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.

1. (20 pts) Do this one by
hand. Suppose we measured the height of 5,000 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 using Table
A and show your work:
What proportion of men can be expected to have heights less than
66 inches? Less than 75 inches?
What proportion of men can be expected to have heights greater
than 64 inches? Greater...

Do this one by hand. Suppose we measured the height of 5,000 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 using Table A and show your work:
What proportion of men can be expected to have heights greater
than 73 inches?

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normal with a mean of 69 inches and a standard deviation of 3
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c) What is the 95th (and 99th) percentile of all men’s heights?

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