Questions: Exhibit 7-5 Random samples of size 17 are taken from a population that has 200 elements, a mean of 36, and a standard deviation of 8. Refer to Exhibit 7-5. The mean and the standard deviation of the sampling distribution of the sample means are 8.7 and 1.94 36 and 1.86 36 and 8 36 and 1.94

Solution Steps

To solve this problem, we need to calculate the mean and standard deviation of the sampling distribution of the sample means. The mean of the sampling distribution is the same as the population mean. The standard deviation of the sampling distribution (also known as the standard error) can be calculated using the formula for the standard error of the mean for a finite population: \( \text{SE} = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}} \), where \( \sigma \) is the population standard deviation, \( n \) is the sample size, and \( N \) is the population size.

The mean of the sampling distribution of the sample means is equal to the population mean. Therefore, we have: \[ \mu_{\bar{x}} = \mu = 36 \]

The standard error (SE) of the sampling distribution is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}} \] Substituting the given values: \[ SE = \frac{8}{\sqrt{17}} \times \sqrt{\frac{200-17}{200-1}} \approx 1.8606 \]

Final Answer