Question

in the united states, the mean average height of adult women is approximately 65.5 inches, with a standard deviation of 2.5 inches. if heights is normally distributed, which of the following percentages represents the number of women in this country who are between 63 and 70.5 inches tall?
a. 82%
b. 68%
c. 48%
d. 16%

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. For $x = 63$, $z_1=\frac{63 - 65.5}{2.5}=\frac{- 2.5}{2.5}=-1$. For $x = 70.5$, $z_2=\frac{70.5 - 65.5}{2.5}=\frac{5}{2.5}=2$.

Step2: Use the empirical rule of normal distribution

The empirical rule states that for a normal distribution: about 68% of the data lies within 1 standard - deviation of the mean ($z=\pm1$), about 95% lies within 2 standard - deviations of the mean ($z = \pm2$), and about 99.7% lies within 3 standard - deviations of the mean ($z=\pm3$). The area between $z=-1$ and $z = 2$ can be found by considering the cumulative areas. The area to the left of $z=-1$ is approximately 0.16 (from the standard normal table), and the area to the left of $z = 2$ is approximately 0.98. So the area between $z=-1$ and $z = 2$ is $0.98-0.16 = 0.82$ or 82%.

Answer:

A. 82%