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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{100}} = 0.12$

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

$P(x < 3) = P(z < \displaystyle\frac{3-3.26}{0.12}) = P(z < -2.167)$

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