# Question Video: Finding the Coefficient of Variation for a Discrete Random Variable Mathematics • 10th Grade

Let π denote a discrete random variable that can take the values β1, 0, and 1. Given that π has a probability distribution function π(π₯) = π/(3 β π₯), find the coefficient of variation to the nearest percent.

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### Video Transcript

Let π denote a discrete random variable that can take the values negative one, zero, and one. Given that π has a probability distribution function π of π₯ equals π over three minus π₯, find the coefficient of variation to the nearest percent.

Weβve been given the probability distribution function for this discrete random variable, but it is in terms of an unknown value π. Before we can calculate the coefficient of variation, weβll need to determine the value of π. To do this, we recall that the sum of all probabilities in a probability distribution must be equal to one. So if we can find expressions for the probabilities for the three values in the range of this discrete random variableβwhich are negative one, zero, and oneβand these will be expressions in terms of π, we can then form an equation and solve it to determine the value of π.

First, π of negative one is π over three minus negative one, which is π over four. π of zero is π over three minus zero, which is π over three. And finally, π of one is π over three minus one, which is π over two. As weβve already said, the sum of all probabilities in a probability distribution must be equal to one. So we have the equation π over four plus π over three plus π over two is equal to one.

We can write each of these terms with a common denominator of 12. So we have three π over 12 plus four π over 12 plus six π over 12 is equal to one. This simplifies to 13π over 12 is equal to one. And then dividing both sides by 13 over 12, which is equivalent to multiplying by the reciprocal of this value, 12 over 13, we find that π is equal to 12 over 13.

We can then find each of the probabilities explicitly. π of negative one was π over four. So thatβs 12 over 13 multiplied by four, which simplifies to three over 13. π of zero was π over three. So thatβs 12 over 13 multiplied by three, which simplifies to four over 13. And our final probability for π of one is π over two. So thatβs 12 over 13 multiplied by two, which simplifies to six over 13. We can then confirm that the sum of these three probabilities is 13 over 13, which is indeed equal to one.

So weβve found the probabilities for each value in the range of this discrete random variable. Letβs now write down the probability distribution in a table. We write the values in the range of the discrete random variable in the top row and then their associated probabilities, which we just calculated, in the second row.

Now the question asks us to find the coefficient of variation for this discrete random variable. This gives the standard deviation as a percentage of the expected value of π₯. If a discrete random variable π has a nonzero mean πΈ of π and a standard deviation π sub π, then the coefficient of variation is given by π sub π over πΈ of π multiplied by 100.

We recall first that to find the expected value of a discrete random variable, we multiply each value in its range by its corresponding probability and then find the sum of these values. We can add another row to our table to work out these values. Negative one multiplied by three over 13 is negative three over 13. Zero multiplied by four over 13 is zero. And one multiplied by six over 13 is six over 13. The expected value of π, then, is the sum of these three values, which is three over 13.

We then recall that the standard deviation of π is the square root of its variance. And the variance of π is the expected value of π squared minus the square of the expected value of π. We need to be very clear on the difference in notation here. In the second term, we find the expected value of π, which weβve just done, and then we square it, whereas in the first term, weβre finding the expectation of the squared values of π. So we square the π-values first.

The formula for calculating the expectation of π squared is the sum of the π₯ squared values multiplied by the probabilities, which are inherited directly from the probability distribution of π. We can add another row to our table for the π₯ squared values, which are one, zero, and one again, and then another row in which we multiply each π₯ squared value by its π of π₯ value. First, we have one multiplied by three over 13, which is three over 13, then zero multiplied by four over 13, which is zero, and finally, one multiplied by six over 13, which is six over 13.

The expected value of π squared then is three over 13 plus zero plus six over 13, which is nine over 13. Next, we calculate the variance of π. This is the expected value of π squared. Thatβs nine over 13. And from this we subtract the square of the expected value of π. So weβre subtracting three over 13 squared. Nine over 13 minus three over 13 squared gives the exact fraction 108 over 169.

So weβve calculated the variance of π, and next we need to calculate the standard deviation. This is equal to the square root of the variance. And in exact form, this is six root three over 13.

Weβre nearly finished. Weβve found the standard deviation of π and the expected value of π. So weβre finally able to calculate the coefficient of variation. We have six root three over 13 for the standard deviation divided by three over 13 for the expectation multiplied by 100. Now dividing by three over 13 is equivalent to multiplying by 13 over three. We can then cross cancel a factor of 13, and we can also cross cancel a factor of three. So weβre left with two root three over one multiplied by one over one all multiplied by 100. Evaluating on a calculator, we have 346.41 continuing.

The question specifies that we should give our answer to the nearest percent. So we round down to 346 percent. Now donβt be concerned that this percentage is greater than 100 percent. A coefficient of variation of 346 percent just means that the standard deviation of π is approximately three and a half times its expected value, which is perfectly possible. We found then that the coefficient of variation of π to the nearest percent is 346 percent.