How would I solve a problem like this
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How would I solve a problem like this

[From: ] [author: ] [Date: 11-08-07] [Hit: ]
-if the standard deviation is 5 the variance is25sincevariance issd^2variance = sum (mean - x)^225 = sum ( mean - x)^2since thepoints are not picked we are free to pick an average since5 is the standarddeviation pick a fairly high meanthat allows 3 standard deviations below it so3*5 = 15 sowe want MAYBE a number over 15 say 25 for easemean is 25one of the points can be 25 so we only needa few more 2 on one side and 2 on the other sqrt(((x1-25)^2 + (x2 - 25)^2+ 0^2 + (x^4 - 25)^2 + (x^5- 25)^2)/5)two of the numbers can be equal distance but opposite sides OF THE MEANto simplifythis has to equal 5 = sqrt(( 2*(x1- 25)^2 + 2*(x2 - 25)^2)/5)25 = (2*(x1-25)^2 + 2*(x2-25)^2)/5125 = (2*(x1 - 25)^2 + 2*(x2 - 25)^2divide by 262.5= (x1 - 25)^2 + (x2 - 25)^2we also knowmeanhas to be 25 which we chose25 = (x1 + x2 + 25 + x3 + x4)/5125 = x1 + x2 + 25 + x4 + x5100 = x1 + x2 + x4 + x50 = x1-25 + x2-25 + x4-25 + x5-25here i subtracted 100 from both sidesbut split the 100 on the right into 4 meansjust to see if the means would work outand the mean would automatically work as long as you pickone at the mean andtwo points on equal sides of the mean62.5= (x1 - 25)^2 + (x2 - 25)^2lets pick x1 = 18 then lets pick i switched the order of the number to pick a lower numberthan 25(25-x2)^2 = 62.5 - 4925-x2 = sqrt(13.5)x2 = 25 - sqrt(13.5)x2 = 21.......
I've already done the first two that were similar, but only through trial and error. I've attempted this next one, but it's never right!

Create a set of at least five data points such that the standard deviation is exactly 5.

Is there a systematic way to solve through problems like that?? Help, please!

-
if the standard deviation is 5

the variance is 25
since variance is sd^2

variance = sum (mean - x)^2

25 = sum ( mean - x)^2

since the points are not picked we are free to pick an average since 5 is the standard deviation pick a fairly high mean that allows 3 standard deviations below it so

3*5 = 15 so we want MAYBE a number over 15 say 25 for ease


mean is 25

one of the points can be 25 so we only need a few more 2 on one side and 2 on the other

sqrt(((x1-25)^2 + (x2 - 25)^2 + 0^2 + (x^4 - 25)^2 + (x^5- 25)^2)/5)

two of the numbers can be equal distance but opposite sides OF THE MEAN to simplify

this has to equal 5 = sqrt(( 2*(x1- 25)^2 + 2*(x2 - 25)^2)/5)

25 = (2*(x1-25)^2 + 2*(x2-25)^2)/5

125 = (2*(x1 - 25)^2 + 2*(x2 - 25)^2

divide by 2

62.5 = (x1 - 25)^2 + (x2 - 25)^2



we also know mean has to be 25 which we chose

25 = (x1 + x2 + 25 + x3 + x4)/5

125 = x1 + x2 + 25 + x4 + x5

100 = x1 + x2 + x4 + x5

0 = x1-25 + x2-25 + x4-25 + x5-25

here i subtracted 100 from both sides but split the 100 on the right into 4 means
just to see if the means would work out and the mean would automatically work as long as you pick one at the mean and two points on equal sides of the mean


62.5 = (x1 - 25)^2 + (x2 - 25)^2
lets pick x1 = 18

then lets pick i switched the order of the number to pick a lower number than 25
(25-x2)^2 = 62.5 - 49

25-x2 = sqrt(13.5)

x2 = 25 - sqrt(13.5)
x2 = 21.32

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