Answer: 21.28%
Step-by-step explanation:
Given : Mean height of females = [tex]\mu=69\text{ in.}[/tex]
Standard deviation : [tex]\sigma=33\text{ in.}[/tex]
Let X be a random variable that represents the heights of females in village .
We assume that that the heights of females in a certain village are normally distributed .
Z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = 60 in.
[tex]z=\dfrac{60-69}{33}\approx-0.27[/tex]
For x = 78 in.
[tex]z=\dfrac{78-69}{33}\approx0.27[/tex]
Now by using standard normal distribution table , the probability that females are between 60 in. and 78 in. tall :-
[tex]P(60<x<78)=P(-0.27<z<0.27)\\\\=P(z<0.27)-P(z<-0.27)\\\\=0.6064198-0.3935801=0.2128397\approx0.2128=21.28\%[/tex]
Hence, the percent of these females are between 60 in. and 78 in. tall =21.28%
