Question

In a study of the life expectancy of 400 people in a certain geographic region, the mean age of 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 less than 68 years?

Answer 1

Solution:

$z=\dfrac{X-\mu}{\frac{\sigma}{\sqrt{n}}}=\dfrac{68-70}{\frac{5.1}{\sqrt{50}}}=-2.77296...\approx-2.77$

Now, $P(X<68)=P(z<-2.77)=0.0028$