Here's my 2cents about deviation, variance, and standard deviation.
Let, set X = {X1,X2,....,Xn} (asscending order)
no matter what the value of a term is sum of each term's deviation is "0"
deaviation of 1st term = X1-mean
deaviation of 2nd term = X2-mean
....
deaviation of nth term = Xn-mean
Sum of deviations = X1+X2+....+Xn - (mean*n) = Total - Total = 0
Range = Xn-X1
Max. deviation = Xn - mean
Min. deviation = X1 - mean
Range > Max. devation > Min. deviation ('cuz mean < Xn)
(Except, when range = 0, Range = deviation = 0)
Standard deviation = sqrt(variance)
Square of 1st term's deviation : (X1-mean)^2
Square of 2nd term's deviation : (X2-mean)^2
.....
Square of nth term's deviation : (Xn-mean)^2
Variance
= [(X1^2-2X1*mean + mean^2)+ .... +(Xn^2-2Xn*mean + mean^2)]/n
= [(X1^2+X2^2...+Xn^2) - 2(X1+X2+..+Xn)(mean) + n*mean^2]n
= (X1^2+X2^2...+Xn^2)/n - 2(X1+X2+..+Xn)(mean)/n + mean^2
= (X1^2+X2^2...+Xn^2)/n - 2[(X1+X2+..+Xn)/n](mean) + mean^2
= (X1^2+X2^2...+Xn^2)/n - 2[mean](mean) + mean^2
= (X1^2+X2^2...+Xn^2)/n - mean^2
= Average of square of each term - mean^2 > 0
Standard devation
= sqrt(variance )
Range > Standard deviation
(Except, when range = 0, Range = varicance = standard deviation = 0)
Range = Xn-X1
Variance = (X1^2+X2^2...+Xn^2)/n - mean^2
Standard deviation = sqrt((X1^2+X2^2...+Xn^2)/n - mean^2)
Range - standard deviation
= Xn-X1 - sqrt((X1^2+X2^2...+Xn^2)/n - mean^2)
Square both side.
(Range - standard deviation)^2
= (Xn-X1 - sqrt((X1^2+X2^2...+Xn^2)/n - mean^2))^2
(Range - standard deviation)^2
= (Xn-X1)^2 - 2(Xn-X1)sqrt((X1^2+X2^2...+Xn^2)/n - mean^2))^2
Stuck here.. will try tomorrow.
Logical proof:
Purpose of those measures - deviation and standard deviation - is to see how far each term is from the mean.
Range is max. distance in a set.
Thus, Range is greater than or equal to those measures.
getzgetzu wrote:
A store sells coats at different prices. Is the deviation of the price less than $120?
1) The median of the price is $90
2) The range of the price is $ 100
1.
the deviation means each deviation.
S1. median of the price is $90. insuff.
S2. range of the price is $100.
tells that Max. deviation is less than $100 suff.
Hence, B.
2.
if question is asking about sum of each deviation.
Then, this question turns to be trivial. Doesn't need to consider S1 and S2.
0 is less than 120.
Hence, D.
3.
if question is asking about standard deviation.
1) The median of the price is $90 insuff.
2) The range of the price is $ 100 suff.
Range is always greater than standard deviation.(Except when range = 0)
Hence, B.
_________________
The only thing that matters is what you believe.