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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

Respuesta :

Answer:

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

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