Answer:
A. 22.5 cm, 30 cm, and 37.5 cm
Step-by-step explanation:
If three sides of a triangle are in the ratio 3:4:5, then the lengths of the sides are 3x cm, 4x cm and 5x cm.
The preimeter of the triangle is the sum of the lengths of all sides, thus
So, the lengths of the triangle's sides are
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!
