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 3.8 minutes. Complete parts (a) through ©.
(a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required?
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 or equal to 30 . The sample size needs to be less than or equal to 30 .
D. Any sample size could be used.
(b) What is the probability that a random sample of oil changes results in a sample mean time less than 20 minutes?
The probability is approximately
(Round to four decimal places as needed.)

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 3.8 minutes. Complete parts (a) through ©.
(a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required?
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 or equal to 30 . The sample size needs to be less than or equal to 30 .
D. Any sample size could be used.
(b) What is the probability that a random sample of

oil changes results in a sample mean time less than 20 minutes?
The probability is approximately
(Round to four decimal places as needed.)

UpStudy AI · Accepted Answer

To answer your questions, let's break them down step by step.

### (a) Sample Size for Normal Model

When dealing with the Central Limit Theorem (CLT), it states that the distribution of the sample mean will be approximately normal if the sample size is sufficiently large, typically $$ n \geq 30 $$, regardless of the shape of the population distribution.

Given that the distribution of the time required to get an oil change is skewed right, we would still rely on the CLT for larger sample sizes.

**Answer:**
B. The sample size needs to be greater than or equal to 30.

### (b) Probability of Sample Mean Less Than 20 Minutes

To find the probability that a random sample of $$ n = 40 $$ oil changes results in a sample mean time less than 20 minutes, we can use the following steps:

1. **Calculate the standard error (SE) of the sample mean:**
   $$
   SE = \frac{\sigma}{\sqrt{n}} = \frac{3.8}{\sqrt{40}} \approx \frac{3.8}{6.32} \approx 0.6006
   $$

2. **Calculate the z-score for a sample mean of 20 minutes:**
   $$
   z = \frac{\bar{x} - \mu}{SE} = \frac{20 - 21.3}{0.6006} \approx \frac{-1.3}{0.6006} \approx -2.162
   $$

3. **Use the z-score to find the probability:**
   We can look up the z-score in a standard normal distribution table or use a calculator to find the probability corresponding to $$ z = -2.162 $$.

Using a standard normal distribution table or calculator:
$$
P(Z < -2.162) \approx 0.0154
$$

**Answer:**
The probability is approximately **0.0154** (rounded to four decimal places).
(a) B. The sample size needs to be greater than or equal to 30. (b) The probability is approximately 0.0154.

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Statistics Questions