After entering the test scores from her Statistics class of 33 students, the instructor calculated the mean, median, IQR and standard deviation of the tes scores. Upon checking, she discovered that she had entered the top score as 85 , but it should have been 95 . a) When she corrects this score, how will the mean and median be affected? b) What effect will correcting the error have on the IQR and the standard deviation? a) Determine how the mean will be affected. Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The mean will increase by \( \square \) points. (Round to two decimal places as needed.) B. The mean will decrease by \( \square \) points. (Round to two decimal places as needed.) C. The mean will not change. D. It is impossible to determine how the mean will change.

To determine how the mean will be affected when the instructor corrects the top score from 85 to 95, we need to calculate the new mean after the correction. Given: - Original top score: 85 - Corrected top score: 95 - Number of students: 33 Let's denote the sum of the original scores as \( S \) and the sum of the corrected scores as \( S' \). The original mean is calculated as: \[ \text{Original Mean} = \frac{S}{33} \] After correcting the top score, the new sum of scores is: \[ S' = S - 85 + 95 \] The new mean is calculated as: \[ \text{New Mean} = \frac{S'}{33} \] Now, let's calculate the difference between the new mean and the original mean to determine how the mean will be affected. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(95-85\right)}{33}\) - step1: Remove the parentheses: \(\frac{95-85}{33}\) - step2: Subtract the numbers: \(\frac{10}{33}\) The difference between the new mean and the original mean is \( \frac{10}{33} \) or approximately 0.30 points. Therefore, the mean will increase by approximately 0.30 points when the instructor corrects the top score from 85 to 95.