Question

The lifespan of a 60-watt lightbulb produced by a company is normally distributed with a mean of 1450 hours and a standard deviation of 8.5 hours. if a 60-watt light bulb produced by this company is selected at random, what is the probability that its lifespan will be between 1440 and 1465 hours

Answer 1

To evaluate the probability that the lifespan will be between 1440 and 1465 hours will be given by:
P(1440<x<1465)
using the z-score formula we obtain:
z=(x-μ)/σ
where:
μ=1450
σ=8.5
hence
when x=1440
z=(1440-1450)/8.5
z=-1.18
P(z<-1.18)=0.1190

when x=1465
z=(1465-1450)/8.5
z=1.77
P(z<1.77)=0.9625

hence:
P(1440<x<1465)
=0.9625-0.1180
=0.8445