Question

The heights of women in the U.S. have been found to be approximately normally distributed with a mean of 63.02 inches and the variance to be 9.00 inches.

a) What percent of women are taller than 64.43 inches?
probability =

b) What percent of women are shorter than 61.7 inches?
probability =

c) What percent have heights between 61.7 and 64.43 inches?
probability =

Note: Do NOT input probability responses as percentages; e.g., do NOT input 0.9194 as 91.94

Answer 1

Z score normal distribution formula:

z = (x - μ) / σ

variance = 9, so we can calculate standard deviation from variance is = sqrt(9) = 3

a) What percent of women are taller than 64.43 inches?

z = (64.43-63.02)/3 = 0.47

Probability (Z > 0.47) = 0.3192

b) What percent of women are shorter than 61.7 inches?

z = (61.7-63.02)/3 = -0.44

Probability (Z < -0.44) = 0.3300

c) What percent have heights between 61.7 and 64.43 inches?

z = (61.7-63.02)/3 = -0.44

z = (64.43-63.02)/3 = 0.47

P(-0.44 < z < 0.47) = 0.3509