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:0.015 is the approximate probability that the mean salary of the 100 players was less than $3.0 millionStep-by-step explanation:We are given the following information in the question: Mean, μ =$3.26 millionStandard Deviation, σ = $1.2 million 100 We assume that the distribution of salaries is a bell shaped distribution that is a normal distribution. Formula: [tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]Standard error due to sampling = [tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{100}} = 0.12[/tex]P(mean salary of the 100 players was less than $3.0 million) [tex]P(x < 3) = P(z < \displaystyle\frac{3-3.26}{0.12}) = P(z < -2.167)[/tex] Calculating the value from the standard normal table we have, [tex]P(Z < -2.167) = 0.015 \\P( x < 3) = 1.5\%[/tex]0.015 is the approximate probability that the mean salary of the 100 players was less than $3.0 million