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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.

Respuesta :

luisejr77

Answer:

P = 21%

Step-by-step explanation:

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

This is:

[tex]P = \frac{x}{n} *100\%[/tex]

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 [tex]P = \frac{x}{n}[/tex]

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

This is:

[tex]P(X< 30)[/tex]

Then we find the z-score

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

We know that:

μ = 38 years

[tex]\sigma = 10[/tex] years

So

[tex]Z = -0.8[/tex]

Then

[tex]P (X<30) = P (\frac{X- \mu}{\sigma} < \frac{30-38}{10})\\\\P (X<30) = P(Z<-0.8)[/tex]

By symmetry of the distribution

[tex]P(Z<-0.8)=P(Z>0.8)[/tex]

Looking in the normal standard tables

[tex]P(Z>0.8)=0.211[/tex]

Finally P = 21%