A normal population has a mean of 63 and a standard deviation of 14. You select a random sample of 9.Compute the probability the sample mean is: **(Round your *z* values to 2 decimal places and final answers to 4 decimal places)**:

Greater than 67.

Less than 62.

Between 62 and 67.

We will solve these probability questions using the **sampling distribution of the sample mean**. For a normally distributed population, the distribution of the sample mean will also be normally distributed, with a mean equal to the population mean and a standard deviation (standard error) given by:

\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]

where:
- \(\sigma_{\bar{x}}\) is the standard error of the sample mean,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.

Given:
- Population mean (\(\mu\)) = 63,
- Population standard deviation (\(\sigma\)) = 14,
- Sample size (\(n\)) = 9.

### 1. Standard error of the mean:
\[
\sigma_{\bar{x}} = \frac{14}{\sqrt{9}} = \frac{14}{3} = 4.67
\]

### Now, we compute the probabilities using the **z-score formula**:

\[
z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}
\]

where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, and \(\sigma_{\bar{x}}\) is the standard error.

### 1. Probability that the sample mean is greater than 67:

First, calculate the z-score for \(\bar{x} = 67\):
\[
z = \frac{67 - 63}{4.67} = \frac{4}{4.67} = 0.86
\]

Now, using the z-table (or normal distribution calculator), find the probability that \(z > 0.86\). The cumulative probability for \(z = 0.86\) is approximately **0.8051**.

So, the probability that the sample mean is greater than 67 is:
\[
P(\bar{x} > 67) = 1 - 0.8051 = 0.1949
\]

### 2. Probability that the sample mean is less than 62:

Now, calculate the z-score for \(\bar{x} = 62\):
\[
z = \frac{62 - 63}{4.67} = \frac{-1}{4.67} = -0.21
\]

Using the z-table, the cumulative probability for \(z = -0.21\) is approximately **0.4168**.

So, the probability that the sample mean is less than 62 is:
\[
P(\bar{x} < 62) = 0.4168
\]

### 3. Probability that the sample mean is between 62 and 67:

We already have the z-scores for 62 and 67 as:
- For \(\bar{x} = 62\), \(z = -0.21\),
- For \(\bar{x} = 67\), \(z = 0.86\).

The cumulative probabilities are:
- \(P(z < -0.21) = 0.4168\),
- \(P(z < 0.86) = 0.8051\).

Thus, the probability that the sample mean is between 62 and 67 is:
\[
P(62 < \bar{x} < 67) = P(z < 0.86) - P(z < -0.21) = 0.8051 - 0.4168 = 0.3883
\]

### Final Answers:
1. \(P(\bar{x} > 67) = 0.1949\)
2. \(P(\bar{x} < 62) = 0.4168\)
3. \(P(62 < \bar{x} < 67) = 0.3883\)

Would you like further clarification or details on any step?

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**Related Questions:**
1. What is the probability the sample mean is less than 60?
2. How would increasing the sample size affect the probability calculations?
3. How would changing the population standard deviation to 10 affect the results?
4. What is the probability the sample mean is greater than 70?
5. What z-score corresponds to the 95th percentile of the sample mean?

**Tip:** When dealing with sampling distributions, remember that larger sample sizes reduce the standard error, resulting in more precise estimates of the population mean.

A normal population has a mean of 63 and a standard deviation of 14. You select a random sample of 9. Compute the probability the sample mean is: Greater than 67, Less than 62, Between 62 and 67. (Round your z values to 2 decimal places and final answers to 4 decimal places).

Compute the probability of sample means from a normal population with mean 63 and standard deviation 14. Learn how to use z-scores and the standard error formula to find the probabilities for greater than 67, less than 62, and between 62 and 67. Suitable for high school students in Grades 10-12.

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