Here are some summary statistics for all of the runners in a local 12 kilometre race: slowest $\text{ time }=129$ minutes, $\text{ mean }=79$ minutes, $\text{ median }=79$ minutes, $\text{ range }=99$ minutes, $\mathrm{IQR}=64\text{, }$ $\mathrm{Q}1=32$ standard $$\text{ deviation }=11$$ minutes.Suppose last year's race results were better by 8%.Find last year's mean and standard deviation.Express your answer in exact decimals.

A) Mean: 6.32 minutes,SD: 0.88 minutes
B) Mean: 85.32 minutes,SD: 11 minutes
C) Mean: 85.32 minutes,SD: 11.88 minutes
D) Mean: 72.68 minutes,SD: 10.12 minutes
E) Mean: 72.68 minutes,SD: 11 minutes

Since the slowest time this year was 11 minutes, we know that the slowest time last year would have been 8% better, or 10.168 minutes (11 - 0.88). We can use this to find the mean time for last year's race:

mean = (10.168 + 8.68 + 8.68 + 7.28 + 7.28 + 6.32)/6 = 7.69 minutes

Next, we need to find the standard deviation for last year's race. Since the race results were 8% better last year, we can use the formula for scaling standard deviation:

new SD = old SD / scaling factor

scaling factor = 1 - percent increase as a decimal = 1 - 0.08 = 0.92

So, the new SD is:

SD = 0.88 / 0.92 = 0.9565 minutes

Rounding to two decimal places, we get:

Mean: 7.69 minutes
SD: 0.96 minutes

Therefore, the best choice is (D) with a mean of 72.68 minutes and standard deviation of 10.12 minutes.