Question

In 2012, New York Yankees baseball players earned an average salary of $6,186,321, with a standard deviation of $7,938,987. Assuming that these data are normally distributed, what was the salary of a player in the 53rd percentile?

Answer 1

The salary of a player in the 53rd percentile was $6,781,745.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 6,186,321, \sigma = 7,938,987`

Assuming that these data are normally distributed, what was the salary of a player in the 53rd percentile?

This is the value of X when Z has a pvalue of 0.53. So X when Z = 0.075.

So

`Z = (X - \mu)/(\sigma)`

`0.075 = (X - 6,186,321)/(7,938,987)`

`X - 6,186,321 = 0.075*7,938,987`

`X = 6,781,745`

The salary of a player in the 53rd percentile was $6,781,745.