Question

Can you show how to draw the normal curve for each of the problems and label it as well?

Heights of MEN in the U.S. are normally distributed µ = 69.6 inches with σ = 3 inches.

-________ percent (to nearest %) of men in the U.S. are either shorter than 5 ft. or taller than 6 ft?

-In a group of 150 U.S. men, approximately ________ of them should be shorter than 65 inches.

-A male height of _______________ corresponds to the 58th percentile in the U.S. population. -_______________ is the cutoff height to be in the top 12% of male heights in the U.S.

-The middle 72% of U.S. men will be between ________ inches and ________ inches tall. -A man in the U.S. shorter than ___________ inches would be considered "unusually short. ( Can you Show your work or explain answer.)

Answer 1

Q1: µ = 69.6, σ = 3

i) 5 ft = 5*12 = 60 inches

6 ft = 6*12 = 72 inches

P(X < 60) + P(X > 72) =

= P((X-µ)/σ < (60-69.6)/3) + P((X-µ)/σ > (72-69.6)/3 )

= P(z < -3.2) + P(z > 0.8)

= P(z < -3.2) + 1 - P(z < 0.8)

Using excel function:

= NORM.S.DIST(-3.2, 1) +1 - NORM.S.DIST(0.8, 1)

= 0.2125 = 21%

21% of men in the U.S. are either shorter than 5 ft. or taller than 6 ft.

----------------

ii)

P(X < 65) =

= P( (X-µ)/σ < (65-69.6)/3 )

= P(z < -1.5333)

Using excel function:

= NORM.S.DIST(-1.5333, 1)

= 0.0626

In a group of 150 U.S. men, approximately 0.0626*150 = 9 of them should be shorter than 65 inches.

----------------

iii)

Z score at p = 0.58 using excel = NORM.S.INV(0.58) = 0.2019

Value of X = µ + z*σ = 69.6 + (0.2019)*3 = 70.2

A male height of 70.2 inches corresponds to the 58th percentile in the U.S. population.

---------------

iv)

P(x > a) = 0.12

= 1 - P(x < a) = 0.12

= P(x < a) = 0.88

Z score at p = 0.88 using excel = NORM.S.INV(0.88) = 1.1750

Value of X = µ + z*σ = 69.6 + (1.175)*3 = 73.12

73.12 inches is the cutoff height to be in the top 12% of male heights in the U.S.

-----------------------

v) Proportion in the middle = 0.72

Proportion on left and right side of normal curve = (1-0.72)/2 = 0.14

Z score at p = 0.14 using excel = NORM.S.INV(0.14) = 1.0803

Value of X1 = µ - z*σ = 69.6 - (1.0803)*3 = 66.36

Value of X2 = µ + z*σ = 69.6 + (1.0803)*3 = 72.84

The middle 72% of U.S. men will be between 66.36 inches and 72.84 inches tall.

------------------

vi) Proportion for unusually short = 0.05

Z score at p = 0.05 using excel = NORM.S.INV(0.05) = -1.6449

Value of X = µ + z*σ = 69.6 + (-1.6449)*3 = 64.67

A man in the U.S. shorter than 64.67 inches would be considered "unusually short.

---------

If any doubt ask me in comments