Question

In human resource management, performance of employees is measured as a numerical score which is assumed to be normally distributed. The mean score is 150 and the standard deviation 13. What is the probability that a randomly selected employee will have a score less than 120?

Answer 1

`P(X<120)=P((X-\mu)/(\sigma)<(120-\mu)/(\sigma))=P(Z<(120-150)/(13))=P(z<-2.308)`

And we can find this probability using the normal standard table or excel and we got:

`P(z<-2.308)=0.0105`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

`X \sim N(150,13)`

Where
`\mu=150` and
`\sigma=13`

We are interested on this probability

`P(X<120)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<120)=P((X-\mu)/(\sigma)<(120-\mu)/(\sigma))=P(Z<(120-150)/(13))=P(z<-2.308)`

And we can find this probability using the normal standard table or excel and we got:

`P(z<-2.308)=0.0105`