Question Video: Variance of a Discrete Random Variable | Nagwa Question Video: Variance of a Discrete Random Variable | Nagwa

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

Let 𝑋 denote a discrete random variable. Given that 𝐸(𝑋) = 15 and Var(𝑋) = 26, find 𝐸(𝑋²).

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Video Transcript

Let 𝑋 denote a discrete random variable. Given that 𝐸 of 𝑋 equals 15 and Var of 𝑋 equals 26, find 𝐸 of 𝑋 squared.

Let’s begin by recalling what each of these pieces of notation mean. 𝐸 of 𝑋, first of all, is the expectation or expected value of the discrete random variable 𝑋. It is its average value, and we often denote this using the Greek letter 𝜇. Var of 𝑋 stands for the variance of 𝑋, which is a measure of spread of the probability distribution. We denote this using the Greek letter 𝜎 squared or sometimes 𝜎 sub 𝑋 squared if there are multiple variables in the same problem. 𝐸 of 𝑋 squared is the expected value of 𝑋 squared. That is, we square the values of the discrete random variable and then find their expectation.

These three quantities are related by the following formula. The variance of 𝑋 is equal to the expected value of 𝑋 squared minus the expected value of 𝑋 squared. As we know the expected value of 𝑋 — it’s 15 — and the variance of 𝑋 — it’s 26 — we can substitute these values into this formula to find the expectation of 𝑋 squared. We have then 26 is equal to the expectation of 𝑋 squared minus 15 squared. 15 squared is 225. And then, we can solve this equation for 𝐸 of 𝑋 squared by adding 225 to each side. That gives 251 is equal to 𝐸 of 𝑋 squared.

So, by recalling the definition of the variance of a discrete random variable and then forming and solving an equation, we found that 𝐸 of 𝑋 squared is 251.

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