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:
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Mean of the sampling distribution (\( \mu_{\bar{x}} \)): \[ \mu_{\bar{x}} = \mu \] So, in this case, since the population mean is 50: \[ \mu_{\bar{x}} = 50 \]
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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 \( \bar{x} \): \( 1.5 \)