in 2002, the average height of a woman aged 20-74 years was 64 inches with an in

Question

in 2002, the average height of a woman aged 20-74 years was 64 inches with an increase approximately 1 inch from 1960 Qhtip/nusgovinfo about.com/od/healthcare) Suppose the height of a woman is normally distributed with a standard deviation of inches. (a) What is the probability that a randomly selected woman in this population is between 58 inches and 70 inches? (b) What are the quartiles of this distribution? (c) Determine the height that is symmetric about the mean that includes 90% of this population. d) What is the probability that five women selected at random from this population all exceed 68 inches? .75

Explanation / Answer

A) P(58 < X < 70)

= P((58 - mean)/SD < (X - mean)/SD < (70 - mean)/SD)

= P((58 - 64)/1.75 < Z < (70 - 64)/1.75)

= P(-3.43 < Z < 3.43)

= P(Z < 3.43) - P(Z < -3.43)

= 0.9997 - 0.0003

= 0.9994

B) P(X < x) = 0.25

Or, P(Z < (x - 64)/1.75) = 0.25

Or, (x - 64)/1.75 = -0.67

Or, x = -0.67 * 1.75 + 64 = 62.8275

P(X < x) = 0.75

Or, P((X - mean)/SD < (x - 64)/1.75 ) = 0.75

Or, P(Z < (x - 64)1.75) = 0.75

Or, (x - 64)/1.75 = 0.67

Or, x = 0.67 * 1.75 + 64 = 65.1725

C) P(- x < X < x) = 0.9

Or, P(-z < Z < z) = 0.9

Or, P(Z <Z ) - P(Z < -z) = 0.9

Or, 2P(Z < z) = 1.9

Or, P(Z < z) = 0.95

Or, Z = 1.645

Or, (x - 64)/1.75 = 1.645

Or, x = 1.645 * 1.75 + 64

Or, x = 66.88

66.88 - 64 = 2.88

So the other x value = 64 - 2.88 = 61.12

D) P(X > 68)

= P((X - mean)/sd > (68 - 64)/1.75)

= P(Z > 2.29)

= 1 - P(Z < 2.29)

= 1 - 0.9890

= 0.011

P(X = 5) = 5C5 * (0.011)^5 * (0.989)^0

= 1.61 * 10-10

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