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

13 students took an exam, 8 students got 6 marks, 3 students got 10 marks, and 2 students got 2 marks. Given that π denotes the number of marks received, find the coefficient of variation of π as a percentage to the nearest hundredth.

06:13

### Video Transcript

13 students took an exam. Eight students got six marks, three students got 10 marks, and two students got two marks. Given that π denotes the number of marks received, find the coefficient of variation of π as a percentage to the nearest hundredth.

The coefficient of variation of a discrete random variable π gives its standard deviation as a percentage of its expected value. If a discrete random variable π has standard deviation π sub π and a nonzero mean πΈ of π, then the coefficient of variation of π is given by π sub π over πΈ of π multiplied by 100. To answer this problem then, we need to calculate both the expected value and stand deviation of π. And in order to do this, letβs first write down its probability distribution. We can do this in a table in which weβll write the values the discrete random variable can take, otherwise known as the values in the range of the discrete random variable, in the top row and then their probabilities, which we will describe as π of π₯ for the probability distribution function of π₯, in the second row.

Now, π₯ represents the number of marks received. Looking at the information in the question carefully, we can see that some students got six marks, some students got 10 marks, and some students got two marks. So the values in the range of this discrete random variable are two, six, and 10. We can work out the probabilities for each of these values using the other information in the question. There are 13 students in total, and eight of them got six marks. So the probability that π₯ is equal to six is eight out of 13. Three of the students got 10 marks, so the probability that π₯ is equal to 10 is three out of 13. And finally, two of the students got two marks. So the probability that π₯ is equal to two is two out of 13.

The sum of all probabilities in a probability distribution function should be equal to one. And we can confirm that the sum of our three values is indeed equal to one. So now that weβve got the probability distribution of π₯, we can calculate its expected value and standard deviation. We recall first that the expected value of π is equal to the sum of each π₯-value multiplied by its π of π₯ value, or its probability. We can add another row to our table to find these values. Two multiplied by two over 13 gives 4 over 13. Six multiplied by eight over 13 gives 48 over 13. And 10 multiplied by three over 13 gives 30 over 13. The expected value of π is then the sum of these three values, which is 82 over 13.

Next, we need to calculate the standard deviation of π, which we recall is equal to the square root of the variance of π. And the variance of π is equal to the expected value of π squared minus the expected value of π squared. The difference in notation is important here. The second term is the expected value of π which we then square. So this is the value weβve just found, 82 over 13, and weβll square it in the variance formula. The first term, however, is the expected value of π squared. So we square the π₯-values first and then find their expectation. The formula for this is the sum of each π₯ squared value multiplied by its π of π₯ value, which is inherited directly from the probability distribution of π₯.

We can add another row to our table for the π₯ squared values, which are four, 36, and 100, and then one final row in which we multiply the π₯ squared values by the π of π₯ values. Four multiplied by two over 13 is eight over 13. 36 multiplied by eight over 13 is 288 over 13. And 100 multiplied by three over 13 is 300 over 13. The expected value of π₯ squared is the sum of these three values, which is 596 over 13.

Next, weβll calculate the variance of π₯, which is the expected value of π squared, so 596 over 13, minus the square of the expected value of π, so minus 82 over 13 squared. As a fraction, this is equal to 1024 over 169. The standard deviation of π is then the square root of this value. And as both the numerator and denominator are square numbers, this simplifies easily to 32 over 13. So we found both the standard deviation and the expected value of this discrete random variable π. And all that remains is to substitute these two quantities into the formula for the coefficient of variation. We divide the standard deviation of π, thatβs 32 over 13, by the expected value of π, which is 82 over 13, and multiply by 100.

Now, of course, dividing by 82 over 13 is the same as multiplying by the reciprocal of this fraction, which is 13 over 82. We can then cross cancel a factor 13 and at the same time cross cancel a factor of two. So we have 16 over one multiplied by one over 41 all multiplied by 100. Thatβs 16 over 41 multiplied by 100, which as a decimal is equal to 39.0243 continuing. The question specifies that we should give our answer to the nearest one hundredth, so we round down to 39.02 percent. This tells us that for this discrete random variable π, its standard deviation is approximately 39 percent of its expected value.

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