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1. A Set if numers 1x,...,nx has a mean of one.
Show that the standard deviation could be as large as √(n-1).
I can get as far as √((∑x^2-n)/(n-1))
2. Ex^2 = 1/6[(n+1)(2n+1)

I think that may be useful. Possibly. I can't remember the formula for SD when you do all the n-1 crap.
3. (Original post by AnonyMatt)
Ex^2 = 1/6[(n+1)(2n+1)

I think that may be useful. Possibly. I can't remember the formula for SD when you do all the n-1 crap.
Oh the formula is √((∑x^2-n(the mean)^2)/(n-1)). Do yo mean 1/6(n+1)(2n+1) or does the square bracket represent something different?
4. (Original post by XShmalX)
Oh the formula is √((∑x^2-n(the mean)^2)/(n-1)). Do yo mean 1/6(n+1)(2n+1) or does the square bracket represent something different?

Yeah that's what I meant. I forgot to but the square bracket at the end.

(n+1)(2n+1) all over 6.

It's in the formula booklet, if you're doing A levels.
5. (Original post by AnonyMatt)
Yeah that's what I meant. I forgot to but the square bracket at the end.

(n+1)(2n+1) all over 6.

It's in the formula booklet, if you're doing A levels.
cool, I thought so. I still can't seem to get to √(n-1) though..
6. bump :/
7. I end up with √((2n-1)/3)
9. Right.

I dunno if this will help, but let's have a go.

Var(X) = (Ex^2)/(n-1) - (E^2x)/n(n-1)

= [n(n+1)(2n+1)]/6(n-1) - 1/(n-1)
= (2n^3 + 3n^2 + n)/6(n-1) - 1/(n-1)
= (2n^3 + 3n^2 + n - 6)/6(n-1)
= (2n^2 + 5n + 6)/6

Yeah I dunno where I'm going with this. It's irritating me now.

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Updated: January 26, 2010
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