Question

Men's heights are normally distributed with mean of 69.3 in and standard deviation of 2.4 in. Women's heights are normally distributed with mean of 63.8 in and standard deviation of 2.6 in. The U.S. Navy requires that fighter pilots have heights between 62 in and 78 in.

a) Find the percentage of women meeting the height requirement. __________ percent (round to the nearest tenth)

b) Find the percentage of men meeting the height requirement. __________ percent (round to the nearest tenth)

Answer 1

a)

Here, μ = 63.8, σ = 2.6, x1 = 62 and x2 = 78. We need to compute P(62<= X <= 78). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (62 - 63.8)/2.6 = -0.69
z2 = (78 - 63.8)/2.6 = 5.46

Therefore, we get
P(62 <= X <= 78) = P((78 - 63.8)/2.6) <= z <= (78 - 63.8)/2.6)
= P(-0.69 <= z <= 5.46) = P(z <= 5.46) - P(z <= -0.69)
= 1 - 0.2451
= 0.7549

= 75.49%

b)

Here, μ = 69.3, σ = 2.4, x1 = 62 and x2 = 78. We need to compute P(62<= X <= 78). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (62 - 69.3)/2.4 = -3.04
z2 = (78 - 69.3)/2.4 = 3.63

Therefore, we get
P(62 <= X <= 78) = P((78 - 69.3)/2.4) <= z <= (78 - 69.3)/2.4)
= P(-3.04 <= z <= 3.63) = P(z <= 3.63) - P(z <= -3.04)
= 0.9999 - 0.0012
= 0.9987
= 99.87%