Final answer:
The 91st percentile of a normal distribution with a mean of 30 and a standard deviation of 5 is approximately 35.90.
Step-by-step explanation:
To find the 91st percentile of a normal distribution with a mean of 30 and a standard deviation of 5, you can use the inverse normal function (invNorm) on a calculator. The invNorm function takes the desired percentile (in decimal form) and the mean and standard deviation of the distribution as inputs.
For this problem, you would use invNorm(0.91, 30, 5) to find the value that corresponds to the 91st percentile.
The result of invNorm(0.91, 30, 5) is approximately 35.90. Therefore, the 91st percentile of this distribution is approximately 35.90.