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Answer:
The radius of a sphere is a random number between 2 and 4. What is the expected value of its volume? What is the probability that its volume is at most 36π?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
E[g(X)]=∫∞−∞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)=43πx3,
and f(x)=12 between 2 and 4, as you noted, and zero everywhere else. So the expected value is going to be
E[g(X)]=∫4243πx3×12dx,
which you should be able to solve.