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%

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*12+9\text{ in.}=57\text{ in.}`

Maximum height = 6ft. 2 in.=
`6*12+2\text{ in.}=74\text{ in.}`

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

z-score :
`z=(x-\mu)/(\sigma)`

For x=57, we have

`z=(57-63.2)/(2.4)\approx-2.58`

For x=74, we have

`z=(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%