Question Video: Finding the Standard Deviation of a Discrete Random Variable | Nagwa Question Video: Finding the Standard Deviation of a Discrete Random Variable | Nagwa

# Question Video: Finding the Standard Deviation of a Discrete Random Variable Mathematics • Third Year of Secondary School

## Join Nagwa Classes

Let π denote the discrete random variable that can take values 0, 2, 4, and 6. Given that π(π = 0) = 1/7, π(π = 2) = 2/7, and π(π = 4) = 2/7, find the standard deviation of π, giving your answer to two decimal places.

05:06

### Video Transcript

Let π denote the discrete random variable that can take values zero, two, four, and six. Given that the probability π equals zero is one-seventh, the probability π equals two is two-sevenths, and the probability π equals four is two-sevenths, find the standard deviation of π, giving your answer to two decimal places.

Weβve been told that this discrete random variable can take the values zero, two, four, and six. But weβve only been given the probabilities for three of them. Before we can proceed with calculating the standard deviation, we need to find the probability that π is equal to the fourth value, which is six. We can work this out because the sum of all the values in a probability distribution π of π must be equal to one. So one-seventh plus two-sevenths plus two-sevenths plus the probability π is equal to six is equal to one. This simplifies to five-sevenths plus the probability π equals six equals one. And then we can subtract five-sevenths from each side to find that the probability π is equal to six is two-sevenths.

We may now find it helpful to express this probability distribution in a table, with the values in the range of the discrete random variable in the top row and their associated probabilities in the second row. So the probability distribution of π can be represented as shown. We are asked to find the standard deviation of π, which is a measure of spread of its probability distribution. We use the Greek letter π, or sometimes π subscript π, to denote the standard deviation. And itβs equal to the square root of the variance.

The variance of a discrete random variable π can be calculated using the formula shown. Itβs equal to the expected value of π squared minus the expected value of π squared. The difference in notation is really important here. In the second term, we find the expected or average value of π and then we square it, whereas in the first term, we square the π-values first and then find their expected or average value. Thereβs quite a lot of work to be done here, so weβll break it down into stages. And weβll begin by calculating the expectation of π. This is the sum of each π-value in the probability distribution multiplied by its associated probability. And we can add a row to our table to work out these values.

First, we have zero multiplied by one-seventh, which is zero; then two multiplied by two-sevenths, which is four-sevenths; four multiplied by two-sevenths, which is eight-sevenths; and finally six multiplied by two-sevenths, which is twelve-sevenths. The expected value of π is then the sum of these four values, which is 24 over seven. Next, we need to calculate the expected value of π squared. The formula for this is the sum of each π squared value multiplied by its corresponding probability. And the probabilities for π squared are inherited directly from the probability distribution of π.

We can add another row to our table for the π squared values, which are zero, four, 16, and 36. And then weβll add another row in which we multiply these values by the probabilities. Zero multiplied by one-seventh is zero. Four multiplied by two-sevenths is eight-sevenths. 16 multiplied by two-sevenths is thirty-two sevenths. And finally, 36 multiplied by two-sevenths is seventy-two sevenths. The expectation of π squared is the sum of these four values, which is 112 over seven.

So having calculated both the expectation of π and the expectation of π squared, weβre now able to calculate the variance of π. Itβs equal to 112 over seven for the expectation of π squared minus 24 over seven squared for the expectation of π squared. And this is equal to 208 over 49. The final step in calculating the standard deviation is to take the square root of this value. π then is the square root of 208 over 49, which in simplified surd form is four root 13 over seven. We can then evaluate this as a decimal, and it is 2.0603 continuing.

The question specifies that we should give our answer to two decimal places. So we round down to 2.06. The standard deviation of π to two decimal places is 2.06, which means that, on average, observations of this discrete random variable π are 2.06 units away from their mean.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions