Question

question 7 of 10
the heights of u.s. adult men can be modeled by a normal distribution with mean μ = 69 inches and standard deviation σ = 2.5 inches.
use the empirical rule to answer the following questions.
(a) what percentage of u.s. adult men are greater than 74 inches tall? % (do not round.)
(b) what proportion of u.s. adult men have heights between 64.0 and 71.5 inches? (enter your answer in decimal form.) (do not round.)

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$. For part (a), $x = 74$, $\mu=69$, $\sigma = 2.5$. So $z=\frac{74 - 69}{2.5}=\frac{5}{2.5}=2$. For part (b), for $x_1 = 64$, $z_1=\frac{64 - 69}{2.5}=\frac{- 5}{2.5}=-2$ and for $x_2 = 71.5$, $z_2=\frac{71.5 - 69}{2.5}=\frac{2.5}{2.5}=1$.

Step2: Apply the empirical rule

The empirical rule for a normal distribution states that 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 ($z=\pm3$).
For part (a), since 95% of the data lies within $z=-2$ and $z = 2$, the percentage of data outside of this range is $100 - 95=5\%$. Since the normal distribution is symmetric, the percentage of data greater than $z = 2$ is $\frac{100 - 95}{2}=2.5\%$.
For part (b), the percentage of data between $z=-2$ and $z = 2$ is 95% and the percentage of data between $z=-1$ and $z = 1$ is 68%. The percentage of data between $z=-2$ and $z = 1$ is $\frac{95+(68)}{2}\%$. The proportion of data between $z=-2$ and $z = 1$ is $\frac{0.95 + 0.68}{2}=0.815$.

Answer:

(a) 2.5
(b) 0.815