Answer: 0.002718
Step-by-step explanation:
Given : The population mean annual salary for environmental compliance specialists is about $62,000.
i.e. [tex]\mu=62000[/tex]
Sample size : n= 32
[tex]\sigma=6200[/tex]
Let x be the random variable that represents the annual salary for environmental compliance specialists.
Using formula [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex], the z-value corresponds to x= 59000 will be :
[tex]z=\dfrac{59000-62000}{\dfrac{6200}{\sqrt{32}}}\approx\dfrac{-3000}{\dfrac{6200}{5.6568}}=-2.73716129032\approx-2.78[/tex]
Now, by using the standard normal z-table , the probability that the mean salary of the sample is less than $59,000 :-
[tex]P(z<-2.78)=0.002718[/tex]
Hence, the probability that the mean salary of the sample is less than $59,000= 0.002718
