Which fraction is NOT equivalent to 4/9 ?
2 answers:
You might be interested in
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!
