Question

the shape of the distribution of the time required to get an oil change at a 10 - minute oil - change facility is skewed right. however, records indicate that the mean time is 11.1 minutes, and the standard deviation is 3.3 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 = 45 ) oil changes results in a sample mean time less than 10 minutes?
the probability is approximately
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The standard error formula is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 3.3$ and $n = 45$, then $\sigma_{\bar{x}}=\frac{3.3}{\sqrt{45}}\approx0.491$

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x} = 10$, $\mu=11.1$. So $z=\frac{10 - 11.1}{0.491}\approx - 2.24$

Step3: Find the probability

Using the standard normal table or a calculator with a normal - distribution function (e.g., in Excel: NORM.S.DIST(-2.24,TRUE)), we find $P(Z\lt - 2.24)$

Answer:

The probability is approximately \(0.0125\)