Question

A study of a specific grade of steel suggests that yield strength (ksi) is normally distributed with

and mu = 42 and sigma = 4.3 a. (10 points) What is the probability that yield strength is at most 40? b. (5 points) What is the probability that yield strength is greater than 50?

Answer 1

a. $P(X\le40)=P(X<40)=P(Z<\frac{40-42}{4.3})=P(Z<-0.47)=0.3192.$

b. $P(X>50)=P(Z>\frac{50-42}{4.3})=P(Z>1.86)=1-P(Z<1.86)=0.0314.$