Question

 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 71 years?

Answer 1

$P(\bar X<71)=P(Z<\frac{71-70}{\frac{5.1}{\sqrt{50}}})=P(Z<1.39)=0.9177.$