Answer:
[tex]0.9480\ or\ 94.80\%[/tex]
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,
[tex]z=\dfrac{X-\mu}{\sigma}[/tex]
We have to calculate the probability that a light bulb will last between 1,965 and 2,165 hours.
i.e [tex]P(1965<X<2165)[/tex]
[tex]=P(1965-\mu<X-\mu<2165-\mu)[/tex]
[tex]=P\left(\dfrac{1965-\mu}{\sigma}<\dfrac{X-\mu}{\sigma}<\dfrac{2165-\mu}{\sigma}\right)[/tex]
[tex]=P\left(\dfrac{1965-2100}{40}<z<\dfrac{2165-2100}{40}\right)[/tex]
[tex]=P\left(-3.38<z<1.63\right)[/tex]
[tex]=P\left(z<1.63\right)-P(z<-3.38)[/tex]
[tex]=0.9484-0.0004[/tex]
[tex]=0.9480\ or\ 94.80\%[/tex]
