During the summer months,the prices of nonsmoking rooms with a king-sized bed in hotels in a certain area are roughly Normally distributed with a mean of $131.80 and a standard deviation of $29.12.A travel agent randomly selects prices of nonsmoking rooms with a king-sized bed from 15 hotels in the area.What is the probability that their average cost will be more than $150?

A) 0.0077
B) 0.1125
C) 0.2660
D) 0.3678

To find the probability that the average cost of the rooms is more than $150, we use the formula for the standard error of the mean: ${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}$ , where $\sigma$ is the standard deviation of the population and $n$ is the sample size. Here, $\sigma =29.12$ and $n=15$ , so ${\sigma }_{\bar{x}}=\frac{29.12}{\sqrt{15}}\approx 7.52$ .Next, we calculate the z-score to find how many standard deviations $150 is from the mean ($131.80): $z=\frac{X-\mu }{{\sigma }_{\bar{x}}}=\frac{150-131.80}{7.52}\approx 2.42$ .Looking up the z-score of 2.42 in the standard normal distribution table or using a calculator, we find that the area to the left of this z-score is approximately 0.9923. Therefore, the probability of the average cost being more than $150 is 1 - 0.9923 = 0.0077.