40}}\approx\frac{3.2}{6.3246}\approx0.506\)

Then, \(\bar{x}=21.3+(-1.28)\times0.506\)

\(\bar{x}=21.3 - 1.28\times0.506\)

\(1.28\times0.506 = 0.64768\)

\(\bar{x}=21.3 - 0.64768=20.65232\approx20.7\) (rounded to one decimal place)

Answer:

\(20.7\)