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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?

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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=(X-100)/(10)


P(X>115)=P\left(Z>(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.

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