The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds.
Approximately what percent of the participants lose more than 20.5 pounds?
A) 2.5 percent
B) 16 percent
C) 34 percent
D) 50 percent
E) 84 percent

Question

Quizplus · Accepted Answer

To find the percentage of participants who lose more than 20.5 pounds, we calculate the z-score for 20.5 pounds using the formula $z = \frac{X - \mu}{\sigma}$, where $X$ is the value of interest (20.5 pounds), $\mu$ is the mean (14.5 pounds), and $\sigma$ is the standard deviation (3 pounds). The z-score is $z = \frac{20.5 - 14.5}{3} = 2$. Looking up a z-score of 2 in a standard normal distribution table or using a calculator, we find that approximately 97.5% of the area under the curve is to the left of this z-score, meaning that 2.5% is to the right. Therefore, approximately 2.5% of the participants lose more than 20.5 pounds.

The Weight Losses for Participants in a Large Exercise and Diet