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What is the SD of the set: a, b, c, d 1) a+b+c+d=50 2)

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Manager
Joined: 07 May 2007
Posts: 178
What is the SD of the set: a, b, c, d 1) a+b+c+d=50 2) [#permalink]

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17 Jun 2007, 09:36
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What is the SD of the set: a, b, c, d

1) a+b+c+d=50
2) a^2+b^2+c^2+d^2=200
Director
Joined: 26 Feb 2006
Posts: 899
Re: DS - Standard Deviation [#permalink]

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17 Jun 2007, 13:32
iamba wrote:
What is the SD of the set: a, b, c, d

1) a+b+c+d=50
2) a^2+b^2+c^2+d^2=200

Which one is correct?
a^2+b^2+c^2+d^2 = 200 or
a^2+b^2+c^2+d^2 = 2000?
Manager
Joined: 28 Aug 2006
Posts: 160

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19 Jun 2007, 20:53
SD can be measured if you have sum of the numbers and sum of the sqaures of the numbers using the following formula

SD= SQRT ((SumX^2/N) -(SumX/N)^2)
Applying this SQRT(2000/4-(50/2)^2)

I think Sum of sqaures should be 2000
Director
Joined: 26 Feb 2006
Posts: 899

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19 Jun 2007, 20:58
iamba wrote:
a^2 + b^2 + c^2 + d^2 = 200

if (a + b + c + d) = 50, how (a^2 + b^2 + c^2 + d^2) = 200? cuz a or b or c or d could be +ve or -ve, but a^2, b^2, c^2, and d^2 can not be -ve. they all are positive.

the minimum value = 4 (12.5)^2 = 625 when a = b = c = d. if a, b, c and d are different from 12.5, the value of (a^2 + b^2 + c^2 + d^2) will be grater than 625.

can you explain where i am i missing?
Manager
Joined: 07 May 2007
Posts: 178

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19 Jun 2007, 21:18
I guess it should be 2000, even though my question booklet has 200. But does this info really required to answer the Q?
Manager
Joined: 23 May 2007
Posts: 108

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19 Jun 2007, 22:11
Is this formula required ???
SD= SQRT ((SumX^2/N) -(SumX/N)^2)

or theres anoher way
Manager
Joined: 07 May 2007
Posts: 178

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19 Jun 2007, 23:18
This is how I approached it

Mean = 50/4 = 12.5

SD = SQRT{(12.5-a)^2+...........)/4}

which takes the form of

SD = SQRT{ (4*12.5^2-12.5(a+b+c+d)+a^2+b^2+c^2+d^2)/4}

Plug in the values to get SD
Director
Joined: 26 Feb 2006
Posts: 899

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20 Jun 2007, 10:08
vijay2001 wrote:
SD can be measured if you have sum of the numbers and sum of the sqaures of the numbers using the following formula

SD= SQRT ((SumX^2/N) -(SumX/N)^2)
Applying this SQRT(2000/4-(50/2)^2)

I think Sum of sqaures should be 2000

SD = SQRT ((SumX^2 / N) - (SumX / N)^2)
SD = SQRT[(2,000/4) - (50 / 4)^2]
SD = sqrt (343.75)
SD = 18.54

so C.
20 Jun 2007, 10:08
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What is the SD of the set: a, b, c, d 1) a+b+c+d=50 2)

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