Question Video: Finding the Coefficient of Variation for a Discrete Random Variable Mathematics

Let 𝑋 denote a discrete random variable that can take the values 2, 3, 5, and 7. Given that 𝑃(𝑋 = 2) = 1/12, 𝑃(𝑋 = 3) = 1/4, 𝑃(𝑋 = 5) = 1/3, and 𝑃(𝑋 = 7) = 1/3, find the coefficient of variation to the nearest percent.

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Video Transcript

Let 𝑋 denote a discrete random variable that can take the values two, three, five, and seven. Given that the probability 𝑋 equals two is one twelfth, the probability 𝑋 equals three is one-quarter, the probability 𝑋 equals five is one-third, and the probability 𝑋 equals seven is one-third, find the coefficient of variation to the nearest percent.

In this problem, we’re asked to find the coefficient of variation for a discrete random variable 𝑋. This gives the standard deviation as a percentage of the expected value of 𝑋. If the discrete random variable has a nonzero mean 𝐸 of 𝑋 and a standard deviation 𝜎 sub 𝑋, then the coefficient of variation is given by 𝜎 sub 𝑋 over 𝐸 of 𝑋 multiplied by 100. We’re given the probability distribution of our discrete random variable 𝑋 in the question. But let’s write it down in a table. In the top row, we put the values in the range of the discrete random variable, which are two, three, five, and seven. In the second row, we put the probabilities for each of these values, which are one twelfth, one-quarter, one-third, and one-third. Notice that the sum of these four probabilities is equal to one, as it should be for the sum of all probabilities in the probability distribution.

We now need to calculate both the standard deviation and the expectation of 𝑋. Let’s begin with the expectation. The formula for this is the sum of 𝑋 multiplied by 𝑓 of 𝑋. We multiply each value in the range of the discrete random variable by its corresponding probability and then add all of these values together. We can add an extra row to our table to find these values. The first value is two multiplied by one twelfth. That’s two twelfths, and we won’t simplify this value for now. The next value is three multiplied by one-quarter, which is three-quarters, then five multiplied by one-third, which is five-thirds, and finally seven multiplied by one-third, which is seven-thirds.

To find the expectation of 𝑋, we need to find the sum of these four values which we can do using a common denominator of 12. We have two twelfths plus nine twelfths plus twenty twelfths plus twenty-eight twelfths. That’s fifty-nine twelfths. And this fraction can’t be simplified any further. Next, we need to calculate the standard deviation of 𝑋. Now, the standard deviation is the square root of the variance. And the variance of 𝑋 is calculated using the formula the expectation of 𝑋 squared minus the expectation of 𝑋 squared.

We need to be clear on the difference in notation here. In the first term, we square the 𝑋-values first and then find their expected or average value. In the second term, we find the expected or average value of 𝑋, which we’ve just done. It’s 59 over 12. And then we square this value. To find the expectation of 𝑋 squared, we multiply each 𝑋 squared value by its corresponding probability, which is inherited directly from the probability distribution of 𝑋. We can add an extra row to our table for the 𝑋 squared values which are four, nine, 25, and 49 and then a row in which we multiply each 𝑋 squared value by each 𝑓 of 𝑋 value.

So we have four multiplied by one twelfth which is four twelfths, nine multiplied by one-quarter which is nine-quarters 25 multiplied by one-third which is 25 over three, and 49 multiplied by one-third which is 49 over three. The expected value of 𝑋 squared then is four twelfths plus nine over four plus 25 over three plus 49 over three. And we can use a calculator to help with this. It’s 327 over 12, which simplifies to 109 over four.

As we now have both the expectation of 𝑋 and the expectation of 𝑋 squared, we can calculate the variance of 𝑋. This is 109 over four minus 59 over 12 squared, which, again using a calculator to help, is 443 over 144. Next, we find the standard deviation of 𝑋 by square rooting the variants. The denominator is a square number. So we have that the standard deviation of 𝑋 is equal to the square root of 443 over 12. And we’ll keep this as an exact value for now.

We’ve found both the standard deviation and the expected value of 𝑋. So the final step is to substitute these two values into the formula for the coefficient of variation. We need to divide the standard deviation of 𝑋, which is the square root of 443 over 12, by the expected value of 𝑋, which is 59 over 12, and then multiply by 100. Of course, dividing by 59 over 12 is the same as multiplying by its reciprocal of 12 over 59. We can then cross cancel a factor of 12. So we have the square root of 443 over 59 multiplied by 100, which is 35.67 continuing.

The question specifies that we should give our answer to the nearest percent. So as the number in the first decimal place is a six, we’ll be rounding up. We found then that the coefficient of variation of 𝑋 to the nearest percent is 36 percent. This tells us that the standard deviation of 𝑋 is approximately 36 percent of its average value.

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