Question Video: Calculating the Variance of a Discrete Random Variable Mathematics

In a final exam taken by 25 students, 12 students got 7 points, 6 students got 8 points, and 7 students got 9 points. Given that 𝑋 denotes the number of points scored, find the variance of 𝑋.

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Video Transcript

In a final exam taken by 25 students, 12 students got seven points, six students got eight points, and seven students got nine points. Given that 𝑋 denotes the number of points scored, find the variance of 𝑋.

The number of points scored by the students in this exam is a discrete random variable, which we’re told to denote by 𝑋. We can write down the probability distribution function of this discrete random variable using the information given in the question. The values this discrete random variable can take are the number of points scored by the students, which was either seven, eight, or nine. The corresponding probabilities, which we denote using the function 𝑓 of π‘₯, are all fractions with denominators of 25, which was the total number of students who took the exam.

12 students got seven points, so the probability that π‘₯ equals seven is 12 out of 25. Six students got eight points, so the probability that π‘₯ equals eight is six out of 25. And finally, seven students got nine points, so the probability that π‘₯ equals nine is seven out of 25. We can confirm that the sum of these three probabilities is one as it should be for the sum of all the probabilities in a probability distribution.

We’ve been asked to calculate the variance of this discrete random variable 𝑋, so let’s recall the formula for doing so. It’s the expected value of 𝑋 squared minus the square of the expected value of 𝑋. We need to be clear on the difference in notation here. For the expected value of 𝑋 squared, we’re squaring the variable first and then finding its expectation, whereas for the second term, we’re calculating the expected value of 𝑋 and then squaring this value.

The formulae for calculating each of these statistics are as follows. The expected value of 𝑋 is the sum of each π‘₯-value multiplied by its corresponding probability. The expected value of 𝑋 squared is the sum of each π‘₯-value squared multiplied by the probability for that π‘₯-value. We’ll add some rows to our table to work the values we need out.

In the first new row, we’ll multiply each π‘₯-value by its probability. Seven multiplied by 12 over 25 is 84 over 25. Eight multiplied by six over 25 is 48 over 25. And nine multiplied by seven over 25 is 63 over 25. The expected value of 𝑋 is the sum of these three values, which is 195 over 25, or 39 over five in simplified form. The next row we add to the table is for the π‘₯ squared values, which are 49, 64, and 81. And in the final row of the table, we’ll multiply these values by the corresponding probabilities, giving 588 over 25, 384 over 25, and 567 over 25. The expected value of 𝑋 squared is the sum of the three values in the final row of the table, which is 1,539 over 25.

We’ve now calculated the expected value of 𝑋 and the expected value of 𝑋 squared. So we’re ready to substitute these values into the variance formula. Doing so gives 1,539 over 25 minus 39 over five squared. Evaluating the square gives 1,539 over 25 minus 1,521 over 25, which is 18 over 25. So by first writing down the probability distribution function of the discrete random variable 𝑋, which represents the number of points scored by the students, we’ve found that the variance of 𝑋 is 18 over 25.

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