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. click here to view the standard normal distribution table (page 1). click here to view the standard normal distribution table (page 2). (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 \\(\square\\). (round to four decimal places as needed.)

Explanation:

Part (a)

The Central Limit Theorem (CLT) states that for a skewed population distribution, the sampling distribution of the sample mean becomes approximately normal when the sample size \( n \) is sufficiently large (usually \( n \geq 30 \)). So, to use the normal model for the sample mean when the population is skewed right, the sample size needs to be greater than 30.

Step 1: Identify the parameters

The population mean \( \mu = 21.3 \) minutes, the population standard deviation \( \sigma = 3.2 \) minutes, and the sample size \( n = 40 \). We need to find \( P(\bar{x} < 20) \), where \( \bar{x} \) is the sample mean. By the Central Limit Theorem, the sampling distribution of the sample mean has mean \( \mu_{\bar{x}}=\mu = 21.3 \) and standard deviation (standard error) \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \).

First, calculate the standard error:

$$ \sigma_{\bar{x}}=\frac{3.2}{\sqrt{40}} \approx \frac{3.2}{6.3246} \approx 0.506 $$

Step 2: Calculate the z-score

The z-score for \( \bar{x} = 20 \) is given by:

$$ z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}=\frac{20 - 21.3}{0.506}\approx\frac{- 1.3}{0.506}\approx - 2.57 $$

Step 3: Find the probability from the z-table

We need to find \( P(Z < - 2.57) \), where \( Z \) is a standard normal variable. Looking up \( z=-2.57 \) in the standard normal table, we find that \( P(Z < - 2.57)=0.0051 \) (approximate value from the z-table: for \( z = - 2.57 \), the cumulative probability is 0.0051).

Answer:

B. The sample size needs to be greater than 30.

Part (b)