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]
1) a+b+c+d=50
2) a^2+b^2+c^2+d^2=200
[spoiler]OA: C how??[/spoiler]
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
