Question

An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean equal to 800 hours and a standard deviation of 40 hours. Find the probability that a bulb burns between 778 and 843 hours.

Answer 1

$P(778<X<843)=P(\frac{778-800}{40}<Z<\frac{843-800}{40})=P(-0.55<Z<1.08)=$

$=P(Z<1.08)-P(Z<-0.55)=0.8599-0.2912=0.5687.$