Question

Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million in a the past year. Suppose a sample of 100 major league players was taken this year. What is the approximate probability that the mean salary of the 100 players was less than $3.0 million?

Answer 1

0.015 is the approximate probability that the mean salary of the 100 players was less than $3.0 million

Explanation:

We are given the following information in the question:

Mean, μ =$3.26 million

Standard Deviation, σ = $1.2 million 100

We assume that the distribution of salaries is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling =

`\displaystyle(\sigma)/(√(n)) = (1.2)/(√(100)) = 0.12`

P(mean salary of the 100 players was less than $3.0 million)

`P(x < 3) = P(z < \displaystyle(3-3.26)/(0.12)) = P(z < -2.167)`

Calculating the value from the standard normal table we have,

`P(Z < -2.167) = 0.015 \\P( x < 3) = 1.5\%`

0.015 is the approximate probability that the mean salary of the 100 players was less than $3.0 million