Question

the probability that the mean salary of the sample is less than $60,000 is? round 4 decimals if needed.

Answer 1

0.26599

Explanation:

We will assume a normal distribution for the salaries

We have the following information

Mean μ = 64000

Standard deviation σ = 6400

And we are asked to find P(X < 60000)

First find the z score corresponding to X = 60000

The z score is given by (X - μ) / σ

z score for 60000 with μ = 64000 and σ = 6400 is

z = (60000 - 64000)/6400
z = -40000/6400 = -0.625

We can either look up the P value from the z-tables or simply use a calculator

P(X < 60000) and this works out to 0.26599