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:

The probability your mean age will be at least 28 is 83%

Step-by-step explanation:

Let X denote the ages of all new employees hired during the last 10 years , then X is normally distributed with;

a mean of 31

a standard deviation of 10.

The sample size obtained is 10 employees. This implies that the sampling distribution of the sample mean will be normal with;

a mean of 31

a standard deviation of $\sqrt{10}$

The sample mean is a statistic and thus has its own distribution. Its mean is equal to the population mean, 31 and its standard deviation is equal to $\frac{sigma}{\sqrt{n} }$

where sigma is the population standard deviation, 10 and n the sample size which in this case is 10. $\frac{10}{\sqrt{10} }=\sqrt{10}$

We are required to find the probability that this sample mean age will be at least 28;

P(sample mean ≥ 28)

Since we know the distribution of the sample mean we simply standardize it to obtain the z-score associated with it;

P(sample mean ≥ 28)

= $P(Z\geq \frac{28-31}{\sqrt{10} } )=P(Z\geq -0.9487)$

=0.8286

As a percentage this is equivalent to, 83%