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 height is normally distributed, which of the following percentages of women in the country are between 63 and 70.5 inches tall?
a. 16%
b. 68%
c. 47%
d. 18%

Explanation:

Step1: Calculate z - scores

The formula for the z - score 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. Given $\mu = 65.5$ inches and $\sigma=2.5$ inches.
For $x = 63$ inches, $z_1=\frac{63 - 65.5}{2.5}=\frac{- 2.5}{2.5}=-1$.
For $x = 70.5$ inches, $z_2=\frac{70.5 - 65.5}{2.5}=\frac{5}{2.5}=2$.

Step2: Use the standard normal distribution table

The standard - normal distribution table gives the cumulative probability $P(Z < z)$.
$P(Z < - 1)=0.1587$ and $P(Z < 2)=0.9772$.
The probability $P(-1$P(-1

Answer:

None of the provided options (A. 10%, B. 68%, C. 48%, D. 16%) are correct. There may be an error in the question or options as the calculated probability of women with heights between 63 and 70.5 inches is approximately 82%.