Question

Suppose x is distributed with a mean of 25.1 and a standard deviation of 3.2. What is the probability of randomly selecting an individual from the distribution with value that is greater than 21.2 and less than 23.4

Answer 1

$\mu=25.1$

$\sigma=3.2$

$P(21.2<X<23.4)=P(\frac{21.2-\mu}{\sigma}<Z<\frac{23.4-\mu}{\sigma})=$

$=P(\frac{21.2-25.1}{3.2}<Z<\frac{23.4-25.1}{3.2})=P(-1.22<Z<-0.53)=$

$=P(Z<-0.53)-P(Z<-1.22)=0.2981-0.1112=0.1869$

Answer: 18.69% of individuals are selected between 21.2 and 23.4.