Question

The monthly utility bills in a city are normally distributed, with a mean of $100 and a standard deviation of $10. If 300 utility bills are randomly selected from this city, approximately how many of them will be more than $115?

Answer 1

First compute the probability that a bill would exceed $115. Let $X$ be the random variable representing the value of a monthly utility bill. Then transforming to the standard normal distribution we have

$Z=\dfrac{X-100}{10}$

$P(X>115)=P\left(Z>\dfrac{115-100}{10}\right)=P(Z>1.5)\approx0.0668$

Then out of 300 randomly selected bills, one can expect about 6.68% of them to cost more than $115, or about 20.