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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?

User Mkataja
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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%

User Nexevis
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