Standard Deviation of Uniform DistributionDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/