Question

Do this one by hand. Suppose we measured the height of 10,000 men and found that the data were normally distributed with a mean of 70.0 inches and a standard deviation of 4.0 inches. Answer the questions and show your work:

A.      What proportion of men can be expected to have heights less than: 66, 70, 72, 75 inches?

B.      What proportion of men can be expected to have heights greater than: 64, 66, 73, 78 inches?

C.      What proportion of men can be expected to have heights between: 65 and 75 inches, 71 and 72 inches?  

D.      How many men (of the 10,000) can be expected to have heights between: 65 and 71 inches, 72 and 75 inches?

E.      What height corresponds to the 43rd percentile? What height corresponds to the 55th percentile? What height corresponds to the 99th percentile? (If necessary, round the numbers, but keep at least two decimal places.)

Answer 1

a-i) Since μ=70 and σ=4 we have:

Proprtion ( X<66 )= ( X−μ<66−70 )=((X−μ)/σ<(66−70)/4)

Since (x−μ)/σ=Z and (66−70)/4=−1 we have:

(X<66)=P (Z<−1)

Use the standard normal table to conclude that:

proportion of men can be expected to have heights less than 66 is 0.1587

a-ii)Proportion ( X<70 )= ( X−μ<70−70 )= ((X−μ)/σ<(70−70)/4)

Since (x−μ)/σ=Z and (70−70)/4=0 we have:

(X<70)= (Z<0)

Use the standard normal table to conclude that:

(Z<0)=0.5

proportion of men can be expected to have heights less than 70 is 0.5

a-iii) : proportion of men can be expected to have heights less than 72

( X<72 )=( X−μ<72−70 )=((X−μ)/σ<(72−70)/4)

Since (x−μ)/σ=Z and (72−70)/4=0.5 we have:

(X<72)= (Z<0.5)

Use the standard normal table to conclude that:

(Z<0.5)=0.6915 is  proportion of men can be expected to have heights less than 72.

a-iv) Proportion of men can be expected to have heights less than 75

( X−μ<75−70 )= ((X−μ)/σ<(75−70)/4)

Since (x−μ)/σ=Z and (75−70)/4=1.25 we have:

(X<75)=(Z<1.25)

Use the standard normal table to conclude that:

(Z<1.25)=0.8944 is proportion of men can be expected to have heights less than 75.

b-1)proportion of men can be expected to have heights greater than: 64

( X>64 )=P ( X−μ>64−70 )=P ( (X−μ)/σ>(64−70)/4)

Since Z=(x−μ)/σ and (64−70)/4=−1.5 we have:

( X>64 )= ( Z>−1.5 )

Use the standard normal table to conclude that:

(Z>−1.5)=0.9332

b-ii) proportion of men can be expected to have heights greater than 66

( X>66 )= ( X−μ>66−70 )= ( X−μ/σ>66−70/4)

Since Z=x−μ/σ and 66−70/4=−1 we have:

( X>66 )=( Z>−1 )

Use the standard normal table to conclude that:

(Z>−1)=0.8413

b-iii) proportion of men can be expected to have heights greater than: 73

Z=(x−μ)/σ and (73−70)/4=0.75 we have:

( X>73 )= ( Z>0.75 )

Use the standard normal table to conclude that:

(Z>0.75)=0.2266

b-iv) proportion of men can be expected to have heights greater than: 78

Z=x−μ/σ and (78−70)/4=2 we have:

( X>78 )= ( Z>2 )

Use the standard normal table to conclude that:

(Z>2)=0.0228

c-1) proportion of men can be expected to have heights between: 65 and 75 inches is 4931

Z=(x−μ)/σ , (65−70)/4=−1.25 and (75−70)/4=1.25 we have:

( 65<X<75 )= ( −1.25<Z<1.25 )

Use the standard normal table to conclude that:

P ( −1.25<Z<1.25 )=0.7888

c-2) proportion of men can be expected to have heights between: 71 and 72 inches

Z=(x−μ)/σ , (71−70)/4=0.25 and (72−70)/4=0.5 we have:

( 71<X<72 )=( 0.25<Z<0.5 )

Use the standard normal table to conclude that:

( 0.25<Z<0.5 )=0.0928

D-1) Number of men (of the 10,000) can be expected to have heights between: 65 and 71 inches

Z=(x−μ)/σ , (65−70)/4=−1.25 and (71−70)/4=0.25 we have:

( 65<X<71 )= ( −1.25<Z<0.25 )

Use the standard normal table to conclude that:

( −1.25<Z<0.25 )=0.4931

10000*0.4931= 4931

D-2) Number men (of the 10,000) can be expected to have heights between: 72 and 75 inches

Since Z=(x−μ)/σ , (72−70)/4=0.5 and (75−70)/4=1.25 we have:

( 72<X<75 )= ( 0.5<Z<1.25 )

Use the standard normal table to conclude that:

( 0.5<Z<1.25 )=0.2029

10,000*0.2029= 2029 men (of the 10,000) can be expected to have heights between: 72 and 75 inches.

NOTE: PLEASE REPOST QUESTION E . Thank you.