Question

According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.
1. Find the probability that the person is between 64 and 68 inches. (Round to four decimal places) and Sketch the Graph.

2. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically.

No, because the probability that an Asian male is over 72 inches tall is_______

3. The middle 40% of heights fall between what two values?

Write the probability statement.

P(x₁ < X < x₂) =

x1=

x2=

Answer 1

X ~ N(66, 2.5)

1. We need to find: P(64 < X < 68)
When X = 64, z = (64 - 66)/2.5 = -0.8
When X = 68, z = (68 - 66)/2.5 = 0.8

Hence, P(-0.8 < z < 0.8)
= P(z < 0.8) - P(z < -0.8)
From the z-table,
= 0.7881 - 0.2119
= 0.5762

2. We need to find: P(X > 72)
When X = 72, z = (72 - 64)/2.5 = 3.2
Hence, P(z > 3.2)
= 1 - P(z < 3.2)
= 1 - 0.9993
= 0.0007

No, because the probability that an asian male is over 72 inches tall is 0.0007

3. We need to find:
P(z1 < z < z2) = 0.40
By symmetry of normal distribution,
P(0 < z < z2) = P(z1 < z < 0) = 0.20

P(z < z2) = 0.5 + 0.20 = 0.70
P(z < 0) - P(z < z1) = 0.20
P(z < z1) = 0.30

From z-table,
z2 = 0.525
z1 = -0.525

X2 = 67.3125
X1 = 64.6875

P(64.69 < X < 67.31)