Question Video: Determining the Variance for a Discrete Random Variable Mathematics

Video Transcript

Let 𝑋 denote a discrete random variable which can take the values one, 𝑎, and seven. Given that 𝑋 has probability distribution function 𝑓 of 𝑥 equals 𝑥 plus two over 18, find the variance of 𝑋. Give your answer to two decimal places.

We recall first that the variance of a discrete random variable is a measure of the extent to which values of that discrete random variable differ from their expected value. It’s calculated using the formula the expected value of 𝑋 squared minus the expected value of 𝑋 squared. And we need to be clear on the difference in notation here. In the second term, we find the expected value of 𝑋 first and then we square it, whereas in the first term we’re finding the expectation of the squared values of 𝑋. So, we square the 𝑥-values first and then find their expectation.

We recall also the formulae for finding the expected value of 𝑋 and the expected value of 𝑋 squared. For the expected value of 𝑋, we multiply each 𝑥-value in the range of the discrete random variable by its 𝑓 of 𝑥 value — that’s its probability — and then we find the sum of these values. For the expectation of 𝑋 squared, we square each 𝑥-value first, multiply it by its probability, and then find their sum. We’ve been given the probability distribution function 𝑓 of 𝑥. It’s 𝑥 plus two over 18. And there are three values in the range of this discrete random variable: one, 𝑎, and seven.

Before we can find the variance of 𝑋, we need to determine the value of the unknown 𝑎. We can do this by recalling that the sum of all probabilities in a probability distribution function must be equal to one. We can therefore find the values of the probabilities for 𝑥 equals one and 𝑥 equals seven and an expression for the probability when 𝑥 equals 𝑎 and form an equation. 𝑓 of one first of all is one plus two over 18, which is three over 18. And we won’t simplify this fraction for now. 𝑓 of 𝑎 is 𝑎 plus two over 18. And 𝑓 of seven is seven plus two over 18, which is nine over 18.

As the sum of these three probabilities must be equal to one, we have an equation: three over 18 plus 𝑎 plus two over 18 plus nine over 18 is equal to one. As we didn’t simplify any of the fractions so they have a common denominator of 18, we can add them by adding the numerators, leaving 𝑎 plus 14 over 18 is equal to one. We can then multiply each side of this equation by 18 to give 𝑎 plus 14 is equal to 18 and then subtract 14 from each side to find that 𝑎 is equal to four. So, we found the value of this unknown 𝑎. We also know that 𝑓 of one is three over 18. 𝑓 of 𝑎, which is now 𝑓 of four, is 𝑎 plus two over 18. So that’s six over 18. And 𝑓 of seven is nine over 18.

We’ll now write this probability distribution of 𝑥 in a table. We write the values in the range of this discrete random variable in the top row and then their corresponding probabilities in the second row. To find the expected value of 𝑥, we need to multiply each 𝑥-value by its 𝑓 of 𝑥 value. So, we’ll add a row to our table in order to do this. This gives three over 18, 24 over 18, and 63 over 18. We’ll keep all of these fractions in an unsimplified form for now. The expected value of 𝑋 is then the sum of these three values, which is 90 over 18, and this is exactly equal to five.

So, we found the expected value of 𝑋. And next, we need to calculate the expected value of 𝑋 squared. To do this, we’ll first add a row to our table in which we calculate each 𝑥 squared value. These values are the squares of one, four, and seven, which are one, 16, and 49. The probabilities for these 𝑥 squared values are inherited directly from the probability distribution of 𝑥. So, we add one final row to our table in which we multiply each 𝑥 squared value by each 𝑓 of 𝑥 value. That gives three over 18, 96 over 18, and 441 over 18. Each of these fractions can be simplified to have a common denominator of six, giving one-sixth, 32 over six, and 147 over six. The expected value of 𝑋 squared is the sum of these three values, which is 180 over six, and this is exactly equal to 30.

Finally, we calculate the variance of 𝑋. It’s equal to the expected value of 𝑋 squared, that’s 30, minus the expected value of 𝑋 squared. So, we’re subtracting five squared. We have 30 minus 25, which is equal to five. We were asked to give our answer to two decimal places. Our answer in this case is an integer, but we can write this as 5.00. So, we found that the variance of 𝑋 to two decimal places is 5.00 or simply five.