Question

At a local company, the ages of all new employees hired during the last 10 years are normally distributed. The mean age is 35 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 37? Round to the nearest percent.

Answer 1

Answer:

P =26%

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 $\mu = 35$ 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

$\mu_m = 35\ years$

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 37.

This is

$P({\displaystyle{\overline {x}}}\geq 37)$

To find this probability we find the Z-score

$Z = \frac{{\displaystyle{\overline{x}}} -\mu}{\frac{\sigma}{\sqrt{n}}}$

$Z = \frac{37 -35}{\frac{10}{\sqrt{10}}} = 0.63$

So

$P({\displaystyle{\overline {x}}}\geq 37) = P(\frac{{\displaystyle{\overline {x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}\geq\frac{37-35}{\frac{10}{\sqrt{10}}}) = P(Z\geq0.63)$

We know that

$P(Z\geq0.63)=1-P(Z<0.63)$

Looking in the normal table we have:

$P(Z\geq0.63)=1-0.736\\\\P(Z\geq0.63) = 0.264$

Finally P = 26%