Question

Suppose that the distance of fly balls hit to the outfielder ( in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomlysample 49 fly balls. c. Find the 90thpercentile of the distribution of the average of 49 fly balls.

Answer 1

The values provided in the above question are as below

Mean = = 250

Standard deviation = = 50

sample size = = 49

We find the 90th percentile of the distribution of the average of 49 fly balls using following formula

-------------(1)

We find the z value using following Excel function

=NORMSINV(probability)

Here, 90th percentile = 0.90

That is , Probability = 0.90

=NORMSINV(0.90) then press Enter we get z as below

=1.281552

Using the above value of z in equation (1) we get

The 90th percentile of the distribution of the average of 49 fly balls is 259.15394

Summary :-

The 90th percentile of the distribution of the average of 49 fly balls is 259.15394

Note :-

(Dear student, In the above question it is not mention that the value of the 90th percentile of the distribution of the average of 49 fly balls is round to integer or round to which decimal places, so if you have to need the answer in the form of interger or specific decimal places then convert it.)