Here are some summary statistics for last year's basketball team scoring output: lowest $\text { score } = 18$ points, $\text { mean } = 58$ points, $\text { median } = 52$ points, $\text { range } = 97$ points, $\mathrm { IQR } = 46$ $\mathrm { Q } 1 = 23 \text {, }$ standard $\text { deviation } = 9$ points.Suppose the opponents' scoring output was 10% lower.Find the opponents' mean and standard deviation.
A)Mean: 52.2 points,SD: 8.1 points
B)Mean: 52.2 points,SD: 9 points
C)Mean: 63.8 points,SD: 9.9 points
D)Mean: 63.8 points,SD: 9 points
E)Mean: 5.8 points,SD: 0.9 points

Question

Quizplus · Accepted Answer

The opponents' mean score is 10% lower than the team's mean score, so $58 - 0.10 \times 58 = 52.2$ points. The standard deviation, being a measure of dispersion that does not depend on the mean, remains unchanged when all values are scaled by the same factor, so it remains $9$ points. However, because we're applying a percentage decrease to the mean, not the standard deviation, the standard deviation for the opponents' scores actually decreases by the same percentage, making it $9 - 0.10 \times 9 = 8.1$ points.

Here Are Some Summary Statistics for Last Year's Basketball Team score =18\text { score } = 18 score =18

Here Are Some Summary Statistics for Last Year's Basketball Team $\text { score } = 18$