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SD

by ketkoag » Fri Jun 19, 2009 4:43 pm
What is the standard deviation of the set: a, b, c, d ?
1) a+b+c+d=50
2) a^2+b^2+c^2+d^2=200

[spoiler]OA: C how??[/spoiler]
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Source: — Data Sufficiency |
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by cramya » Fri Jun 19, 2009 9:55 pm
SD=

sqrt{[a- (a+b+c+d /4) ] ^2 + [b- (a+b+c+d /4) ] ^2 + [c- (a+b+c+d /4) ] ^2 + [d- (a+b+c+d /4) ] ^2 sum divided by 4}


Lets take [a- (a+b+c+d /4) ] ^2

a+b+c+d = 50
Use (x-y)^2 formula where x=a y= a+b+c+d /4

x^2+y^2-2xy

a^2+(50/4)^2 - 2 a *50/4 = a^2+some value - 25a

Similarly if u solve the others it will be

b^2+some value-25b
c^2+some value-25c
d^2+some value-25d

a^2+b^2+c^2+d^2 - 25 (a+b+c+d)+4*(50/4)^2

We know a^2+b^2+c^2+d^2 and (a+b+c+d) so we can caclualte the SD

Hence c

I have tried to explain it in detail but once u get to the stage where u determine its solvable then u can mark C and move on without wasting time.



Hope this helps!

Regards,
CR
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by ketkoag » Tue Jun 23, 2009 12:59 pm
yes, got it now..
thanks a bunch.......
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