Susan has been on a bowling team for 14 years. After examining all of her scores over that period of time, she finds that they follow a normal distribution. Her average score is 225, with a standard deviation of 13. If during a typical week Susan bowls 16 games, what is the probability that her average score is more than 230?
A) 0.0620
B) 0.3520
C) 0.6480
D) 0.9382

Question

Quizplus · Accepted Answer

To find the probability that Susan's average score is more than 230, we use the formula for the z-score: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean (230), $\mu$ is the population mean (225), $\sigma$ is the standard deviation (13), and $n$ is the sample size (16). Plugging in the values, we get $z = \frac{230 - 225}{\frac{13}{\sqrt{16}}} = \frac{5}{\frac{13}{4}} = \frac{20}{13} \approx 1.54$. Looking up 1.54 on the standard normal distribution table or using a calculator, we find the area to the left of 1.54 is approximately 0.9382. Since we want the probability of being more than 230, we subtract this from 1 to get $1 - 0.9382 = 0.0620$.

Susan Has Been on a Bowling Team for 14 Years