Question

A survey found that​ women's heights are normally distributed with mean 62.4 in. and standard deviation 2.1 in. The survey also found that​ men's heights are normally distributed with mean 67.3 in. and standard deviation 3.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 56 in. and a maximum of 63 in.

1.Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement​ park?

2.Find the percentage of women meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement​ park?

Answer 1

1) We have to find the percentage of men meeting the height requirement.

That is we have to find P( 56 < X < 63)

Mean = = 67.3

Standard deviation = = 3.1

For finding this probability we have to find z score.

That is we have to find P( - 3.65 < Z < - 1.39)

P( - 3.65 < Z < - 1.39) = P(Z < - 1.39) - P(Z < -3.65) = 0.0827 - 0.0001 = 0.0826 ( Using z table)

8.26%

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2) We have to find the percentage of women meeting the height requirement.

That is we have to find P( 56 < X < 63)

Mean = = 62.4

Standard deviation = = 2.1

For finding this probability we have to find z score.

That is we have to find P( - 3.05 < Z < 0.29)

P( - 3.05 < Z < - 0.29) = P(Z < 0.29) - P(Z < -3.05) = 0.6125 - 0.0011 = 0.6113 ( Using z table)

61.13%