Question

At a local company, the ages of all new employees hired during the last 10 years are normally distributed. The mean age is 38 years old, with a standard deviation of 10 years. Find the percent of new employees that are no more than 30 years old. Round to the nearest percent.

Answer 1

P = 21%

Explanation:

We look for the percentage of employees who are not more than 30 years old.

This is:

`P = (x)/(n) *100\%`

Where x is the number of new employees who are not over 30 years old and n is the total number of new employees.

We do not know the value of x or n. However, the probability of randomly selecting an employee that is not more than 30 years old is equal to
`P = (x)/(n)`

Then we can solve this problem by looking for the probability that a new employee is not more than 30 years old.

This is:

`P(X< 30)`

Then we find the z-score

`Z = (X - \mu)/(\sigma)`

We know that:

μ = 38 years

`\sigma = 10` years

So

`Z = -0.8`

Then

`P (X<30) = P ((X- \mu)/(\sigma) < (30-38)/(10))\\\\P (X<30) = P(Z<-0.8)`

By symmetry of the distribution

`P(Z<-0.8)=P(Z>0.8)`

Looking in the normal standard tables

`P(Z>0.8)=0.211`

Finally P = 21%