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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?

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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 $b - a$.

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

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