The annual salary for one particular occupation is normally​ distributed, with a mean of about ​$126 comma 000 and a standard deviation of about ​$21 comma 000. Random samples of 37 are drawn from this​ population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of these sample means.​ Then, sketch a graph of the sampling distribution.
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Part 1
The mean is mu Subscript x overbarequals

enter your response here​, and the standard deviation is sigma Subscript x overbarequals

enter your response here.
​(Round to the nearest integer as needed. Do not include the​ $ symbol in your​ answers.)
Part 2

### Part 1: Mean and Standard Deviation of the Sampling Distribution

For the given scenario, you need to find the mean $\mu_{\bar{x}}$ and the standard deviation $\sigma_{\bar{x}}$ of the sampling distribution of the sample means.

#### **Mean of the Sampling Distribution ($\mu_{\bar{x}}$):**
The mean of the sampling distribution of the sample means is the same as the population mean. Therefore:
\[
\mu_{\bar{x}} = \mu = 126,000
\]

#### **Standard Deviation of the Sampling Distribution ($\sigma_{\bar{x}}$):**
The standard deviation of the sampling distribution (also called the **standard error**) is calculated using the formula:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
Where:
- $\sigma = 21,000$ (population standard deviation),
- $n = 37$ (sample size).

So,
\[
\sigma_{\bar{x}} = \frac{21,000}{\sqrt{37}} \approx \frac{21,000}{6.08} \approx 3,454
\]
(Rounded to the nearest integer.)

### Final answers:
- The mean $\mu_{\bar{x}} = 126,000$,
- The standard deviation $\sigma_{\bar{x}} \approx 3,454$.

### Part 2: Graph of the Sampling Distribution
The graph of the sampling distribution is a **normal distribution**, as the population distribution is normal. The mean of the distribution is 126,000, and the standard deviation is about 3,454. The bell curve would be centered at 126,000, with most of the values falling within the range of $\mu_{\bar{x}} \pm 3\sigma_{\bar{x}}$ (approximately from 115,638 to 136,362).

Would you like more details or have any questions?

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Here are 5 questions that could expand your understanding:
1. How does the sample size affect the standard deviation of the sampling distribution?
2. What is the probability of getting a sample mean greater than $130,000?
3. What is the significance of the Central Limit Theorem in this context?
4. How would the distribution change if the sample size were smaller, say 16?
5. Can you calculate a 95% confidence interval for the sample mean based on this distribution?

**Tip:** Larger sample sizes reduce the standard error, making the sampling distribution more concentrated around the population mean.

The annual salary for one particular occupation is normally distributed, with a mean of about $126,000 and a standard deviation of about $21,000. Random samples of 37 are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of these sample means. Then, sketch a graph of the sampling distribution. Part 1: The mean is mu Subscript x overbarequals, and the standard deviation is sigma Subscript x overbarequals. (Round to the nearest integer as needed. Do not include the $ symbol in your answers.) Part 2

Calculate the mean and standard deviation of the sampling distribution for a population with a mean salary of $126,000 and standard deviation of $21,000, with sample sizes of 37. Learn how to apply the Central Limit Theorem and standard error formula.

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