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 21.1 minutes, and the standard deviation is 3.9 minutes. complete parts (a) through (c)
(a) to compute probabilities regarding the sample mean using the normal model, what size sample would be required?
○ a. the sample size needs to be less than or equal to 30
○ b. any sample size could be used
○ c. the normal model cannot be used if the shape of the distribution is skewed right
○ d. the sample size needs to be greater than or equal to 30
(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
(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, there would be a ( 10 % ) chance of the mean oil - change time being at or below what value? this will be the goal established by the manager
there is a ( 10 % ) chance of being at or below a mean oil - change time of minutes.
(round to one decimal place as needed)

Explanation:

Step1: Central Limit Theorem

The Central Limit Theorem (CLT) states that if the population is not normally distributed, the sampling distribution of the sample mean \(\bar{X}\) will be approximately normal if the sample size \(n\geq30\). So for part (a), when the population is skewed (not normal), we need \(n\geq30\) to use the normal model for the sample mean.

Step2: Calculate \(z -\)score for part (b)

The formula for the \(z -\)score of the sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Given \(\mu = 21.1\), \(\sigma=3.9\), \(n = 40\), and \(\bar{x}=20\).

$$ z=\frac{20 - 21.1}{\frac{3.9}{\sqrt{40}}}=\frac{- 1.1}{\frac{3.9}{6.3246}}\approx\frac{-1.1}{0.6166}\approx - 1.78 $$

Using the standard normal table \(P(Z\lt - 1.78)=0.0281\)

Step3: Find the value for part (c)

We know that if \(P(\bar{X}\leq x)=0.10\), then from the standard normal table, the \(z -\)score corresponding to a left - tail probability of \(0.10\) is \(z=-1.28\)
Using the formula \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), and solving for \(\bar{x}\):

$$ -1.28=\frac{\bar{x}-21.1}{\frac{3.9}{\sqrt{40}}}, \quad\frac{3.9}{\sqrt{40}}\times(-1.28)=\bar{x}-21.1 $$

$$ 0.6166\times(-1.28)=\bar{x}-21.1,\quad - 0.79=\bar{x}-21.1 $$

$$ \bar{x}=21.1-0.79 = 19.1 $$

Answer:

A. D. The sample size needs to be greater than or equal to 30
B. 0.0281
C. 19.1