A survey found that women\'s heights are normally distributed with mean 62.7 in.

Question

A survey found that women's heights are normally distributed with mean 62.7 in. and standard deviation 2.7 in. The survey also found that men's heights are normally distributed with a mean 67.1 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.

b. Find the percentage of men meeting the height requirement.

The percentage of men who meet the height requirement is ?? %

c. If the height requirements 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

as z score =(X-mean)/std deviaiton

a) here 4 ft 9 in =12*4+9=57 in and 6 ft 2 in =12*6+2 =74 in

hence  percentage of women meeting the height requirement =P(57<X<74)=P(-2.111<Z<4.1852)=0.9999-0.0174

=0.9826~ 98.26%

b) percentage of men meeting the height requirement =P(57<X<74)=P(-3.7407<Z<2.556)=0.9947-0.0001

=0.9946

c) for bottom and top 5% ; z= 1.6449

hence  height requirements are at least =62.7-1.6449*2.7=58.3 inch

at most =67.1+1.6449*2.7 =71.5 inch

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