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?
Answers
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?
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
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.