Question Video: Calculating the Expectation and Variance of a Discrete Random Variable | Nagwa Question Video: Calculating the Expectation and Variance of a Discrete Random Variable | Nagwa

# Question Video: Calculating the Expectation and Variance of a Discrete Random Variable Mathematics • Third Year of Secondary School

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A single, fair, four-sided die is tossed twice. The random variable π expresses the arithmetic mean of the numbers that appear on the lower face. Calculate the expectation of π. Calculate the variance of π.

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### Video Transcript

A single fair, four-sided die is tossed twice. The random variable π expresses the arithmetic mean of the numbers that appear on the lower face. Calculate the expectation of π. Calculate the variance of π.

Before we can calculate the expectation and variance of π, which is a discrete random variable, we first need to determine its probability distribution. Weβre told that π is the discrete random variable representing the arithmetic mean of the numbers that we get when we roll the die twice. The arithmetic mean of two numbers is their mean average. So if the two numbers are π and π, then their arithmetic mean is π plus π over two.

Letβs consider then what the possible values of π₯ are. And in order to do this, letβs consider the possible outcomes for each roll. Weβre told that the die is four-sided. So weβll assume that it has the numbers one, two, three, and four on its four faces. The possible outcomes from the first roll then are the numbers one, two, three, and four. And as the same die is rolled twice, the same is true for the possible outcomes for the second roll.

We then need to work out the arithmetic mean for each possible pair of numbers. For example, if we get a one on the first roll and then a one on the second roll, the arithmetic mean is one plus one over two. Thatβs two over two, which is equal to one. If, however, we roll a two on the first roll and a one on the second roll, the arithmetic mean is two plus one over two. Thatβs three over two, which is equal to 1.5.

We continue to complete the first row of the table. Three plus one over two, thatβs four over two, is two. And four plus one over two, thatβs five over two, is 2.5. In fact, we may observe that there is a pattern. Each value is 0.5 more than the previous value. In fact, this pattern continues throughout the table. As we go across each row and also as we go down each column, the values increase by 0.5. Now, the numbers inside the table are the possible values that this discrete random variable π can take. They are the values from one to four, increasing by 0.5 each time.

We can start to write the probability distribution of π₯ in a table with the values π₯ can take or the values in the range of this discrete random variable in the top row. In the second row, we write the probability that π₯ is equal to each of these values, which we can also write as π of π₯, where the function π is the probability distribution function of π₯. There are 16 cells in the center of this table. So we can write each of these probabilities as fractions with denominators of 16. To determine the numerator of each fraction, we just need to count up the number of times that that value appears inside the table.

The value one only appears once. The value of 1.5 appears twice. The value of two appears three times. 2.5 appears four times. Three appears three times. 3.5 appears twice. And four appears once. We can in fact see that there is a diagonal pattern throughout the table which helps us in determining how many times each value appears. So now we have the probability distribution function of π₯, each value in its range, together with the probability that π₯ is equal to that value. We can now calculate the expectation of π.

To do so, we find the sum of each π₯-value multiplied by its individual probability. We can add a row to our table to find these values. And we have one over 16, three over 16, six over 16, 10 over 16, nine over 16, seven over 16, and four over 16. The expected value of π is the sum of these seven values, which is 40 over 16. And this can be simplified to five over two. Or we can give our answer as a decimal of 2.5.

So weβve calculated the expectation of π. And now we need to calculate the variance. The formula for this is the expectation of π squared minus the expectation of π squared. And we can think of this as the expectation of the squared values of π₯ minus the square of the expected value of π. The expected value of π squared is found by summing each π₯ squared value multiplied by π of π₯. And the π of π₯ values are inherited directly from the probability distribution of π₯.

So we can add a row to our table for the π₯ squared values, which are one, 2.25, four, 6.25, nine, 12.25, and 16, and then one final row in which weβre going to multiply the π₯ squared values by the π of π₯ values. That gives the simplified fractions one over 16, nine over 32, three over four, 25 over 16, 27 over 16, 49 over 32, and one.

The expectation of π squared is the sum of these seven values, which we could work out using a common denominator of 32. Or using a calculator, we find that the expectation of π squared is equal to 55 over eight. The variance of π is then this value minus the square of the expected value of π. Itβs 55 over eight minus five over two squared. Thatβs 55 over eight minus 25 over four, which is 55 over eight minus 50 over eight. And this is equal to five over eight, or as a decimal 0.625.

So by first finding the probability distribution of this discrete random variable π, weβve found that its expectation is 2.5 and its variance is 0.625.

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