Question

At a local company, the ages of all new employees hired during the last 10 years are normally distributed. The mean age is 31 years old, with a standard deviation of 10 years. If you were to take a sampling of 10 employees, what is the probability your mean age will be at least 28? Round to the nearest percent.

Answer 1

Answer:

P = 83%

Step-by-step explanation:

In this problem we have the ages of all new employees hired during the last 10 years of normally distributed.

We know that the mean is $\mu = 31$ years and standard deviation is $\sigma = 10$ years

By definition we know that if we take a sample of size n of a population with normal distribution, then the sample will also have a normal distribution with a mean

$\mu_m = \mu$

And with standard deviation

$\sigma_m = \frac{\sigma}{\sqrt{n}}$

Then the average of the sample will be

$\mu_m = 31\ years$

And the standard deviation of the sample will be

$\sigma_m =\frac{10}{\sqrt{10}} = 3.1622$

Now we look for the probability that the mean of the sample is greater than or equal to 28.

This is

$P ({\displaystyle{\overline {x}}}\geq 28)$

To find this probability we find the Z-score

$Z = \frac{X -\mu}{\frac{\sigma}{\sqrt{n}}}$

$Z = \frac{28 -31}{\frac{10}{\sqrt{10}}} = -0.95$

So

$P({\displaystyle{\overline {x}}}\geq 28) = P(\frac{{\displaystyle {\overline {x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}\geq\frac{28-31}{\frac{10}{\sqrt{10}}}) = P(Z\geq-0.95)$

We know that

$P(Z\geq-0.95)=1-P(Z<-0.95)$

Looking in the normal table we have:

$P(Z\geq-0.95)=1-0.1710\\\\P(Z\geq-0.95) = 0.829$

Finally P = 83%