Now, lets talk about an example of obtaining a population expected value. So suppose a coin is flipped and X is declared 0 or 1 corresponding to a head or a tail. What is the expected value of X? So again, the expected value is the property of the population. If we plug into our formula expected value of X, is the probability 0.5 times the value 0 plus the probability 0.5 times value 1, so all the values that the coin can take 0, 1 times the probability that it can take them, added up. This winds up to be 0.5. So notice the expected value is a value that the coin can't even take. However, if we were to think about this geometrically, the answer is quite obvious. If we have two bars, they are equal height, one at 0 and one at 1. And we wanted to balance them out. It's clear where exactly we'd put our finger if we wanted to. It would be right here at 0.5. Suppose that a random variable X is such that P, the probability that it takes the value 1 as p, and the probability it takes the value 0 is 1 minus p. This is a biased coin, where the probability of the head is not necessarily 0.5. The probability of the head is now p. What is it's expected value? We'll plug in directly and then formula 0 times 1 minus p plus 1 times p which works out to be p. So the expected value of a coin flip, even a potentially biased coin is exactly the true long run proportions of heads you would obtain in infinitely many flips of the coin. What about a die? Suppose that a die is rolled, and X is the number that is face up, what is the expected value of X? So, here we take the values 1, 2, 3, 4, 5, 6 and multiply them the, times the probability that the random variable takes them. One-sixth, one-sixth, and so on. You get 3.5. Again, a number that the die can't actually obtain. So again, the geometric argument makes this obvious. We have six bars, all of height one-sixth, and if we had to balance them out, it's clear that we would balance them out at 3.5.