Question

The distribution of heights of adult American women is approximately normal with a mean of 64 inches and standard deviation of 2 inches. What percent of women is taller than 68 inches?

a) 0.0014

b) 0.025

c) 0.01

d) 0.05

Answer 1

$\boldsymbol{SOLUTION}$

$\boldsymbol{{\color{Blue} Answer}}:\;\;{\color{Red} \mathbf{\;b) \;0.025}}$

$\boldsymbol{given:}\;Mean\;(\mu )=64,\;standard\;deviation\;(\sigma)=2$

$according\; to\; empirical\; rule\; or\; 68-95-99.7 \;rule$

$\mu-3\sigma=64-(3*2)=58$

$\mu-2\sigma=64-(2*2)=60$

$\mu-1\sigma=64-(1*2)=62$

$\mu=64$

$\mu+1\sigma=64+(1*2)=66$

$\mu+2\sigma=64+(2*2)=68$

$\mu+3\sigma=64+(3*2)=70$

$Taller \;than \;68\Rightarrow P(X>68)$

$=2.35\%+0.15\%$

$=2.5\%$

$={\color{Red} \mathbf{0.025}}$