Question

The population mean annual salary for environmental compliance specialists is about ​$60 comma 500. A random sample of 32 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than ​$57 comma 500​? Assume sigmaequals​$6 comma 400.

Answer 1

0.004 is the probability that the mean salary of the sample is less than ​$57,500.

Explanation:

We are given the following information in the question:

Mean, μ = $60,500

Standard Deviation, σ = $6,400

Sample size, n = 32

Formula:

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

Standard error due to sampling =

`=(\sigma)/(√(n)) = (6400)/(√(32)) =1131.37`

P(mean salary of the sample is less than $57,500)

`P( x < 57500) = P( z < \displaystyle(57500 - 60500)/(1131.37)) = P(z <-2.6516)`

Calculation the value from standard normal z table, we have,

`P(x < 57500) = 0.004`

0.004 is the probability that the mean salary of the sample is less than ​$57,500.