MATH SOLVE

2 months ago

Q:
# 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?

Accepted Solution

A:

Answer: 99.51%Step-by-step explanation:Given : A survey found that women's heights are normally distributed.Population mean : [tex]\mu =63.2 \text{ inches}[/tex]Standard deviation: [tex]\sigma= 2.4\text{ inches}[/tex]Minimum height = 4ft. 9 in.=[tex]4\times12+9\text{ in.}=57\text{ in.}[/tex]Maximum height = 6ft. 2 in.=[tex]6\times12+2\text{ in.}=74\text{ in.}[/tex]Let x be the random variable that represent the women's height. z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x=57, we have [tex]z=\dfrac{57-63.2}{2.4}\approx-2.58[/tex]For x=74, we have [tex]z=\dfrac{74-63.2}{2.4}\approx4.5[/tex]Now, by using the standard normal distribution table, we haveThe probability of women meeting the height requirement :-[tex]P(-2.58<z<4.5)=P(z<4.5)-P(z<-2.58)\\\\= 0.9999966- 0.00494=0.9950566\approx0.9951=99.51\%[/tex]Hence, the percentage of women meeting the height requirement = 99.51%