A survey found that women\'s heights are normally distributed with mean 63.4 in

Question

A survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 29 in. The survey also found that men's heights are normally distributed with a mean 67.8 in and standard deviation 2.8. 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. The percentage of women who meet the height requirement is %. (Round to two decimal places as needed.) b. Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is %. (Round to two decimal places as needed) c. If the height requirement are changed to exclude only the tallest 5% of men and the shortest 5% of women. What are the new height requirements? The new height requirements are at least in and at most in. (Round to one decimal place as needed.)

Explanation / Answer

Answer:

a).

Requirement minimum height: 57 in.   maximum: 74 in.

z value for 57, z =(57-63.4)/2.9 =-2.21

z value for 74, z=(74-63.4)/2.9 =3.66

P( 57<x<74)= P( -2.21 <z<3.66)= P( z <3.66)- P( z < -2.21)

=0.9999- 0.0136 =0.9863

The required percentage=98.63%

b).

z value for 57, z=(57-67.8)/2.8 =-3.86

z value for 74, z=(74-67.8)/2.8 =2.21

P( 57<x<74)= P( -3.86 <z<2.21)= P( z <2.21)- P( z < -3.38)

=0.9864-0.0004 =0.9860

The required percentage=98.60%

c).

z value for top and bottom 5% are (1.645, -1.645)

the required minimum value =63.4-1.645*2.9 =58.6295

=58.6 in

the required maximum value =67.8+1.645*2.8 =72.406

=72.4 in

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