The free throw percentages for the participants of a basketball tournament were compiled.The percents ranged from 40% to 96% with a mean of 53% and a standard deviation of 8%.The quartiles and median percentages were 50%,84%,and 53%.This year's tournament percentages are 3% higher than last year's tournament percentages.Find last year's tournament mean,median,and standard deviation of the free throw percentages.
A)Mean: 56%,median: 56%,SD: 11%
B)Mean: 53%,median: 56%,SD: 11%
C)Mean: 51.5%,median: 53%,SD: 7.8%
D)Mean: 51.5%,median: 54.4%,SD: 7.8%
E)Mean: 50%,median: 50%,SD: 5%

Question

Quizplus · Accepted Answer

Since this year's percentages are 3% higher, we need to add 3% to the mean and the quartiles of last year's percentages to get this year's mean and quartiles. 
Therefore, the median of last year's percentages would still be 53%. 
Next, we can use the formula to find the new standard deviation of the percentages: 
New SD = Old SD * (1 + Percent Increase as a Decimal) 
New SD = 8 * (1 + 0.03) 
New SD = 8.24 
Therefore, last year's tournament mean, median, and standard deviation of the percentages are: 
Mean: 53% + 3% = 56% 
Median: 53% 
Standard Deviation: 7.8% (rounded to the nearest tenth) 
Hence, the best choice is D.

The Free Throw Percentages for the Participants of a Basketball