Question

An example of an early application of statistics was in the year 1817. A study of chest circumference among a group of Scottish men exhibited an approximately normal distribution. Their chest circumferences ranged from 33 to 48 inches, with a mean chest measurement of 40 inches and a standard deviation of 2 inches. Make a normal bell curve, then use the Empirical Rule to help you understand the distribution of chest circumferences in the study.
a) What range of chest measurements contains the 58% which fall closest to the mean?


b) What would you expect the chest measurements to be fore the 2.5% of the men with the smallest chest measurements?


c) What would you expect to be the smallest and largest chest measurements of the "middle" 95% of the men?


d) What would you expect to be the measurements of the 16% of the men with the largest chests in the population

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: chest circumference of a Scottish man.

X≈N(μ;δ²)

μ= 40 inches

δ= 2 inches

The empirical rule states that

68% of the distribution lies within one standard deviation of the mean: μ±δ= 0.68

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

99% of the distribution lies within 3 standard deviations of the mean: μ±3δ= 0.99

a)

The 58% that falls closest to the mean can also be referred to as the middle 58% of the distribution, assuming that both values are equally distant from the mean.

P(a≤X≤b)= 0.58

If 1-α= 0.58, then the remaining proportion α= 0.42 is divided in two equal tails α/2= 0.21.

The accumulated proportion until "a" is 0.21 and the accumulated proportion until "b" is 0.21 + 0.58= 0.79 (See attachment)

P(X≤a)= 0.21

P(X≤b)= 0.79

Using the standard normal distribution, you can find the corresponding values for the accumulated probabilities, then using the information of the original distribution:

P(Z≤zᵃ)= 0.21

zᵃ= -0.806

P(Z≤zᵇ)= 0.79

zᵇ= 0.806

Using the standard normal distribution Z= (X-μ)/δ you "transform" the values of Z to values of chest circumference (X):

zᵃ= (a-μ)/δ

zᵃ*δ= a-μ

a= (zᵃ*δ)+μ

a= (-0.806*2)+40= 38.388

and

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (0.806*2)+40= 41.612

58% of the chest measurements will be within 38.388 and 41.612 inches.

b)

The measurements of the 2.5% men with the smallest chest measurements, can also be interpreted as the "bottom" 2.5% of the distribution, the value that separates the bottom 2.5% of the distribution from the 97.5%, symbolically:

P(X≤b)= 0.025 (See attachment)

Now you have to look under the standard normal distribution the value of z that accumulates 0.025 of the distribution:

P(Z≤zᵇ)= 0.025

zᵇ= -1.960

Now you reverse the standardization to find the value of chest circumference:

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (-1.960*2)+40= 36.08

The chest measurement of the 2.5% smallest chest measurements is 36.08 inches.

c)

Using the empirical rule:

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

(μ-2δ) ≤ Xc ≤ (μ+2δ)=0.95 ⇒ (40-4) ≤ Xc ≤ (40+4)= 0.95 ⇒ 36 ≤ Xc ≤ 44= 0.95

d)

The measurements of the 16% of the men with the largest chests in the population or the "top" 16% of the distribution:

P(X≥d)= 0.16

P(X≤d)= 1 - 0.16

P(X≤d)= 0.84

First, you look for the value that accumulates 0.84 of probability under the standard normal distribution:

P(Z≤zd)= 0.84

zd= 0.994

Now you reverse the standardization to find the value of chest circumference:

zd= (d-μ)/δ

zd*δ= d-μ

d= (zd*δ)+μ

d= (0.994*2)+40= 41.988

The measurements of the 16% of the men with larges chess are at least 41.988 inches.

I hope this helps!