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

$z= \frac{1440-1450}{8.5}$
$z=-1.77$

$z= \frac{1465-1450}{8.5}$
$z=1.77$

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