Discrete Random Variables - Example

okay so in this video we will consider an example of a random variable construct a probability distribution table its histogram find its mean its variance and its standard deviation so here's the experiment consider that we have an urn that contains three blue balls and five red balls and the experiment is we will choose from this urn randomly three balls and well what is our random variable let's say here that X our random variable is the number of blue balls selected so X is obviously here a random variable Y well if you choose three balls from this urn randomly the number of blue balls that you have will vary right sometimes you may get 0 blue balls sometimes you may get one other times two other times three and each time you have a number the number of blue balls selected so here X is a discrete random variable the first thing we construct or at least let's state what are the possible values for the random variable since we are choosing three balls at random and there are three blue balls well X could be zero so maybe none of our balls are blue maybe exactly one or exactly two or exactly three so there are four possible values here for our random variable and always remember use uppercase X for the random variable use lowercase X with the actual values that the random variable can take and all we need now is to build what's called the probability distribution of X and once we have this everything will be done very easily from this distribution table and what a probability distribution is is simply an assignment of our abilities so the probability distribution of X well we will assign to each value of x its corresponding probability and of course lowercase X contains the actual values that the variable can take so X can be 0 or it can be 1 or it can be 2 or it can be 3 and to each possible value the X can take we assign its probability so what is the probability that the random variable equals the corresponding x value so for each value let's find the corresponding probability let's do it here on the side so first probability that x equals 0 well X is the number of blue balls and we're asking if we choose three balls at random what is the probability that the number of blue balls is zero well here look at the experiment the set that we have is without replacement once we pick a ball from the urn we don't put it back when we pick the second so it is without replacement and we do not care about the order of the selection all we are interested in is out of the three balls selected how many are blue how many are red so we have a setup that is without replacement and unordered that is the setup of combination and because each selection is equally likely we can use the equiprobable ax T theorem so and how many ways can we choose zero blue balls well out of the three blue balls we choose none and x because we are choosing three balls in total out of the five red balls we choose three over the total number of possible selections the size of the sample space well out of eight balls three plus five is eight out of eight balls we are choosing three and if you compute the numerator and denominator what you have is ten out of 56 and now there you have it the probability x equals zero hence that the number of blue balls is equal to zero is 10 out of 56 and if you compute a decimal for this you will get approximately zero point one seven nine let's keep going the probability now that X is equal to one so the number of blue balls is equal to one well this means out of the three blue balls we choose one and therefore out of the five red we choose two over again the total number of possible selections a truce 3 and if you compute this you'll get third out of 56 if you compute this approximately you'll have point five three six let's keep going what is the probability that x equals two that the number of blue balls selected is equal to two out of the three blue balls we choose to hence out of the five red we choose one over the total number of possibilities a true is three the exact fraction is fifteen out of 56 and you find the decimal expansion you'll get approximately point two six eight and finally what is the probability that x equals three so the number of blue balls is equal to three well out of the three blue balls we choose three and out of the five red we choose none over the total number of possibilities eight root 3 this gives us one out of 56 and that's approximately zero point zero one seven nine and now we have the probability distribution of X for each of its possible value we have the corresponding probability and if you notice if you add all these up you get 10 plus 30 40 plus 15 45 a 55 sorry plus 156 56 out of 56 so the total is equal to one so when you build your probability distribution since you will consider all possible values of x when you add the probability of each case it has to add up to one because you're adding over all possibilities if doesn't happen to add up to one you've made a mistake somewhere so double check your work and now we have the probability distribution of X from this we can easily find the mean the variance and standard deviation let's find the mean first so the mean is the expectation of X and if you remember all we have to do is sum over all possible values of X the value times the probability that the random variable equals is given value well the first term is x equals 0 so it's 0 times its corresponding probability 10 out of 56 plus when X is 1 1 times its probability 30 out of 56 plus when X is 2 2 times its probability 15 out of 56 plus finally the third variable the throw the value sorry 3 times its probability one out of 56 and here when you compute this use the fraction command on your calculator to get the exact value as a fraction and if you do you will get that the mean is equal to 63 out of 56 right 0 plus 30 plus 30 is 60 plus 3 63 oops sorry it's not 1 out of 15 it's 1 out of 56 right 3 times 1 over 56 3 times 1 over 56 if you add those up exactly again you will find that the mean the average value of our random variable is 63 out of 56 and if you find the decimal expansion it happens to be exactly 1.125 but keep you as a fraction for now so think of what this is saying mu is the mean the expectation the average value of x so if you say out of this urn where there are three blue balls and five red balls if I choose three balls at random and X is the number of blue balls what do I expect X to be equal to well you should expect X to be equal to one point one through five but X is the number of blue balls so you will round off to one so on average as you pick from this urn three balls on average approximately one ball will be blue and the other two will be red let's now look at how much variability there is in this random variable and this is done of course by computing the so called variance and we denote this by Sigma squared or V of X and this is if you remember the sum over all X values the difference between X and the mean squared times the probability of observing this particular x value so let's go and again we'll find the exact value of the variance so we look up and we say what's the first value it is zero so zero minus the mean squared so zero minus the mean 63 out of 56 all squared so that's this term times the probability of observing an x value of zero well the probability of this was 10 out of 56 so times 10 over 56 let's keep going Plus the next X value when X is 1 so 1 minus the mean squared so 1 minus 63 out of 56 squared times the probability that x equals 1 well this is 30 out of 56 plus the other value of X when X is 2 so 2 minus the mean squared times the probability of observing x equals 2 and that is 15 out of 56 plus our last value of X when X is 3 so 3 minus the mean 63 out of 56 squared times the probability of observing a value of x equals 3 and that is one out of 56 and there you have it right so you sum over all possible values of x x minus the mean squared 1 minus the mean squared 2 minus the mean squared 3 minus the mean squared and you have to multiply each term by the probability of observing the particular x value 0 is a probability of 10 out of 56 130 out of 56 2 was 1556 and 3 was 1 out of 56 when you compute this again use the fraction command on your calculator to get the exact value and you should find the value of 225 divided by 448 and that is the variance of the random variable if you want an approximate value to get a feeling for what the value is equal to in decimals you can compute you'll get approximately 0.5 or 2 and finally the easiest quantity to compute is the standard deviation it is simply the root of the variance the variance is 225 over 448 and if you compute this you get approximately 0.7 oh right and you have your standard deviation so you see it's very easy to find the mean variance and deviation once you build your probability distribution table which is an assignment of probability to each possible value that your random variable can take finally we'd like to graphically visualize this distribution table by building the histogram for X and all we do is we plot the probabilities along the y axis over the values of x along the x axis and then we have our histogram for X let's do this as we've just said the y-axis contains the probability so this will be the probability that the random variable equals the given particular x value over all possible X values well we only had if you remember four possible values of x x was allowed to be 0 1 2 and 3 so x equals 0 x equals 1 x equals 2 x equals 3 and now along our y-axis will include the probabilities for each possible value of x we don't have to go up to 100% simply go up to about the largest probability so roughly point eighteen point fifty three point twenty seven point o-- 18 so we'll go up to and say point six 60% so we can go in slices of 10% so we'll go with one two three four five and six and besides each right the probability so 0.1 0.2 0.3 0.4 0.5 and finally zero point six and now all we have to do is above each value of x include a vertical bar whose height corresponds to the probability so here it's roughly point eighteen that's a little below point two and put the bar right in the middle of the given x value so you see the probability of X being equal to zero is approximately from our histogram 18 percent what about x equals one it's a bad point fifty-four percent so it's about in the middle of 0.5 and 0.6 and once again the height of the vertical bar about x equals one represents the probability that X indeed is equal to one what about two now the probability is about point twenty seven that's slightly below point three about here and finally when x equals 3 the probability is about two percent so it's going to be really low about this and now let's see with respect to the histogram of X let me write this here this is the histogram of X let's see how the mean and the variance fit in with this picture if you remember the mean was exactly 63 out of 20 63 out of 56 we'll use for the decimal one point one two five the variance was exactly to 25 over 4 48 which is about point 502 the deviation was roughly 0.7 oh nine so if you look at the mean about 1.25 1.125 and you see it's about here and the mean is again the average value of the random variable so you can look at the histogram and say the average is about a little bigger than 1 and here it's 1 point 1 through 5 the deviation gives you how much variability there is from the mean and you see the deviation here is not too big so most values are about the mean if Sigma was much bigger you'd have higher bars away from the mean if the Sigma was even smaller you'd have even higher bars about the mean value and this completes our example and once again if you remember the mean is just your average value what do you expect X to be equal to so if we were to pick three balls of our own we would expect to get one blue ball and two red balls

Channel: slcmath@pc

Views: 174,575

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Id: 9ycJNH0F3OM

Length: 20min 54sec (1254 seconds)

Published: Sat Sep 21 2013