The population mean annual salary for environmental compliance specialists is about $63 comma 500. A random sample of 31 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than $60 comma 500? Assume sigmaequals$6 comma 200.4

0.0035289

Step-by-step explanation:

From the question;

mean annual salary = $63,500

n = sample size = 31

Standard deviation = $6,200

Firstly, we calculate the z-score of $60,500

Mathematically;

z-score = x-mean/SD/√n = (60500-63500)/6200/√(31) = -2.6941

So we want to find the probability that P(z < -2.6941)

We can get this from the standard normal table

P( z < -2.6941) = 0.0035289

care to share were the trapezoid shown is ?

$6.42

step-by-step explanation: