Maths Problem - OcUK Rule!

badgermonkey

badgermonkey

badgermonkey

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Oke, so my mate (it's not me, you all know I'm doing a computer degree, not A-Level maths :p ) needs a bit of help answering a maths question.

The letter u is being used to represent mew (the mean) and the o- is used to represent the standard deviation. So sorry about how odd it looks ;)

Anyway:

Suppose that a list of data-points X1, X2,....Xn has mean u (mew) and standard deviation o-. Prove that the set of data points Y1, Y2,...Yn is given by calculating:

Yi = ( Xi - u ) / o- for each i from l



to n, has mean 0 and standard deviation 1.


u = mean
o- = standard deviation
Click to expand...

Can anyone help? Unfortunately I did mechanics not stats in 6th form :(
 
I did stats(+data analysis last year at uni) but I dont recall to have ever been asked to prove the calculation for Yi. We just had it as sort of given in order to find certain probabilities in hypothesis tests. This has always applied to normal distributions i.e. where u=0 and o-=1.

If I get the time later, ill try to think of a way of proving it but,as I said, I dont recall to have ever been asked to do that.
 
All you have to do is find the mean and standard deviation of the Y_is using the usual formulae. I'll do the mean and leave the standard deviation as an exercise :p

I'm going to use m (mu) for the mean and s (sigma) for the std dev of the X_is.

mean of Y_i = Sum[Y_i, i = 1..n]/n = Sum[(X_i - m)/s, i = 1..n]/n
= (Sum[X_i, i = 1..n] - nm)/ns = 0 since m = Sum[X_i, i = 1..n]/n.

Now, you can do a similar thing for the std dev of the Y_is.

I hope that helps.
 
Its simple definitions. I'll use u for the mean and s for the standard deviation.The mean of Xi is

mean(Xi) = SUM(Xi)/n = u

and the variance (the square of the standard deviation is)

var(Xi) = SUM(Xi^2)/n - (SUM(Xi)/n)^2 = SUM(Xi^2)/n - u^2 = s^2

You can then easily work out:

mean(Yi) = mean[(Xi - u)/s] = 1/s * mean(Xi - u) = 1/s * (mean(Xi) - u) = 1/s * (u - u) = 0.

var(Yi) = var[(Xi - u)/s] = SUM[(Xi-u)^2/s^2]/n - {SUM[(Xi - u)/s]/n}^2 = 1/s^2 * SUM[(Xi - u)^2]/n - 1/s^2 * {SUM(Xi) - u}^2 = 1/s^2 * SUM[Xi^2 - 2nuXi + (nu)^2]/n = 1/s^2 * {SUM[Xi^2] - 2uSUM[Xi] + u^2} = 1/2^2 * {SUM[Xi]^2 - u^2} = 1/s^2 * s^2 = 1.

and the standard deviation is the (positive) square root of the variance, so stdev(Yi) = 1.

Apologies for the awful notation!
 
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