# Continuous Uniform Distribution Problem

Let $X$ be a continuous random variable on the interval $(a,b)$. The mean of $X$ is $800$ and the variance of $X$ is $120,000$. Calculate the range of $(a,b)$.

My default approach was to proceed as follows:

$$E(X) = \frac{(a+b)}{2} = 800 .$$

$$Var(X) = \frac{(b-a)^2}{12} = 120,000.$$

Therefore

$$\frac{Var(X)}{E(X)} = \frac{2(b-a)}{12} = 150,$$

and so $b-a = 900$.

However, the solutions in the back of the book take an alternative approach, and attain different results.

$$b + a = 1,600.$$

Therefore

$$\frac{(1600 - 2a)^2}{12} = 120,000 ,$$

and so

$$4a^2 -6,400a +1,120,000 = 0 .$$

Solving with the quadratic equation we get $a = 200$ and $b = 1,400$, giving a range OF 1,200 $. Can anyone see why we get diferent results? ## 1 Answer Your mistake is in the computation of $$\frac{\operatorname{Var}(X)}{E(X)} = \frac{(b - a)^2 / 12}{(a + b)/2} = \frac{2(b - a)^2}{12 (a + b)}$$ Had it been$b^2 - a^2$, you could simplify this quotient. But as it stands, this does not simplify to$b - a\$.

• Ah of course. There goes an hour and a half I'll never get back. – cfairwea Apr 11 '16 at 20:48