The population mean annual salary for environmental compliance specialists is about $63,500. A random sample of 35 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than $60,000? Assume σ = $6100

0.003

Step-by-step explanation:

By formula we know that:

z (x) = (x - m) / [sd / sqrt (n)]

where x is the value we want to know (60,000), m is the mean (63500), sd is the standard deviation (6100) and n is the sample size (35).

Replacing we have:

z (60000) = (60000 - 63500) / [6100 / sqrt (35)]

z = -3.39

If we look in the normal distribution table (attached), we have that the probability is 0.0003.