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So I am studying for a final, and I am doing this problem that says: The radius of a sphere is a random number between $2$ and $4$. What is the expected value of its volume? My first thought was that since were given the probability distribution of the radius $\left(\text{which is}\frac12\right)$ we would just plug $\frac12$ into the radius value of the volume of a sphere formula. But I don't think this is right. Am I doing something wrong?

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First, how does the radius being between $2$ and $4$ lead you to $r=1/2$ as some kind of expected value for $r$? Second, do you assume that the distribution of $r$ over $[2,4]$ is uniform? Third, if so then you would answer this by integrating the volume over $r$ in $[2,4]$ and dividing by $\|[2,4]\|$. – alex.jordan Dec 21 '12 at 2:35
@alex.jordan I was saying that the probability distribution of the radius was $\frac12$ because the distribution of it is indeed uniform, it's coming from the section of the text that covers uniform distribution. I should have mentioned that, my appologies. – TheHopefulActuary Dec 21 '12 at 2:43
Kyle, if the formula was linear in $r$, you could get the expected value by setting $r$ to the average radius ($3$), but here larger radii correspond to much larger volumes, so substituting the average radius will significantly underestimate the average volume. – copper.hat Dec 21 '12 at 3:15
up vote 4 down vote accepted

Expected values (almost) always correspond to sums or integrals, depending on whether the random variable is discrete or continuous. In this case, the random variable (the radius) is continuous--it can take on any value between 2 and 4--so it's going to be an integral. The form of the integral is $$\mathbb E[g(X)] = \int_{-\infty}^{\infty} g(x) f(x) \; dx,$$ where $g(X)$ is an arbitrary function of the random variable $X$ and $f(x)$ is the density function. In this case, $g(x)$ is the formula for the volume of a sphere, $$g(x) = \frac43 \pi x^3,$$ and $f(x) = \frac12$ between 2 and 4, as you noted, and zero everywhere else. So the expected value is going to be $$\mathbb E[g(X)] = \int_2^4 \frac43 \pi x^3 \times \frac12 \; dx,$$ which you should be able to solve.

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Oh duh... Darn it I don't know how I missed that. I guess the end of finals week really drains a person! Thanks for the reminder! – TheHopefulActuary Dec 21 '12 at 2:45
And the answer is $40\pi$. – copper.hat Dec 21 '12 at 2:59

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