Question Video: Determining the Variance for a Discrete Random Variable | Nagwa Question Video: Determining the Variance for a Discrete Random Variable | Nagwa

Question Video: Determining the Variance for a Discrete Random Variable Mathematics • Third Year of Secondary School

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Let 𝑋 denote a discrete random variable which can take the values βˆ’2, βˆ’1, 𝑀, and 2. Given that 𝑋 has probability distribution function 𝑓(π‘₯) = (π‘₯ + 4)/16, find the variance of 𝑋.

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Video Transcript

Let 𝑋 denote a discrete random variable which can take the values negative two, negative one, 𝑀, and two. Given that 𝑋 has probability distribution function 𝑓 of π‘₯ equals π‘₯ plus four over 16, find the variance of 𝑋.

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 variance of 𝑋 is equal to the expectation of 𝑋 squared minus the expected value of 𝑋 squared. We can think of this as the expectation of the squares minus the square of the expectation.

The expectation of 𝑋 is found by multiplying each π‘₯-value in the range of the discrete random variable by its probability, denoted by 𝑓 of π‘₯, and then finding the sum of these values, whereas the expectation of 𝑋 squared is found by squaring each π‘₯-value, multiplying this by the probability, and then finding the sum of these values.

We’ve been given the probability distribution function 𝑓 of π‘₯. It’s equal to π‘₯ plus four over 16. And there are four values in the range of this discrete random variable: negative two, negative one, an unknown value 𝑀, and two. Before we can calculate the expectation of 𝑋 and the expectation of 𝑋 squared, we need to find the value of this unknown 𝑀. We can do this by recalling that the sum of all probabilities in a probability distribution function must be equal to one. So we can find the probabilities for π‘₯ equals negative two, π‘₯ equals negative one, and π‘₯ equals two and an expression for the probability when π‘₯ equals 𝑀 and then form an equation.

Substituting π‘₯ equals negative two into the probability distribution function first, we have 𝑓 of negative two is negative two plus four over 16, which is two over 16. 𝑓 of negative one is negative one plus four over 16, which is three over 16. 𝑓 of 𝑀 is 𝑀 plus four over 16. And finally 𝑓 of two is two plus four over 16, which is six over 16. Summing these four values or expressions, we have the equation two over 16 plus three over 16 plus 𝑀 plus four over 16 plus six over 16 is equal to one. Adding the terms on the left-hand side, we have 𝑀 plus 15 over 16 is equal to one. We can then multiply each side of this equation by 16 to give 𝑀 plus 15 equals 16. And then subtracting 15 from each side, we find that 𝑀 is equal to one. The probability for 𝑓 of 𝑀 or 𝑓 of one will then be equal to five over 16.

So we found the value of this unknown 𝑀. We may find it helpful to now write the probability distribution of π‘₯ in a table. We write the values π‘₯ can take in the top row and then the probabilities, which we found to be two sixteenths, three sixteenths, five sixteenths, and six sixteenths, in the second row. To calculate the expectation of π‘₯, we need to multiply each π‘₯-value by its 𝑓 of π‘₯ value. So we add an extra row to our table to do this. This gives the values negative four sixteenths, negative three sixteenths, five sixteenths, and twelve sixteenths. And we’ll keep each of these with a common denominator of 16. The expectation of 𝑋 is then the sum of these four values, which is ten sixteenths. And we’ll simplify this to five over eight.

Next, we need to calculate the expected value of π‘₯ squared. So we add a row to our table to find the π‘₯ squared values, which are four, one, one, and four. Now, as there are repeated values for π‘₯ squared, we could create a new probability distribution for π‘₯ squared by combining the probabilities for the π‘₯-values that lead to each repeated π‘₯ squared value. But if we want to use the same table we’re already working with, we don’t need to do this. We can just create a new row in this table, in which we multiply the π‘₯ squared values by 𝑓 of π‘₯, giving eight over 16, three over 16, five over 16, and 24 over 16. The expectation of 𝑋 squared is the sum of these four values, which is 40 over 16. And this simplifies to five over two.

Finally, to calculate the variance of 𝑋, we take the expectation of 𝑋 squared and from this we subtract the square of the expectation of 𝑋. So we have five over two minus five over eight squared. That’s five over two minus 25 over 64. By multiplying both the numerator and denominator of the fraction five over two by 32, we can express this as the equivalent fraction 160 over 64. And then subtracting 25 over 64 from this value gives 135 over 64.

So by first determining the value of this unknown 𝑀, which we did by recalling that the sum of all probabilities in a probability distribution function must be equal to one, we found that the variance of this discrete random variable π‘₯ is equal to 135 over 64.

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