Questions: The variance of a distribution of means is less than the original population variance because: - it is based on fewer individuals than of the original population - it is an estimate of the sample parameters rather than the original population - extreme scores are less likely to affect a distribution of means

Solution Steps

To determine why the variance of a distribution of means is less than the original population variance, we need to understand the concept of the Central Limit Theorem. The theorem states that the distribution of sample means will have a smaller variance than the original population because it averages out the extreme values. This is because the variance of the sampling distribution of the mean is equal to the population variance divided by the sample size.

The variance of the distribution of means, denoted as \( \sigma^2_{\bar{x}} \), is calculated using the formula:

\[ \sigma^2_{\bar{x}} = \frac{\sigma^2}{n} \]

where \( \sigma^2 \) is the population variance and \( n \) is the sample size.

Given the population variance \( \sigma^2 = 100 \) and the sample size \( n = 30 \), we substitute these values into the formula:

\[ \sigma^2_{\bar{x}} = \frac{100}{30} \]

Performing the division gives:

\[ \sigma^2_{\bar{x}} = 3.3333 \]

This indicates that the variance of the distribution of means is \( 3.3333 \).

Final Answer