Statistics & probability

Question

Heights of anAmerican woman Ages 18-29 are normally distributed with a mean of 64.3 inches and a standard deviation of 3.8 inches. an American woman in this age is bracket randomly selected.

A. What is the probability she is more than 68 inches tall?

b. What is the probability she is less than 70 inches tall?

c what is the probability she is less than 59 or greater than 71 inches tall?

Answer 1

x = height is N(64.3,3.8)

z = (x - mean)/SD is N(0,1), can use z value to look up associated probabilities from standard normal probability table

P(x > 68) = P(z > (68-64.3)/3.8) = P(z > 0.97) = P(z < -0.97) = 0.166

P(x < 70) = P(z < (70 - 64.3)/3.8) = P(z < 1.5) = 0.9332

P(59 < x < 71) = P( (59 - 64.3)/3.8 < z < (71 - 64.)/3.8) = P(-1.39 < z < 1.84) = P(z < 1.84) - P(z < -1.39) = 0.9671 - 0.0823 = 0.8851