Question

Assume the heights of men 18 to 24 are approximately normally distributed with μ=70 inches, 3.0 is the standard deviation. A. What percent of men in this age group are taller than 74 inches? Z-score is _____ P-value__________ (hint give the percentage). B. What percent of men in this age group are taller than 65 inches? Z-score is _____ P-value__________ (hint give the percentage). C. What percent of men in this age group are shorter than 69 inches? Z-score is _____ P-value__________ (hint give the percentage). D. What percent of men in this age group is between 67-70 inches? Z-score is _____ P-value__________ (hint give the percentage).

Answer 1

Z = (X - mean)/ standard deviation

P(X < A) = P(Z < (A - mean)/ standard deviation)

Here, mean = 70 inches

Standard deviation = 3.0

A. For X = 74, corresponding Z = (74 - 70)/3 = 1.33

P(taller than 74 inches) = 1 - P(X < 74)

= 1 - P(Z < 1.33)

= 1 - 0.9082

= 0.0918

= 9.18%

Z score is 1.33. P-value = 9.18%

B. For X = 65, Z = (65 - 70)/3 = -1.67

P(taller than 65 inches) = 1 - P(X < 65)

= 1 - P(Z < -1.67)

= 1 - 0.0475

= 0.9525

= 95.25%

Z score is -1.67. P-value = 95.25%

C. For X = 69, Z = (69 - 70)/3 = -0.33

P(shorter than 69 inches) = P(X < 69)

= P(Z < - 0.33)

= 0.3707

Z score is -0.33 and P-value is 37.07%

D. Z score corresponding to 67 = (67 - 70)/3 = -1

Z score corresponding to 70 = 0

P(between 67-70 inches) = P(X < 70) - P(X < 67)

= P(Z < 0) - P(Z < -1)

= 0.5 - 0.1587

= 0.3413

Z-scores are -1 and 0, and P-value is 34.13%