How would I solve a problem like this

[From: ] [author: ] [Date: 11-08-07] [Hit: ]

5)-25)^2 + (25 + 7-25)^2 + (25 - 7-25)^2 + (25-25)^2)we see that the variance = (13.5+ 13.5 + 7^2 + 7^2)/5 5^2=125/5so thats true thisworked out the trick was to know that whendoing the final square root the value inside could be(25 - x2)^2 or (x2 - 25)^2 but since we were picking a number smaller than 25 for x2 the way for me to ensure this was to make the equation inside the square as shown.(25 - x2)^2that made themath pick a number lower than 25 not higher unless it was a fluke it seems you can pick randomlyone number for the second point and thenplug it into the equation for the variance simplified and getanother valid number that will makethe standard deviation the value you wish .........

lets see if the mean is 25
our numbers would be +/- sqrt(13.5) from mean and +/- the chosen number 7 from mean
and 25 of course the mean for the 5th value

mean = ((25 + sqrt(13.5)) + (25-sqrt(13.5)) + (25 + 7) + (25 - 7) + 25)/5

we see that the added values cancel so we get

mean = (25 + 25 + 25 + 25 + 25)/5

so mean = 25

we got our mean lets confirm we got the sd too

variance = ((25 + sqrt(13.5) - 25)^2 + (25-sqrt(13.5)-25)^2 + (25 + 7-25)^2 + (25 - 7-25)^2 + (25-25)^2)

we see that the variance = (13.5 + 13.5 + 7^2 + 7^2)/5

5^2 = 125/5

so thats true

this worked out the trick was to know that when doing the final square root the value inside could be (25 - x2)^2 or (x2 - 25)^2 but since we were picking a number smaller than 25 for x2 the way for me to ensure this was to make the equation inside the square as shown.

(25 - x2)^2 that made the math pick a number lower than 25 not higher

unless it was a fluke it seems you can pick randomly one number for the second point and then plug it into the equation for the variance simplified and get another valid number that will make the standard deviation the value you wish ...

and since you are picking equal number of values on either side of the mean plus the mean for the 5 points the difference cancel and produce a mean automatically

if the number of points had been even you would have chosen ONLY the points on either side of the mean again the math would have worked out the rest and you would have your 4 points say ..

hope this is not too confusing I wanted to be VERY sure it worked out and It worked fantastically ...

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If you know the standard deviation formula, you will see the data points deviate from the mean ... you have 5 data points, but you don't know the mean...you can enter what you know into the formula, but you will still end up solving it by trial and error with the limited information you have...

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sd^2 = sigma(Xi - Xbar)^2 over n
Let n = 6(say) and Xbar = 50(arbitrary). Then,
5^2 = sigma(Xi - 50)^2 over 6
sigma(Xi - 50)^2 = 150
Now, split 150 into six non-negative parts (two each parts may be equal), say, 25,25, 49, 49, 1 and1. Their square roots are 5, -5, 7,-7, 1 and -1.
Now the required numbers are
55,45, 57, 43, 51 and 49 (mean is 50).
They can as well be
40, 30, 42, 28, 36 and 34 (mean is 35)

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I wonder if you could pick any four points and solve for the 5th.

x1 = 1 x2 = 2 x3 = 3 x4 = 4 Average = 2 + x5/5
25*4 = 100
100 = (1 - (2 + x5/5) )^2 etc. It looks pretty cumbersome.

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