If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:8:07
AP.STATS:
VAR‑5 (EU)
,
VAR‑5.E (LO)
,
VAR‑5.E.2 (EK)
,
VAR‑5.E.3 (EK)

Video transcript

so we've defined two random variables here the first random variable X is the weight of the cereal in a random box of our favorite cereal Mathies or random closed box of our favorite cereal Matthies and we know a few other things about it we know what the expected value of X is it is equal to 16 ounces in fact we tell it to us on a box they say you know net weight 16 ounces now when you see that on a cereal box it doesn't mean that every box is going to be exactly 16 ounces remember you have a discrete number of these flakes in here they might have slightly different densities slightly different shapes depending how they get packed into this volume so there is some variation which we can measure with standard deviation so the standard deviation let's just say for the sake of argument for the random variable X is zero point eight ounces and just to build our intuition a little bit later in this video let's say that this the random variable X is always stays constrained within a within a range that if it goes above a certain weight or below a certain weight then the company that produces it just throws out that box and so let's say that our random variable X is always greater than or equal to fifteen ounces and it is always less than or equal to 17 ounces just for argument this will help us build our intuition later on now separately let's consider a bold we're always going to sitt consider the same sized bowl let's consider this a 4 ounce bowl because the expected value of y if you took a random one of these bowls always the same Bowl and if or if you took the same bowl and you someone filled it with Mathies there's expected weight of the Mathies in that bowl is going to be 4 ounces but once again there's going to be some variation depends who filled it in how it packed in did they shake it before while they were filling it there could be all sorts of things that could make some variation here and so for the sake of argument let's say that variation can be measured by standard deviation and it's 0.6 ounces and let's say whoever the bowl fillers are they are also they don't like bowls that are too heavy or too light and so they'll also throw out bowls we can say that why can its maximum value that'll ever take on is five ounces and the minimum value that it could ever take on let's say it is three ounces so given all of this information what I want to do is let's just say I take a random box of Mathies and I take a random filled Bowl and I want to think about the combined weight in the closed box and the fill Bowl so what I want to think about is really X plus y I want to think about the sum of the random variables so in previous videos we already know that the expected value of this is just going to be the sum of the expected values of each of the random variables so it would be the expected value of X plus the expected value of y and so it would be 16 plus 4 ounces in this case this would be equal to 20 ounces but what about the variation can we just add up the standard deviations if I want to figure out the standard deviation of X plus y how can I do this well it turns out that you can't just add up the standard deviations but you can add up the variances so it is the case that the variance of X plus y is equal to the variance of X plus the variance of Y and so this is going to have an X right over here X and then we have plus y and our Y and actually both of these assume independent random variables so it assumes assumes x and y are independent I'm going to write it in caps in a future video I'm going to get a give you a hopefully a better intuition for why this must be true that they're independent in order to make this claim right over here I'm not going to prove it in this video but we could build a little bit of intuition here for each of these random variables we have a range of two ounces over which this random variable can take and that's true for both of them but what about this sum well this sum here could get as high as so let me write it this way so X plus y X plus y what's the maximum value that it could take on well if you get a heavy version of each of these then it's going to be seventeen plus five so this has to be less than 22 ounces that's going to be greater than or equal to well what's the lightest possible scenario well you could get a 15 ounce er here and a three answer here and it is 18 ounces and so notice now the variation for the sum is larger we have a range that this thing can take on now of four while the range for each of these was just two or another way you could think about it is these upper and lower ends of the range are further from the mean then these upper and lower ends of the range were from their respective means so hopefully this gives you an intuition for why this makes sense but let me ask you another question what if I were to say what about the variance what about the variance of X minus y what would this be would you would you subtract the variances of each of the random variables here well let's just do the exact same exercise let's take X minus y X minus y and think about it what would be the lowest value that X minus y could take on well the lowest value is if you have a low X and you have a high Y so it'd be 15 minus 5 so this would be 10 right over here that would be the lowest value that you could take on and what would be the highest value well the highest values if you have a high X and a low Y so 17 minus 3 is 14 so notice just in the case of just as if we saw in this case of the sum even in the difference your variability seems to have increased this is still going to be this the end of the extremes are still further than the mean of the difference the mean of the difference would be 16 minus 4 is 12 these these extreme values are 2 away from the 12 and this is just to give us an intuition once again it's not a rigorous proof so it actually turns out that in either case when you're taking the variance of X plus y or X minus y you would sum the variance assuming X and y are independent variables now with that out of the way let's just calculate the standard deviation of X plus y well we know this let me just write it using this Sigma notation so another way of writing the variance of X plus y is to write the standard deviation of X plus y squared and that's going to be equal to the variance of X plus the variance of Y now what is the variance of X what's the standard deviation of x squared 0.8 squared this is zero point six four zero point six for the standard deviation of Y is zero point six use square to get the variance that's zero point three six you add these two up and you are going to get one so the variance of the sum is one and then if you take the square root of both of these you get the standard deviation of the sum is also going to be one and that just happened to work out because we're dealing with the scenario where the variance where the square root of one is well one well so this hopefully builds your intuition whether we are adding or subtracting two independent random variables the variance of that sum or the difference the variability will increase in the next video we'll go into some depth talking about getting an intuition for why independence is an important condition for making this statement this claim
AP® is a registered trademark of the College Board, which has not reviewed this resource.