Question 639183
Start with the set 



20,20,20,20,20



That has a mean of 20, but a standard deviation of 0. Then increase one value to get 21, but decrease another value (to balance things out) to get 19.


So we might have


19, 20, 20, 20, 21


Now find the standard deviation. It will be roughly 0.70710678 (use a calculator here). 


Repeat the last 3 steps to find another standard deviation.


We can keep going and do this by trial and error, but there's a much easier way


This way involves using the definition of sample standard deviation.



sigma = sqrt(sum((x-mu)^2)/(n-1))


6 = sqrt(sum((x-mu)^2)/(n-1))


6^2 = sum((x-mu)^2)/(n-1)


36 = sum((x-mu)^2)/(n-1)


36 = sum((x-mu)^2)/(5-1)


36 = sum((x-mu)^2)/4


36*4 = sum((x-mu)^2)


144 = sum((x-mu)^2)


sum((x-mu)^2) = 144


(x1-mu)^2+(x2-mu)^2+(x3-mu)^2+(x4-mu)^2+(x5-mu)^2 = 144


(x1-20)^2+(x2-20)^2+(x3-20)^2+(x4-20)^2+(x5-20)^2 = 144


(20-x-20)^2+(20-20)^2+(20-20)^2+(20-20)^2+(20+x-20)^2 = 144


(20-x-20)^2+(0)^2+(0)^2+(0)^2+(20+x-20)^2 = 144


(-x)^2+0+0+0+(x)^2 = 144


x^2 + x^2 = 144


2x^2 = 144


x^2 = 144/2


x^2 = 72


x = sqrt(72)


x = 8.48528137423857


So the data set of 


20,20,20,20,20

becomes


20-8.48528137423857,20,20,20,20+8.48528137423857


11.5147186257614,20,20,20,28.48528137423857


So the data set


11.5147186257614,20,20,20,28.48528137423857


has a mean of 20 and a standard deviation of (roughly) 6


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If you want to be exact, then the data set


20-sqrt(72), 20, 20, 20, 20+sqrt(72)


has a mean of 20 and a standard deviation of exactly 6