A small population of quiz scores has values of: \( 68,74,76,80,88 \), and 92 . After converting each score into a \( Z \) score, what is the mean of the \( Z \) scores?

To find the mean of the \( Z \) scores for the given quiz scores, we first need to calculate the mean and standard deviation of the original scores. The quiz scores are: \( 68, 74, 76, 80, 88, 92 \). **Step 1: Calculate the mean (\( \mu \)) of the scores.** \[ \mu = \frac{68 + 74 + 76 + 80 + 88 + 92}{6} = \frac{478}{6} \approx 79.67 \] **Step 2: Calculate the standard deviation (\( \sigma \)) of the scores.** First, we calculate the variance. The variance is given by: \[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{N} \] where \( N \) is the number of scores and \( x_i \) are the individual scores. Calculating each \( (x_i - \mu)^2 \): - For \( 68 \): \( (68 - 79.67)^2 \approx 136.11 \) - For \( 74 \): \( (74 - 79.67)^2 \approx 32.11 \) - For \( 76 \): \( (76 - 79.67)^2 \approx 13.49 \) - For \( 80 \): \( (80 - 79.67)^2 \approx 0.11 \) - For \( 88 \): \( (88 - 79.67)^2 \approx 67.11 \) - For \( 92 \): \( (92 - 79.67)^2 \approx 148.11 \) Now, summing these squared differences: \[ \sum (x_i - \mu)^2 \approx 136.11 + 32.11 + 13.49 + 0.11 + 67.11 + 148.11 \approx 396.12 \] Now, we can calculate the variance: \[ \sigma^2 = \frac{396.12}{6} \approx 66.02 \] Now, take the square root to find the standard deviation: \[ \sigma \approx \sqrt{66.02} \approx 8.12 \] **Step 3: Calculate the \( Z \) scores.** The \( Z \) score for each value is calculated using the formula: \[ Z = \frac{x - \mu}{\sigma} \] **Step 4: Calculate the mean of the \( Z \) scores.** The mean of the \( Z \) scores is always 0, because the \( Z \) score transformation standardizes the data around the mean. Therefore, regardless of the original scores, the mean of the \( Z \) scores will be: \[ \text{Mean of } Z \text{ scores} = 0 \] Thus, the mean of the \( Z \) scores is: \[ \boxed{0} \]