Question

In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, and the standard deviation was 5.1 years.

If a sample of 50 people from this region is selected, what is the probability that the mean life expectancy will be more than 90 years?

Answer 1

Let the random variable X denotes the age at death.

Given that X has a distribution with $\mu = 70$ years and $\sigma=5.1$ years

Population size

N = 400

Sample size

n = 50

The sample drawn is 12.5% of the population from which it is being selected. It is relatively large sample.

So, we should use the correction factor.

The z value corresponding to 90 is

$z = \frac{X- \mu}{ \frac{\sigma}{\sqrt{n}} \times \sqrt{ \frac{N-n}{N-1} }} \\ = \frac{90- 70}{ \frac{5.1}{\sqrt{50}} \times \sqrt{ \frac{400-50}{400-1} }} \\ = 29.63 \\ P(X>90) = 1 -P(Z<29.63) \\ = 1 -0.999968 \\ = 0.000032$