Answer:
0.004 is the probability that the mean salary of the sample is less than $57,500.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $60,500
Standard Deviation, σ = $6,400
Sample size, n = 32
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling =
[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{6400}{\sqrt{32}} =1131.37[/tex]
P(mean salary of the sample is less than $57,500)
[tex]P( x < 57500) = P( z < \displaystyle\frac{57500 - 60500}{1131.37}) = P(z <-2.6516)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 57500) = 0.004[/tex]
0.004 is the probability that the mean salary of the sample is less than $57,500.
