According to FedSmithcom the average salary for a federal em

According to FedSmith.com, the average salary for a federal employee in 2012 was $74,804. Assume the population standard deviation is $15,300. A random sample of 34 federal employees is selected. A. what is the probability that the sample mean will be less than $70,000. B. What is the probability that the sample mean will be more than $74,000. C. What is the probability that the sample mean will be between $76,000 and $81,000.

Solution

u = mean = 74804

SD = 15300

n = 34

A. what is the probability that the sample mean will be less than $70,000.

Answer :

z = (x - u) / (SD/sqrt(n))

z = (70,000 - 74804) / (15300/sqrt(34))

z = -1.8308426733880277

Use the following link for the z-table to P-value calculator :
https://www.easycalculation.com/statistics/p-value-for-z-score.php

P(z < -1.8308426733880277) = 0.0336 --> ANSWER

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B. What is the probability that the sample mean will be more than $74,000

z = (74000 - 74804) / (15300/sqrt(34))

z = -0.3064108054546158

P(z > -0.3064108054546158) = 0.6204 --> SECOND ANSWER

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C. What is the probability that the sample mean will be between $76,000 and $81,000.

z = (76000 - 74804) / (15300/sqrt(34)) = 0.4558051285120902

P(z < 0.4558051285120902) = 0.6757

z = (81000 - 74804) / (15300/sqrt(34)) = 2.361344963428855

P(z < 2.361344963428855) = 0.9909

So, in between, we have :

0.9909 - 0.6757

0.3152 ---> THIRD ANSWER