First compute the probability that a bill would exceed $115. Let [tex]X[/tex] be the random variable representing the value of a monthly utility bill. Then transforming to the standard normal distribution we have
[tex]Z=\dfrac{X-100}{10}[/tex]
[tex]P(X>115)=P\left(Z>\dfrac{115-100}{10}\right)=P(Z>1.5)\approx0.0668[/tex]
Then out of 300 randomly selected bills, one can expect about 6.68% of them to cost more than $115, or about 20.