A survey found that womens heights are normally distributed

A survey found that women?s heights are normally distributed with mean 63.4 in. and standard deviation 2.2 in. The survey also found that men\'s heights are normally distributed with a mean 67.5 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 3 in. Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is 99.81 %. (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 99.79 %. (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 D in. and at most D (Round to one decimal place as needed.)

Solution

10.
Z0.10
= 1.28

12.
a.
population mean,u = 63.5
standard deviation,sigma = 2.6

4ft 9in = 4*12 + 9 = 57 in

P( 57 <X< 74 )
= P( ((57-63.5) / 2.6) <Z< ((74-63.5) / 2.6) )
= P( -2.50 <Z< 4.04 )
= P(Z<4.04) - P(Z<-2.5)
= P(Z<4.04) - (1 - P(Z<2.5))
= 0.9999 - ( 1 - 0.9938)
= 0.9999 - 0.0062
= 0.9937
= 99.37% (answer may be 99.38%)

b.
population mean,u = 69.2
standard deviation,sigma = 2.9

P( 57 <X< 74 )
= P( ((57-69.2) / 2.9) <Z< ((74-69.2) / 2.9) )
= P( -4.21 <Z< 1.66 )
= P(Z<1.66) - P(Z<-4.21)
= P(Z<1.66) - (1 - P(Z<4.21))
= 0.9515 - ( 1 - 0.9999)
= 0.9515 - 0.0001
= 0.9514
= 95.14% (answer may be 95.15%)

c.
z-score corresponding to tallest 5% is 1.645
z-score corresponding to shortest 5% is -1.645
69.2 + 1.645*2.9 = 74.0
63.5 - 1.645*2.6 = 59.2

The new height requirements are at least 59.2 in. ans at most 74.0 in.