Introduction

In this section (Section 4.2), we learn about calculating the mean (what we will start calling the expected value of a PDF) and standard deviation of a PDF of a discrete random variable!

So yes, the term "mean" in statistics can also be called the average or the expected value, depending on what context we are working in (hypothetical or concrete)! But they are all essentially the same concept!

What You Will Do

Make sure you read through all of the information below and watch the videos that are embedded within this page. The information written here is meant to help introduce concepts, and the videos are meant to help show examples and reinforce ideas about these concepts, so both are necessary to help you learn!

Expected Value

If we are dealing with hypothetical situations where we are trying to predict what might happen, we often rely on a concept called expected value. This is the value that we would EXPECT to happen on average if we repeated an experiment an infinite number of times.

Remember my example from before about guessing on every question of a multiple-choice assessment?

Let's say you are taking a 10-question multiple choice assessment where each question has options (a) through (d). If you did just go through and guess like that on that assessment, on average, you would get 2.5 answers correct. This is the expected value of the experiment!

Notice, the expected value does NOT need to be a possible data value (similarly to the mean if we are calculating it from a data set we have). So, we are not saying if you took one multiple choice assessment, you would get 2.5 answers right - that is not even possible! This means that if you were to take more and more and more of these assessments and averaged the number of questions you got right on them all, it would get closer and closer to 2.5!

To calculate the expected value from a discrete probability distribution, it is very similar to how we found an estimated mean in our Challenge assignments in Excel!

Here is a specific example:

Consider the discrete random variable X=the number of peas with green pods among 5 offspring peas (when one pea plant reproduces)

	$LaTeX: P\left(x\right)$
0	0.0010
1	0.0146
2	0.0879
3	0.2637
4	0.3955
5	0.2373

Let's first make some observations to help our brains make sense of this distribution!

It does meet our requirements of a valid PDF (all probabilities are between 0 and 1, and they add up to 1)
The highest probability is 0.3955, which corresponds to 4 pea pods being green.
The second highest probability is 0.2637, which corresponds to 3 pea pods being green.
The lowest probability is 0.0010, which corresponds to 0 pea pods being green.
Based on all of this, I would expect the average number of green pea pods out of 5 would be closer to 3-4 pea pods if we were to conduct this experiment over and over and over again (X=3 and X=4 have the highest likelihood of happening!)

Now, how do we calculate it!

We can take each of our possible random variable values and multiply them by their corresponding probabilities and then add all of them up! You are essentially "weighting" each value heavier or lighter depending on the likelihood that it happens. This formula looks like $LaTeX: \mu=\Sigma\left(x\cdot P\left(x\right)\right)$ , which means the expected value is the "sum of the products of each x and P(x)". We also call this E(x) sometimes (like "E" for "Expected").

Meme of P(X=x) telling E(x) that they are "mean" as a pun

Let's find each $LaTeX: x\cdot P\left(x\right)$ , which is the product (multiplication) of each with its corresponding probability in a third column!

	$LaTeX: P\left(x\right)$	$LaTeX: x\cdot P\left(x\right)$
0	0.0010	0*0.0010 = 0
1	0.0146	1*0.0146 = 0.0146
2	0.0879	2*0.0879 = 0.1758
3	0.2637	3*0.2637 = 0.7911
4	0.3955	4*0.3955 = 1.582
5	0.2373	5*0.2373 = 1.1865

Now, let's add all of those together from that column! (Note that these can be higher than 1 because they are not probabilities anymore once we multiply them with the x)

$LaTeX: \mu=\Sigma\left(x\cdot P\left(x\right)\right)$ = 0+0.0146+0.7911+1.582+1.1865 = 3.75

This means in the long-run, on average, we expect about 3.75 out of 5 pea pods to be green. Again, this does NOT mean we actually expect to collect three pea pods and see 3.75 of them green (that is not how that works!). What it means is we could have many samples with 3, many samples with 4, some samples with 2, etc., and if we averaged all of those values, we would get 3.75.

That is pretty close to what we predicted in our observations since X=3, 4, 5 are the most likely to happen!!

[*Foreshadowing: We can calculate this value in Excel with ONE function!!]

Now you!

Watch this video from Khan Academy about finding the expected value of a discrete PDF:

Watch this Khan Academy video about interpreting the expected value of a distribution:

Standard Deviation (and Variance) of a Discrete PDF

We have already been introduced to standard deviation and variance, but now we are seeing them in reference to a discrete PDF. Similarly to the expected value, if we have a specific data set of observations, we can calculate the standard deviation the way we have before. However, sometimes we are working with situations where we are trying to predict what happens, in which case we cannot calculate that in the same way!

This calculation is a little more intense than the expected value, but take it one step at a time and you got it!

The formula for standard deviation of a PDF is $LaTeX: \sigma=\sqrt{\Sigma\left\lbrack\left(x-\mu\right)^2\cdot P\left(x\right)\right\rbrack}$

Let's break that down step by step (and make as many columns as you need to!)

Find $LaTeX: \left(x-\mu\right)^2$ for each x value.
Multiply each of the values from #1 by the corresponding probability of that x value
Add up all of those things from #2 (this is the variance)
Take the square root of #3

Recall that before, we found $LaTeX: \mu=3.75$ .

Steps #1-2 (I rounded the last column to 4 decimal places):

	$LaTeX: P\left(x\right)$	$LaTeX: \left(x-\mu\right)^2$	$LaTeX: \left(x-\mu\right)^2\cdot P\left(x\right)$
0	0.0010	$LaTeX: \left(0-3.75\right)^2$ = 14.0625	14.0625*0.0010 $LaTeX: \approx$ 0.0141
1	0.0146	$LaTeX: \left(1-3.75\right)^2$ = 7.5625	7.5625*0.0146 $LaTeX: \approx$ 0.1104
2	0.0879	$LaTeX: \left(2-3.75\right)^2$ = 3.0625	3.0625*0.0879 $LaTeX: \approx$ 0.2692
3	0.2637	$LaTeX: \left(3-3.75\right)^2$ = 0.5625	0.5625*0.2637 $LaTeX: \approx$ 0.1483
4	0.3955	$LaTeX: \left(4-3.75\right)^2$ = 0.0625	0.0625*0.3955 $LaTeX: \approx$ 0.0247
5	0.2373	$LaTeX: \left(5-3.75\right)^2$ = 1.5625	1.5625*0.2373 $LaTeX: \approx$ 0.3708

Step 3: Now, we need to add all of the values up in that last column. When you do this, we get 0.0141+0.1104+0.2692+0.1483+0.0247+0.3708 = 0.9375.

Step 4: We found the variance in Step 3, so now we take the square root of that to find the standard deviation! When we do this, we get 0.9682 [rounded to 4 decimal places].

So, $LaTeX: \sigma$ =0.9682. Remember this helps us to describe the spread of the distribution! We can use this value to estimate how much of the data would be between particular values (using the expected value) like we did before when we learned about mean and standard deviation!

Now you!

Watch this Khan Academy video about finding the standard deviation of a discrete PDF:

Make Connections in our Textbook

Read through Section 4.2 and the examples provided to help your understanding!

Read about Expected Value and Standard Deviation in our text

What's Next

You will learn about the Binomial Distribution and then complete a Progress Check and complete a CYU about the material so far.