This Question: 1 pt 5 of 25 (4 complete) This Test: 25 pts possible A survey fou
ID: 3326848 • Letter: T
Question
This Question: 1 pt 5 of 25 (4 complete) This Test: 25 pts possible A survey found that women's heights are normally distributed with mean 62.5 in. and standard deviation 2.2 in. The survey also found that men's heights are normally distributed with a mean 67.9 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 4 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 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
Solution:- GIven information :
Women : mean = 62.5in , sd = 2.2in
men : mean = 67.9in , sd = 2.8in
a) minimum = 4ft 9in = 4(12)+9 = 57in
maximum = 6ft 4in = 6(12)+4 = 76 in
Z = (x - mu)/sd
P(57 < X < 76) = P((57 - 62.5)/2.2 < Z < (76 - 62.5)/2.2)
= P(-2.5 < Z < 6.1364)
= 0.9938
=> The percentage of women who meet the height requirement is 99.38%
b. For men :
P(57 < X < 76) = P((57 - 67.9)/2.8 < Z < (76 - 67.9)/2.8)
= P(-3.8929 < Z < 2.8929)
= 0.998
=> The percentage of men who meet the height requirement is 99.80%
c. Z = (x - mu)/sd
Tallest :
=> (x - 67.9)/2.8 = 1.645
= x - 67.9 = 1.645*2.8
= x = 67.9 + 4.606
x = 72.506
shorst :
x = 67.9 - 4.606 = 63.294
=> The new hight requirement are at least 63.3 in and most 72.5 in