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

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

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