Question

Consider the population with mean 50 and standard deviation 15. We draw a sample of size 100. According to the
theorem a sampling distribution x̄ is approximately normal with mean
and standard deviation
.

Answer 1

According to the Central Limit Theorem, the sampling distribution of the sample mean (\( \bar{x} \)) will be approximately normal if the sample size is sufficiently large (typically \( n \geq 30 \) is considered sufficient).

For a population with a mean (\( \mu \)) and standard deviation (\( \sigma \)), the mean and standard deviation of the sampling distribution (\( \bar{x} \)) can be described as follows:

1. **Mean of the sampling distribution** (\( \mu_{\bar{x}} \)):
\[
\mu_{\bar{x}} = \mu
\]
So, in this case, since the population mean is 50:
\[
\mu_{\bar{x}} = 50
\]

2. **Standard deviation of the sampling distribution** (also known as the standard error, \( \sigma_{\bar{x}} \)):
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
Here, the population standard deviation is 15, and the sample size \( n \) is 100. We can calculate it as follows:
\[
\sigma_{\bar{x}} = \frac{15}{\sqrt{100}} = \frac{15}{10} = 1.5
\]

Putting it all together, we have:

- The mean of the sampling distribution \( \mu_{\bar{x}} = 50 \)
- The standard deviation of the sampling distribution \( \sigma_{\bar{x}} = 1.5 \)

In conclusion:

- Mean of \( \bar{x} \): \( 50 \)
- Standard deviation of \( \bar{x} \): \( 1.5 \)