A survey found that women\'s heights are normally distributed with mean 63.3 in.

Question

A survey found that women's heights are normally distributed with mean 63.3 in. and standard deviation 3.2 in. The survey also found that men's heights are normally distributed with mean 68.4 in. and standard deviation 3.5 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in. Complete parts (a) and (b) below.

a. 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?

b. If the height requirements are changed to exclude only the tallest 50% of men and the shortest 5% of men, what are the new height requirements?

Explanation / Answer

a)

  percentage of men meeting the height requirement:

result suggest that women has higher probability to meet the height requirement than men,

b)

here for 50 percentile ; z score =0.

therefore corresponding height =mean+z*std deviation =68.4+0*3.5 =68.4

and for 5percentile ; z =-1.6449

therefore corresponding height =mean+z*std deviation =68.4-1.6449*3.5 =62.64

therefore height requirement should be from 62.64 to 68.4.

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