Question

A survey found that​ women's heights are normally distributed with mean 63.2 in. and standard deviation 2.4 in. The survey also found that​ men's heights are normally distributed with a mean 67.2 in. and standard deviation 2.7. Complete parts a through c below. a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 2 in. Find the percentage of women meeting the height requirement?

Answer 1

Answer: 99.51%

Step-by-step explanation:

Given : A survey found that​ women's heights are normally distributed.

Population mean : $\mu =63.2  \text{ inches}$

Standard deviation: $\sigma= 2.4\text{ inches}$

Minimum height = 4ft. 9 in.=$4\times12+9\text{ in.}=57\text{ in.}$

Maximum height = 6ft. 2 in.=$6\times12+2\text{ in.}=74\text{ in.}$

Let x be the random variable that represent the women's height.

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x=57, we have

$z=\dfrac{57-63.2}{2.4}\approx-2.58$

For x=74, we have

$z=\dfrac{74-63.2}{2.4}\approx4.5$

Now, by using the standard normal distribution table, we have

The probability of women meeting the height requirement :-

$P(-2.58<z<4.5)=P(z<4.5)-P(z<-2.58)\\\\= 0.9999966- 0.00494=0.9950566\approx0.9951=99.51\%$

Hence, the percentage of women meeting the height requirement = 99.51%