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?
[Math] How to find the expected volume of a sphere
probabilityprobability theory
Best Answer
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.