(part 2)

b. Let k = the 90^th precentile.

Find k, where P($\overline{x}$ < k) = 0.90.

The 90^th percentile for the mean of 75 scores is about 3.2. This tells us that 90 percent of all the means of 75 stress scores are at most 3.2, and that 10 percent are at least 3.2.

invNorm$\left(\text{0}\text{.90,3,}\frac{1.15}{\sqrt{75}}\right)$ = 3.2