Answer
a) 99.88 percent
b) 98.9 percent
c) 59.5 t0 73.4 inches
Work Step by Step
a) We first must find the z scores for each of the ends of the range:
$= \frac{56-63.8}{2.6}=-3.0$
$= \frac{75-63.8}{2.6}=4.3$
Using a table of z-scores and subtracting the two z-scores, we find that $100(1-.0012)=99.88$ percent of women are in this range.
b) We use the same process for men:
$= \frac{56-69.5}{2.4}=-5.625$
$= \frac{75-69.5}{2.4}=2.29$
Using a table of z-scores and subtracting the two z-scores, we find that 98.9 percent of men are in this range.
c) Using a table of z-scores, we find that the value of z is $\pm1.65$. We now consider the tallest men and the shortest women to obtain:
$max=(1.65)(2.4)+69.5=73.4 \ in$
$min=-(1.65)(2.6)+63.8=59.5\ in$
