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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 34 years old, with a standard deviation of 10 years. Find the percent of new employees that are no more than 40 years old. Round to the nearest percent.

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luisejr77

Answer:

P = 73%

Step-by-step explanation:

We look for the percentage of employees who are not more than 40 years old.

This is:

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

Where x is the number of new employees who are not over 40 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 40 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 selected at random is not more than 40 years old.

This is:

[tex]P(X< 40)[/tex]

Then we find the z-score

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

We know that:

μ = 34 years

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

So

[tex]Z = 0.6[/tex]

Then

[tex]P (X<40) = P (\frac{X- \mu}{\sigma} < \frac{40-34}{10})\\\\P(X<40) = P(Z<0.6)[/tex]

So we have

[tex]P(Z<0.6)[/tex]

Looking in the normal standard tables:

[tex]P(Z<0.6)=0.726[/tex]

Finally P = 73%

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