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 2.3 in. The survey also found that men's heights are normally distributed with a mean 67.8 in. and standard deviation 2.6

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

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 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

(Round to one decimal place as needed.)

Explanation / Answer

Question a)

Mean = 63.4

Standard deviation = 2.3

4 feet and 9 inches = (4*12) + 9 = 57 inches

6 feet and 2 inches = (6*12) + 2 = 74 inches

P (57 < x < 74)

z = ( x – Mean ) / Standard deviation

z1 = (57 – 63.4)/2.3

     = -2.78

z1 = (74 – 63.4)/2.3

     = 4.61

P (-2.78 < z < 4.61)

= P ( z < 4.61 ) – P ( z < -2.78 )

= 1 – 0.0027

= 0.9973

Answer: 99.73%

Question b)

Mean = 67.8

Standard deviation = 2.6

4 feet and 9 inches = (4*12) + 9 = 57 inches

6 feet and 2 inches = (6*12) + 2 = 74 inches

P (57 < x < 74)

z = ( x – Mean ) / Standard deviation

z1 = (57 – 67.8)/2.6

    = -4.15

z2 = (74 – 67.8)/2.6

= 2.38

P (-4.15 < z < 2.38)

= P ( z < 2.38 ) – P (z < -4.15)

= 0.9913 – 0.0000

= 0.9913

Answer: 99.13%

Question c)

Tallest 5%

z = 1.645

We know that,

x = Mean + ( z * standard deviation )

    = 67.8 + (1.645*2.6)

= 72.1

Shortest 5%

z = -1.645

x = Mean + ( z * standard deviation )

    = 63.4 - (1.645*2.3)

   = 59.6

Answer:

The new requirement:

59.6 in and 72.1 in

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