Question

Information reference tables 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 lightbulb produced by this company is selected at random, what is the probability that its lifespan will be between 1440 and 1465 hours

Answer 1

First, we need to standardize the value 1440 and 1465 to be able to work out the z-score


`z= (1440-1450)/(8.5)`

`z=-1.77`


`z= (1465-1450)/(8.5)`

`z=1.77`

The z-scores are shown in the diagram below

To work out the probability of
`-1.77\ \textless \ z\ \textless \ 177` we can first read on the table, the probability when
`P(z\ \textless \ 1.77)=0.9616`, then we subtract this value from 1, as we are interested in the area to the right of 1.77.


`1-0.9616=0.0384`

Then the area between
`z=-1.77` and
`z=1.77` is
`1-2(0.0384)=0.9232` which is also the probability of lifespan between 1440 and 1465