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 this distribution with a value that is greater than 21.2 and less than 23.4?

Answer 1

P(21.2 < X < 23.4)=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: 0.1869