Question

The life expectancy of a typical lightbulb is normally distributed with a mean of 2,100 hours and a standard deviation of 40 hours. What is the probability that a lightbulb will last between 1,965 and 2,165 hours?

Answer 1

the answer would be: 0.9480

Hope this helps

Answer 2

`0.9480\ or\ 94.80\%`

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=(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<X<2165)`

`=P(1965-\mu<X-\mu<2165-\mu)`

`=P\left((1965-\mu)/(\sigma)<(X-\mu)/(\sigma)<(2165-\mu)/(\sigma)\right)`

`=P\left((1965-2100)/(40)<z<(2165-2100)/(40)\right)`

`=P\left(-3.38<z<1.63\right)`

`=P\left(z<1.63\right)-P(z<-3.38)`

`=0.9484-0.0004`

`=0.9480\ or\ 94.80\%`