 This is a video about the mean and variance of a Poisson distribution. Suppose that x has the Poisson distribution with parameter lambda. Remember that what lambda represents is the expected value of x, the expected number of events in the particular interval of time or space. So the mean value of x, the expected value, is equal to lambda by definition. But it's also true that the variance of x is equal to lambda. So for the Poisson distribution, both the mean and the variance are equal to lambda. For example, suppose that x has the Poisson distribution with parameter 5, then the mean will be 5, and the variance will also be 5, which means that the standard deviation, which is the square root of the variance, will be the square root of 5. Let's look at a couple of examples. My first example would be to do with radioactive decay. Now over a long period of time, the rate at which a radioactive substance decays will slow down. But if we just look at a short period of time, the rate at which it decays will stay roughly constant, and so we can use the Poisson distribution to model radioactive decay. I'm going to look at an example involving uranium-238. One mole of uranium-238 emits about 3 million alpha particles per second. Assuming that the mass of one mole of uranium-238 is exactly 238 grams, let's find the mean and the variance for the number of alpha particles emitted by 50 grams of uranium-238 in 10 seconds. Well the first thing to notice is that the question is telling us the rate of emission for 238 grams. We're told that 238 grams of uranium emit 3 million alpha particles per second. But we need to know about 50 grams of uranium-238. So obviously the rate of decay for 50 grams will be lower than the rate of decay for 238 grams. In fact, the rate of emission for 50 grams of uranium is going to be 50 over 238 times 3 million per second, which is 150 over 238 times 10 to the power of 6 emissions per second. Secondly, this is the rate of emission per second, and we're interested in an interval of 10 seconds. So the mean number of emissions in a 10 second period will be this number times 10. 150 over 238 times 10 to the power of 6 times 10, which is 150 over 238 times 10 to the power of 7, which is about 6.3 times 10 to the power of 6. I've given the answer to two significant figures because the information in the question seemed fairly approximate. Okay, well this is the mean number of emissions, and the variance for the number of the emissions will be the same. It will also be 6.3 times 10 to the power of 6. My second example is to do with hockey. Suppose that the number of goals in a hockey match follows a Poisson distribution, where the mean number of goals is four and a half. Let's find the probability that the number of goals in a particular match exceeds the mean by at least one standard deviation. Okay, well in this situation, X, the number of goals in a match, follows the Poisson distribution with parameter 4.5. And the question is asking us for the probability that X is greater than or equal to mu plus sigma. Well mu is 4.5, and sigma is the square root of 4.5. So this is actually asking us for the probability that X is greater than or equal to 4.5 plus the square root of 4.5. And if you work that out, that's approximately equal to 6.62. Now obviously X can't equal 6.62 because it can only be a whole number. So this is really the same as the probability that X is greater than or equal to 7. Okay, now we'll want to use the tables to work this out. And remember that the tables only tell us the probability that X is less than or equal to something. So we need to realize that the probability that X is greater than or equal to 7 is the same as 1 minus the probability that X is less than or equal to 6. And this is because being 7 or more is the opposite of being 6 or less. Okay, well this is something that we can look up in the tables. We find where lambda is equal to 4.5. And then we look along the row where X is equal to 6 until we see the probability 0.8311. So the sum that we have to do is 1 take away 0.8311 which is equal to 0.1689. Okay, this is the end of my little video about the mean and variance of a Poisson distribution. Remember that if X has the Poisson distribution with parameter lambda, then both the mean and the variance are equal to lambda. Thank you very much for watching.