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Standard Deviation of Uniform Distribution


Date: 02/16/98 at 23:51:16
From: Darin Parker
Subject: Std. deviation of uniform distribution

My problem is not in the calculation of the standard deviation of a 
uniform distribution, but in where they got the formula.  I find it 
easier for me to learn if I understand the formulas as well as knowing 
where to plug in the numbers.  

In the equation Std. deviation = (b-a)/ square root of 12, where did 
the square root of 12 come from? 
 
I have looked through every statistics text I have, but they all give 
the formula with no explanation for where the square root of 12 came 
from.  My professor doesn't even know.  I would greatly appreciate any 
insight you can give me on the subject.

Thank you,
Darin Parker


Date: 02/17/98 at 15:52:58
From: Doctor Anthony
Subject: Re: Std. deviation of uniform distribution

The uniform distibution on the interval from a to b is given by

            f(x) = 1/(b-a)   for a < x < b

                 = 0  elsewhere.


            E(X) =  INT(from a to b)[x.dx/(b-a)]

                 =  (1/(b-a)) x^2/2 from a to b

      b^2 - a^2      (b+a)
     ----------  =  ------
       2(b-a)          2


          E(X^2) =  INT(from a to b)[x^2.dx/(b-a)]

                 =  (1/(b-a)) x^3/3  from a to b

                     b^3 - a^3      b^2+ab+a^2
                 =  ----------  =  ------------
                      3(b-a)            3

          Var(X) =  E(X^2) - [E(X)}^2

                    b^2+ab+a^2       b^2 + 2ab + a^2
                 =  -----------  -   --------------- 
                         3                  4

                    4b^2 + 4ab + 4a^2 - 3b^2 - 6ab - 3a^2
                 =  ---------------------------------------
                                      12

                     b^2 -2ab + a^2
                 =  ----------------
                           12

                     (b-a)^2
                 =  --------
                       12

                      (b-a)
and so the s.d. is   --------
                     sqrt(12)


You can do this without all the algebra if you work on the interval 
0 to 1.

     E(X) = INT[x.dx]    = x^2/2  from 0 to 1

          = 1/2

   E(X^2) = INT[x^2.dx]  = x^3/3  from 0 to 1

          = 1/3

   Var(X) = 1/3 - 1/4

          = 1/12

     s.d. = 1/sqrt(12)   = length of interval/sqrt(12)
        

-Doctor Anthony,  The Math Forum
Check out our web site http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

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