Question

) One year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million. Suppose a sample of 400 major league players was taken. Find the approximate probability that the average salary of the 400 players exceeded $1.1 million.

Answer 1

Answer:

Probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

Step-by-step explanation:

We are given that one year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million.

Suppose a sample of 400 major league players was taken.

Let $\bar X$ = sample average salary

The z-score probability distribution for sample mean is given by;

                 Z = $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$  ~ N(0,1)

where, $\mu$ = mean salary = $1.5 million

            $\sigma$ = standard deviation = $0.9 million

             n = sample of players = 400

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, probability that the average salary of the 400 players exceeded $1.1 million is given by = P($\bar X$ > $1.1 million)

    P($\bar X$ > $1.1 million) = P( $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ >  $\frac{ 1.1-1.5}{{\frac{0.9}{\sqrt{400} } }} }$ ) = P(Z > -8.89) = P(Z < 8.89)

Now, in the z table the maximum value of which probability area is given is for critical value of x = 4.40 as 0.99999. So, we can assume that the above probability also has an area of 0.99999 or nearly close to 1.

Therefore, probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.