Given a uniform distribution, this means that the probability density function is going to be a constant over the range of possible values. You also know that the integral of the pdf over the entire range of possible values must be 1. So:
f(x)={c where a≤x≤b, else 0}
[a, b]∫c dx=1
cx|[a, b]=1
cb-ca=1
c(b-a)=1
c=1/(b-a)
So for the range 0 to 1, your pdf would be f(x)={1 where 0≤x≤1, else 0}. And for the range 0 to 5, you have f(x)={1/5 where 0≤x≤5, else 0}
Now you find the mean, which is the integral of x*f(x) over the range of possible values. So:
E(x)=[a, b]∫x/(b-a) dx
1/(b-a) [a, b]∫x dx
1/(2b-2a) (x²|[a, b])
1/(2(b-a)) (b²-a²)
(b-a)(b+a)/(2(b-a))
(b+a)/2
(This result should be thoroughly unsurprising, since it simply states that the mean of a uniform distribution between two points is the average of those two points)
Now you remember the formula for variance:
σ²=E((x-E(x))²)
There is a simple computational formula for this, which you may have memorized, but which I will derive here:
E((x-E(x))²)
expand the square:
E(x²-2xE(x)+E(x)²)
Use linearity of expected value:
E(x²) - E(2xE(x)) + E(E(x)²)
Using linearity of expected value and the fact that expected value of a constant is that constant (in particular, E(E(x)²)=E(x)²):
E(x²) - 2E(x)E(x) + E(x)²
Simplifying:
E(x²)-E(x)²
So now we must find E(x²):
[a, b]∫x²/(b-a) dx
1/(b-a) [a, b]∫x² dx
1/(b-a) (x³/3 |[a, b])
1/(3(b-a)) (b³-a³)
(b-a)(b²+ab+a²)/3(b-a)
(b²+ab+a²)/3
Now, using the formula for variance:
σ²=(b²+ab+a²)/3 - ((b+a)/2)²
σ²=(b²+ab+a²)/3 - (b²+2ab+a²)/4
σ²=(4(b²+ab+a²) - 3(b²+2ab+a²))/12
σ²=(b²-2ab+a²)/12
σ²=(b-a)²/12
Now, the standard deviation is just the square root of that:
σ=(b-a)/√12
So for the range 0 to 1, the standard deviation is just 1/√12. And for 0 to 5, it is 5/√12. You can go ahead and memorize this formula if you want, but I think the process used to derive it will be more useful to you in the long run.
2007-02-07 06:19:05
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answer #1
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answered by Pascal 7
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