Question

A construction firm manufactures steel and is normally distributed with mean equal to 800MPa and a standard deviation of 40 MPa before it reaches yielding. Find the probability that the steel is in between 778 MPa and 834 MPa.

Answer 1

$P(778<X<834)=P[\frac{778-800}{40}<\frac{X-800}{40}<\frac{834-800}{40}]$

$=P(-0.55<Z<0.85)$

$=P(Z<0.85)-P(Z<-0.55)$

$=0.80234-0.29116$

$=0.51118$