First compute the probability that a bill would exceed $115. Let
be the random variable representing the value of a monthly utility bill. Then transforming to the standard normal distribution we have
![Z=(X-100)/(10)](https://img.qammunity.org/2020/formulas/mathematics/college/s3zf3gisl9ti20vpp2j7nlo0js2djfej2z.png)
![P(X>115)=P\left(Z>(115-100)/(10)\right)=P(Z>1.5)\approx0.0668](https://img.qammunity.org/2020/formulas/mathematics/college/c0dw54eg58ob5mvn5rcmlvysfw4ixk1hg5.png)
Then out of 300 randomly selected bills, one can expect about 6.68% of them to cost more than $115, or about 20.