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.

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