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

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
Transcript text: 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. \[ 0.0853 \]
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Solution

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Solution Steps

Step 1: Mean of the Distribution of Sample Mean Differences

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

μdifference=μnightμday=2.532.75=0.22 \mu_{\text{difference}} = \mu_{\text{night}} - \mu_{\text{day}} = 2.53 - 2.75 = -0.22

Step 2: Standard Deviation of the Distribution of Sample Mean Differences

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

σdifference=σnight2nnight+σday2nday=0.97250+0.432250.162 \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

Step 3: Probability that the Mean GPA of Night Students is Greater than Day Students

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

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

P(Z>0μdifferenceσdifference)=P(Z>0(0.22)0.162)=P(Z>1.3588) 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 Φ \Phi :

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

Final Answer

  • Mean of the distribution of sample mean differences: μdifference=0.22 \mu_{\text{difference}} = -0.22
  • Standard deviation of the distribution of sample mean differences: σdifference0.162 \sigma_{\text{difference}} \approx 0.162
  • Probability that the mean GPA of night students is greater than that of day students: P0.0871 P \approx 0.0871

Thus, the final answers are: μdifference=0.22 \boxed{\mu_{\text{difference}} = -0.22} σdifference0.162 \boxed{\sigma_{\text{difference}} \approx 0.162} P0.0871 \boxed{P \approx 0.0871}

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