Solve the problem.
-In a certain country, weights of women are normally distributed with a mean of 138 lb and a standard deviation of 15 lb. What percentage of women in that country weigh more than 120 lb?
A)88.49%
B)53.98%
C)11.51%
D)1.20%

Question

Quizplus · Accepted Answer

To find the percentage of women weighing more than 120 lb, we calculate the z-score for 120 lb using the formula $z = \frac{X - \mu}{\sigma}$, where $X$ is 120 lb, $\mu$ is 138 lb (the mean), and $\sigma$ is 15 lb (the standard deviation). This gives us $z = \frac{120 - 138}{15} = -1.2$. Looking up -1.2 on the standard normal distribution table or using a calculator, we find that the area to the right of this z-score (which represents the percentage of women weighing more than 120 lb) is approximately 88.49%.

Solve the Problem. -In a Certain Country, Weights of Women Are Normally Distributed