Questions: The mean GPA of night students is 2.53 with a standard deviation of 0.97. The mean GPA of day students is 2.75 with a standard deviation of 0.43. You sample 50 night students and 25 day students. What is the mean of the distribution of sample mean differences (night GPA - day GPA)? -0.22 What is the standard deviation of the distribution of sample mean differences (night GPA - day GPA)? 0.162 Find the probability that the mean GPA of the sample of night students is greater than the mean GPA of the sample of day students. 0.0853

Solution Steps

The mean GPA of night students is \( \mu_{\text{night}} = 2.53 \) and the mean GPA of day students is \( \mu_{\text{day}} = 2.75 \). The mean of the distribution of sample mean differences (night GPA - day GPA) is calculated as:

\[ \mu_{\text{difference}} = \mu_{\text{night}} - \mu_{\text{day}} = 2.53 - 2.75 = -0.22 \]

The standard deviations of the night and day students' GPAs are \( \sigma_{\text{night}} = 0.97 \) and \( \sigma_{\text{day}} = 0.43 \), respectively. The standard deviation of the distribution of sample mean differences is given by:

\[ \sigma_{\text{difference}} = \sqrt{\frac{\sigma_{\text{night}}^2}{n_{\text{night}}} + \frac{\sigma_{\text{day}}^2}{n_{\text{day}}}} = \sqrt{\frac{0.97^2}{50} + \frac{0.43^2}{25}} \approx 0.162 \]

To find the probability that the mean GPA of the sample of night students is greater than that of the sample of day students, we need to calculate:

\[ P(\text{night GPA} - \text{day GPA} > 0) \]

This can be expressed in terms of the standard normal distribution as:

\[ P(Z > \frac{0 - \mu_{\text{difference}}}{\sigma_{\text{difference}}}) = P(Z > \frac{0 - (-0.22)}{0.162}) = P(Z > 1.3588) \]

Using the cumulative distribution function \( \Phi \):

\[ P(Z > 1.3588) = 1 - \Phi(1.3588) \approx 0.0871 \]

Final Answer