# If we assume that the annual salaries for clerks are normally distributed with a standard deviation of $2,875, find the following probabilities. What is the probability that the sample annual...

If we assume that the annual salaries for clerks are normally distributed with a standard deviation of $2,875, find the following probabilities. What is the probability that the sample annual salary will be greater than $39,000?

Accordig to a website about salaries, the national average salary as of October 2003 for a human resources clerk was $39,932.

A. The probability a randomly chosen human resources clerk will earn betweenÂ $38,000 and $41,000.

B. A sample of 100 human resources clerk is taken, and annual salaries are reported. What is the probability that the sample annual salary will be greater than $39,000.

### 1 Answer | Add Yours

Given `mu=39932,sigma=2875` (The population mean is 39932 and the population standard deviation is 2875):

(1) Find the probability that a random datum is between 38000 and 41000: find `P(38000<x<41000):`

First we convert to standardized normal scores using `z=(x-mu)/sigma`

Thus `z_(38000)=(38000-39932)/2875~~-.67` and `z_(41000)=(41000-39932)/2875~~.37`

Then `P(38000<x<41000)=P(-0.67<z<0.37)`

Using a standard normal table we find that `P(z<0.37)~~.6443` while `P(z<-0.67)~~.2514` . Then `P(-0.67<z<0.37)~~.6443-.2514~~.3929` **My calculator gives .3928798752**

------------------------------------------------------------------

**The probability that a randomly selected clerk earns between $38000 and $41000 is approximately .3929**

------------------------------------------------------------------

(2) Find the probability that the mean of a random sample of 100 clerks will be greater than $39000: `P(bar(x)>39000)` :

Again we convert to a standard normal score. However, this time, since we are dealing with a sample, we must use the standard error; use the formula `z=(bar(x)-mu)/(sigma/sqrt(n))` :

`z_(39000)=(39000-39932)/(2875/sqrt(100))~~-3.24`

So `P(bar(x)>39000)=P(z> -3.24)` and `P(z> -3.24)=1-P(z<=-3.24)`

Using a standard normal table we find `P(z <= -3.24)~~.0006` so `1-P(z <= -3.24)~~.9994` **My calculator gives .9994059245**

-------------------------------------------------------------------

**The probability that a random sample of 100 clerks has a mean salary greater than $39000 is approximately .9994**

------------------------------------------------------------------

**Sources:**