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Q: determine a required future variation to achieve a goal ( No Answer,   2 Comments )
Subject: determine a required future variation to achieve a goal
Category: Science > Math
Asked by: baz229-ga
List Price: $20.00
Posted: 19 May 2006 11:45 PDT
Expires: 18 Jun 2006 11:45 PDT
Question ID: 730450
how do i determine a required future variation given a to date mean &
variation and a known number of future parts?

for example 
.... say I have 20 parts with mean 22.4 units and standard deviation 7.33 units

.... I need to achieve a total goal of better than a mean of 25 AND std dev of 5

.... I have the opportunity to achieve this with 28 more parts (ie
overall 48 parts) ... I know their mean needs to be < 27.5 but what
does their variation need to be & how is this calcualted?


Clarification of Question by baz229-ga on 19 May 2006 11:51 PDT
OK to assume my data is normally distributed to an acceptable
confidence level ... the 20 parts ... and the 28 to come
There is no answer at this time.

Subject: Re: determine a required future variation to achieve a goal
From: berkeleychocolate-ga on 20 May 2006 12:11 PDT
Let X(i) (i=1,2,...,20) be the numbers you already have and Y(i)
(i=1,2,...,28) be the new numbers. So the mean of X is 22.4 and the
standard deviation of X is 7.33.

The sum of X and Y must be 48*25 = 1200 to make the overall mean 25.
Since the sum of X is 20*22.4 = 448, the sum of Y must be 1200 - 448 =
752, which makes the mean of Y equal to 752/28 = 26.86.

Also the standard deviation of X is given to be 7.33. This means 53.73
= 7.33^2 = the [sum of X^2 - (mean of X)^2/20]/20 = [sum of X^2 -
22.4^2/20]/20 . (This is a common formula for the standard deviation.
Here I'm assuming this is a "population mean". If it's a "sample mean"
divide by 19 instead of 20.) So 1074.6 = 53.73*20 = sum of X^2 - 
25.09 . So the sum of X^2 = 1099.7.

We want the standard deviation of X and Y to be 5. So 25 = 5^2 = [sum
of X^2 +sum of Y^2 - 27.5^2/48 ]/48 by the same formula. So 1200 =
25*48 =  1099.7 +sum of Y^2 - 15.76 . So the sum of Y^2 = 116.06 .

Once more using the same standard deviation formula, we get the
standard deviation of Y is sqrt[(sum of Y^2 - (mean of Y)^2/28)/28 ]=
sqrt [ (116.06 - 26.86^2/28 )/28 ] =  2.03 .
Subject: Re: determine a required future variation to achieve a goal
From: berkeleychocolate-ga on 06 Jun 2006 17:57 PDT
I made a common error in the standard deviation formula. Sorry about
that. The corrected formula for a standard deviation is sd^2 = (sum of
squares - n*average^2)/n . The "n" in the formula was in the wrong
place. Making these corrections, I get (check me): sum of X^2 =
11109.8 . This forces sum of Y^2 to be 20090.2 if the overall sd is to
be 5. But then the sd of Y is imaginary since sum of Y^2 is less than
20200.87 which is 28*(26.86)^2.

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