A large corporation employs 15710 individuals. The average income of all employees is $97284, with a standard deviation of $15868 and is skewed to the right. Consider this to be the population distribution.

You are given a data set consisting of the incomes of 70 randomly selected employees.

The population mean is

The population standard deviation is

The sample size is

Since the sample size is relatively large, the Central Limit Theorem tells us that the sample averages should have a sampling distribution that is (skewed to the rightapproximately normal).
The sampling distribution of the sample means is centered at the (populationsample) mean.
The sampling distribution has a standard deviation of

. Round to two

Let's break down the information provided and clarify the points step by step:

- **Population Mean ($ \mu $)**: This is the average income of all employees in the population.
  \[
  \mu = 97284
  \]

- **Population Standard Deviation ($ \sigma $)**: This is the standard deviation of the income distribution for the entire population.
  \[
  \sigma = 15868
  \]

- **Sample Size ($ n $)**: The number of randomly selected employees in the sample.
  \[
  n = 70
  \]

### Central Limit Theorem (CLT) Application:
Since the sample size is relatively large (greater than 30), the Central Limit Theorem applies. According to the CLT, the distribution of the sample means will be approximately **normal**, even if the population distribution is skewed (in this case, skewed to the right). So, we can conclude that the sampling distribution of the sample averages should be **approximately normal**.

- **The Sampling Distribution of the Sample Means** is centered at the **population mean** ($ \mu $), which is $ 97284 $.

### Standard Deviation of the Sampling Distribution (Standard Error):
The standard deviation of the sample means, also known as the **standard error of the mean**, is calculated by the following formula:
\[
\text{Standard Error} = \frac{\sigma}{\sqrt{n}}
\]
Substitute the values for $ \sigma $ and $ n $:

\[
\text{Standard Error} = \frac{15868}{\sqrt{70}} = \frac{15868}{8.3666} \approx 1896.17
\]

Thus, the standard deviation of the sampling distribution is approximately **1896.17**.

### Summary of Results:
1. **Population Mean** = $ 97284 $
2. **Population Standard Deviation** = $ 15868 $
3. **Sample Size** = $ 70 $
4. **The Sampling Distribution is approximately normal** (due to the Central Limit Theorem).
5. **The Sampling Distribution's Standard Deviation (Standard Error)** = **1896.17** (rounded to two decimal places).

Would you like more details or explanations on any part? Here are five related questions to extend this information:

1. How does increasing the sample size affect the standard error?
2. What would happen to the sampling distribution if the sample size were smaller than 30?
3. How does skewness affect the sampling distribution when the sample size is small?
4. What is the probability that a sample mean is within $2000 of the population mean?
5. How does the Central Limit Theorem help when dealing with non-normal population distributions?

**Tip**: The larger the sample size, the smaller the standard error, meaning the sample mean will more closely approximate the population mean.

A large corporation employs 15710 individuals. The average income of all employees is $97284, with a standard deviation of $15868 and is skewed to the right. Consider this to be the population distribution. You are given a data set consisting of the incomes of 70 randomly selected employees. The population mean is The population standard deviation is The sample size is Since the sample size is relatively large, the Central Limit Theorem tells us that the sample averages should have a sampling distribution that is (skewed to the rightapproximately normal). The sampling distribution of the sample means is centered at the (populationsample) mean. The sampling distribution has a standard deviation of Round to two

Explore the application of the Central Limit Theorem to a population of 15710 employees, where the average income is $97284 with a standard deviation of $15868. Learn how to calculate the standard error for a sample size of 70 and understand why the sampling distribution is approximately normal, despite a skewed population distribution.

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