Answer:
\(0.9480\ or\ 94.80\%\)
Step-by-step explanation:
The life expectancy of a typical light bulb is normally distributed with a mean of 2,100 hours and a standard deviation of 40 hours.
So, here,
μ = mean = 2100,
σ = standard deviation = 40,
We know that,
\(z=\dfrac{X-\mu}{\sigma}\)
We have to calculate the probability that a light bulb will last between 1,965 and 2,165 hours.
i.e \(P(1965
\(=P(1965-\mu
\(=P\left(\dfrac{1965-\mu}{\sigma}<\dfrac{X-\mu}{\sigma}<\dfrac{2165-\mu}{\sigma}\right)\)
\(=P\left(\dfrac{1965-2100}{40}
\(=P\left(-3.38
\(=P\left(z<1.63\right)-P(z<-3.38)\)
\(=0.9484-0.0004\)
\(=0.9480\ or\ 94.80\%\)
