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# A store sells coats at different prices. Is the deviation of

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Senior Manager
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A store sells coats at different prices. Is the deviation of [#permalink]  22 Jul 2006, 05:31
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 Intern Joined: 14 May 2006 Posts: 32 Followers: 0 Kudos [?]: 0 [0], given: 0 I will go with B [#permalink] 22 Jul 2006, 12:20 I will go with B 1) Median 90 means (80,90, 100) or (80, 90, 600) deviation can be anything. 2) Range means difference betwen first and last element if it si 100 then devaition can not be more than 100 So ans is B. Pravin Director Joined: 28 Dec 2005 Posts: 758 Followers: 1 Kudos [?]: 8 [0], given: 0 Re: Price of coats [#permalink] 22 Jul 2006, 15:05 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

What is meant by "deviation" here? Is it the same as standard deviation?
If it is then I don't think we can solve this (E).
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Re: Price of coats [#permalink]  23 Jul 2006, 13:16
Futuristic wrote:
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 $100 What is meant by "deviation" here? Is it the same as standard deviation? If it is then I don't think we can solve this (E). we do not need to solve this one for SD I think B too oa please SVP Joined: 30 Mar 2006 Posts: 1739 Followers: 1 Kudos [?]: 46 [0], given: 0 [#permalink] 26 Jul 2006, 01:32 E . If we care looking for Standard deviation. B If we are looking only for the deviation Intern Joined: 21 Jul 2006 Posts: 27 Followers: 0 Kudos [?]: 0 [0], given: 0 [#permalink] 26 Jul 2006, 01:53 jaynayak wrote: E . If we care looking for Standard deviation. B If we are looking only for the deviation Even in terms of STDEV ... Consider Example JUST 3 PRICES ARE USED: 0$ 90$100$

RANGE: 100
MEDIAN: 90
STDEV: 55

SO NIETHER STATEMENT IS SUFFICIENT, whether Its asked STDEV or Just Deviation.... MEANABSDEV ... etc... )
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Senior Manager
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yes the range is max - min and standard deviation will not be > range...i thought...isnt it ? Please correct me...it is not true
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B

Even if SD is asked then also answer should be B. SD is ralative to mean. It doesn't matter values in the set are high or low. SD is the measure of dispersion of values of a set measured from mean.

Correct me if I am wrong.
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ps_dahiya wrote:
B

Even if SD is asked then also answer should be B. SD is ralative to mean. It doesn't matter values in the set are high or low. SD is the measure of dispersion of values of a set measured from mean.

Correct me if I am wrong.

Thanks ps_dahiya, I was also thinking the same...
Ans should be B.
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Re: Price of coats [#permalink]  26 Jul 2006, 22:29
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.
Then, this question turns to be trivial. Doesn't need to consider S1 and S2.
0 is less than 120.
Hence, D.

3.
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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Re: Price of coats   [#permalink] 26 Jul 2006, 22:29
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