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?

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

Step-by-step explanation:

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

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

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:

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

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

s + (a + 178) = 2000

step-by-step explanation:

2000 - 178 = 1822

1822 / 2 = 911

911 + 178 = 1089

1089 + 911 = 2000

abha = 1089

shane = 911