# Chapter 4 Random Variables

## 4.1 Module Overview

In previous modules, we introduced the idea of variables and examined their distributions. We also began our discussion on probability theory. Now, we extend these concepts into what are called random variables. We will introduce the concept of random variables in general and will discuss a specific type of distribution - the binomial distribution. Then we will discuss a continuous probability distribution, the normal distribution. The normal distribution will provide a foundation for much of the inference we will complete throughout the rest of this course.

Module Learning Objectives/Outcomes

1. Discuss discrete random variables using key terminology.
2. Express cumulative probabilities using probability notation.
3. Calculate the expected value and standard deviation of a discrete random variable.
4. Calculate binomial probabilities.
5. Convert normal distributions to standard normal distributions.
6. Calculate probabilities for a normal distribution using area under the curve.
7. Approximate binomial probabilities using the normal curve.

This module’s outcomes correspond to course outcomes (4) use the binomial distribution as a model for discrete variables and (5) use the normal distribution as a model for continuous variables.

## 4.2 Discrete Random Variables

A random variable is a quantitative variable whose values are based on chance. By “chance,” we mean that you can’t know the outcome before it occurs.

A discrete random variable is a random variable whose possible values can be listed.

Notation:

• $$x$$,$$y$$,$$z$$ (lower case letters) denote variables.
• $$X$$, $$Y$$, $$Z$$ (upper case letters) denote random variables.

In contrast to events, where we usually used letters toward the start of the alphabet, (random) variables are typically denoted by letters from the end of the alphabet.

• $$\{X=x\}$$ denotes the event that the random variable $$X$$ equals $$x$$.
• $$P(X=x)$$ denotes the probability that the random variable $$X$$ equals $$x$$.

Recall: a probability distribution is a list of all possible values and their corresponding probabilities. (See Section 3.3 for a refresher.) A probability histogram is a histogram where the heights of the bars correspond to the probability of each value. For discrete random variables, each “bin” is one of the listed values.

Example:

Number of Siblings, $$x$$ 0 1 2 3 4
Probability, $$P(X=x)$$ 0.200 0.425 0.275 0.075 0.025

(Assume for the sake of the example that no one has more than 4 siblings.)

Interpretation: in a large number of independent observations of a random variable $$X$$, the proportion of times each possible value occurs will approximate the probability distribution of $$X$$.

### 4.2.1 The Mean and Standard Deviation

Mean of a Discrete Random Variable

The mean of a discrete random variable $$X$$ is denoted $$\mu_X$$. If it’s clear which random variable we’re talking about, we can drop the subscript and write $$\mu$$. $\mu_X = \Sigma xP(X=x)$ where $$\Sigma$$ denotes “the sum over all values of $$x$$”: $\Sigma xP(X=x) = x_1P(X=x_1) + x_2P(X=x_2) + \dots + x_nP(X=x_n).$

The mean of a random variable is also called the expected value or expectation. Recall that measures of center are meant to identify the most common or most likely, thus the value we can expect to see (most often).

Example: for the Siblings distribution, $\mu = 0(0.200)+1(0.425)+2(0.275)+3(0.075)+4(0.025)=1.3$ Make sure you understand how we used the formula for $$\mu$$ and the probability distribution to come up with this number.

Interpretation: in a large number of independent observations of a random variable $$X$$, the mean of those observations will approximately equal $$\mu$$.

The larger the number of observations, the closer their average tends to be to $$\mu$$. This is known as the law of large numbers.

Example: Suppose I took a random sample of 10 people and asked how many siblings they have. $2,2,2,2,1,0,3,1,2,0$ In my random sample of 10, $$\bar{x}=2$$, which is a reasonable estimate but not that close to the true mean $$\mu=1.3$$.

• A random sample of 30 gave me a mean of $$\bar{x}=1.53$$.
• A random sample of 100 gave me a mean of $$\bar{x}=1.47$$.
• A random sample of 1000 gave me a mean of $$\bar{x}=1.307$$.

We use concepts related to the law of large numbers as a foundation for statistical inference, but note that - although very large samples are nice to have - it’s not necessary to take enormous samples all the time. Often, we can come to interesting conclusions with fewer than 30 observations!

Standard Deviation of a Discrete Random Variable

The variance of a discrete random variable $$X$$ is denoted $$\sigma_X^2$$ (or $$\sigma^2$$ if it’s clear which variable we’re talking about). $\sigma_X^2 = \Sigma[(x-\mu_X)^2P(X=x)]$ OR $\sigma_X^2 = \Sigma[x^2P(X=x)]-\mu_X^2$ These formulas are exactly equivalent and you may use whichever you wish, but note that the second may be a little easier to work with.

As before, the standard deviation is the square root of the variance: $\sigma = \sqrt{\sigma^2}$

Example: Calculate the standard deviation of the Siblings variable.

In general, a table is the best way to keep track of a variance calculation:

$$x$$ $$P(X=x)$$ $$xP(X=x)$$ $$x^2$$ $$x^2P(X=x)$$
0 0.200 0 0 0
1 0.425 0.425 1 0.425
2 0.275 0.550 4 1.100
3 0.075 0.225 9 0.675
4 0.025 0.100 16 0.400
$$\mu$$ = 1.3 Total = 2.6

Then the variance is $\sigma^2 = 2.6 - 1.3^2 = 0.9$ and the standard deviation is $\sigma = \sqrt{0.9} = 0.9539.$

## 4.3 The Binomial Distribution

Think back to replication in an experiment. Each replication is what we call a trial. We will consider a setting where each trial has two possible outcomes.

For example, suppose you want to know if a coin is fair (both sides equally likely). You might flip the coin 100 times (thus running 100 trials). Each trial is a flip of the coin with two possible outcomes: heads or tails.

The product of the first $$k$$ positive integers $$(1, 2, 3, \dots)$$ is called k-factorial, denoted $$k!$$: $k! = k \times (k-1) \times\dots\times 3 \times 2 \times 1$ We define $$0!=1$$.

Example: $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

If $$n$$ is a positive integer $$(1, 2, 3, \dots)$$ and $$x$$ is a nonnegative integer $$(0, 1, 2, \dots)$$ with $$x \le n$$, the binomial coefficient is $\binom{n}{x} = \frac{n!}{x!(n-x)!}$

Example: $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)}$

Sometimes, we may want to simplify a binomial coefficient before taking all of the factorials. Why? Well, $20! = 2432902008176640000$ Most calculators will not print this number. Instead, you’ll get an error or a rounded version printed using scientific notation. Neither will help you accurately calculate the binomial coefficient.

Example: $\binom{20}{17} = \frac{20\times 19\times 18\times 17\times 16\times \dots \times 3\times 2\times 1}{(17\times 16\times \dots \times 3\times 2\times 1)(3\times 2\times 1)}$ but notice that I can rewrite $$20!$$ as $$20\times 19\times 18\times 17!$$, so $\binom{20}{17} = \frac{20\times 19\times 18\times 17!}{17!(3\times 2\times 1)} = \frac{20\times 19\times 18}{3\times 2\times 1} = \frac{6840}{6} = 1140$

Bernoulli trials are repeated trials of an experiment that satisfy 1. Each trial has two possible outcomes: success and failure. 2. Trials are independent. 3. The probability of success (the success probability) $$p$$ remains the same from one trial to the next: $P(X=\text{success})=p$

The binomial distribution is the probability distribution for the number of successes in a sequence of Bernoulli trials.

Fact: in $$n$$ Bernoulli trials, the number of outcomes that contain exactly $$x$$ successes equals the binomial coefficient $$\binom{n}{x}$$.

Binomial Probability Formula

Let $$x$$ denote the total number of successes in $$n$$ Bernoulli trials with success probability $$p$$. The probability distribution of the random variable $$X$$ is given by $P(X=x) = \binom{n}{x}p^x(1-p)^{n-x} \quad\quad x = 0,1,2,\dots,n$ The random variable $$X$$ is called a binomial random variable and is said to have the binomial distribution. Because $$n$$ and $$p$$ fully define this distribution, they are called the distribution’s parameters.

To find a binomial probability formula:

1. Check assumptions.
1. Exactly $$n$$ trials to be performed.
2. Two possible outcomes for each trial.
3. Trials are independent (each trial does not impact the result of the next)
4. Success probability $$p$$ remains the same from trial to trial.
2. Identify a “success.” Generally, this is whichever of the two possible outcomes we are most interested in.
3. Determine the success probability $$p$$.
4. Determine $$n$$, the number of trials.
5. Plug $$n$$ and $$p$$ into the binomial distribution formula.

We can also use the binomial probability formula to calculate probabilities like $$P(X\le x)$$. Notice that we can rewrite this uisng concepts from the previous module $P(X \le k) = P(X=k \text{ or } X=k-1 \text{ or } \dots \text{ or } X=2 \text{ or } X=1 \text{ or } X=0)$ Since $$X$$ is a discrete random variable, each possible value is disjoint. We can use this! $P(X \le k) = P(X=k) + P(X=k-1) + \dots + P(X=2) + P(X=1) + P(X=0)$

Example: $$P(X \le 3) = p(X=3)+P(X=2)+P(X=1)+P(X=0)$$

We can also extend this concept to work with probabilities like $$P(a < X \le b)$$.

Example: $$P(2 < X \le 5)$$

First, notice that if $$2 < X \le 5$$, then $$X$$ can be 3, 4, or 5: $P(2 < X \le 5) = P(X=3)+P(X=4)+P(X=5)$

Note: if going from $$2 < X \le 5$$ to “$$X$$ can be 3, 4, or 5” doesn’t make sense to you, start by writing out the sample space. Suppose $$n=10$$. Then the sample space for the binomial distribution is $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ Then I can check any number in this sample space by plugging it in for $$X$$. So for 1, I can check $$2 < 1 \le 5$$. Obviously this is not true, so we won’t include 1. Checking the number 2, I get $$2 < 2 \le 5$$. Since 2 < 2 is NOT true, we don’t include 2. Etc.

### 4.3.1 Mean and Variance

The shape of a binomial distribution is determined by the success probability:

• If $$p \approx 0.5$$, the distribution is approximately symmetric.
• If $$p < 0.5$$, the distribution is right-skewed.
• If $$p > 0.5$$, the distribution is left-skewed.

The mean of a binomial distribution is $$\mu = np$$. The variance is $$\sigma^2 = np(1-p)$$.

## 4.4 The Normal Distribution

If we can represent a discrete variable with a probability histogram, what can we do with a continuous variable?

We represent the shape of a continuous variable using a density curve. This is like a histogram, but with a smooth curve:

Properties:

1. The curve is always above the horizontal axis (because probabilities are always nonnegative).
2. The total area under the curve equals 1.

For a variable with a density curve, the proportion of all possible observations that lie within a specified range equals the corresponding area under the density curve.

A normal curve is a special type of density curve that has a “bell-shaped” distribution. In fact, all of the density curves I’ve shown so far have been normal curves! We say that a variable is normally distributed or has a normal distribution if its distribution has the shape of a normal curve.

Why “normal?” Because it’s very common! Lots of things are more common around the average and less common as you get farther from the average: height, amount of sleep people get each night, standardized test scores, etc. In practice, these things aren’t exactly normally distributed… instead, they’re approximately normally distributed (and that’s ok).

Normal distributions…

• are fully determined by parameters mean $$\mu$$ and standard deviation $$\sigma$$.
• are symmetric and centered at $$\mu$$.
• have spreads that depend on $$\sigma$$.

Pay close attention to the horizontal axis and how spread out the densities are in each of the following plots:

Notice that the bottom left plot comes to a sharper peak, while the bottom right has a gentler slope. This is what we mean by “spread”: the density on the bottom right is the most spread out.

To check whether a variable is (approximately) normally distributed,

1. Check the histogram to see if it is symmetric and bell-shaped.
2. Estimate the parameters: $$\mu$$ using $$\bar{x}$$ and $$\sigma$$ using $$s$$.

### 4.4.1 The Standard Normal Distribution

In order to make normal distributions easier to work with, we will standardize them. A standard normal distribution is a normal distribution with mean $$\mu=0$$ and standard deviation $$\sigma=1$$. We standardize a variable using $z = \frac{x-\mu}{\sigma}.$ This is also called a z-score. Standardizing using this formula will always result in a variable with mean 0 and standard deviation 1 (even if it’s not normal!). If $$X$$ is approximately normal, then the standardized variable $$Z$$ will have a standard normal distribution.

Note: when we z-score a variable, we preserve the area under the curve properties! If $$X$$ is Normal$$(\mu,\sigma)$$, then $P(X < c) = P\left(Z < \frac{c - \mu}{\sigma}\right) = P(Z < z).$

## 4.5 Area Under the Standard Normal Curve

Properties:

1. Total area under the curve is 1.
2. The curve extends infinitely in both directions, never touching the horizontal axis.
4. Almost all of the area under the curve is between -3 and 3.

We will think about area under the standard normal curve in terms of cumulative probabilities or probabilities of the form $$P(Z < z)$$.

We will use the fact that the total area under the curve is 1 to find probabilities like $$P(Z > c)$$:

Using the graphic to help visualize, we can see that $1 = P(Z < c) + P(Z > c)$ which we can then rewrite as $P(Z > c) = 1-P(Z<c).$

We can also use this concept to find $$P(a < Z < b)$$.

Notice that $1 = P(Z < a) + P(a < Z < b) + P(Z > b),$ which we can rewrite as $P(a < Z < b) = 1 - P(Z > b) - P(Z < a)$ and since we just found that $$P(Z > b) = 1 - P(Z < b)$$, we can replace $$1 - P(Z > b)$$ with $$P(Z < b)$$, and get $P(a < Z < b) = P(Z < b) - P(Z < a).$

Key Cumulative Probability Concepts

• $$P(Z > c) = 1 - P(Z < c)$$
• $$P(a < Z < b) = P(Z < b) - P(Z < a)$$

A final note, because the normal distribution is symmetric, $$P(X < \mu) = P(X > \mu) = 0.5$$. Notice this also implies that, when a distribution is symmetric (and unimodal), the mean and median are the same!

Now that we can get all of our probabilities written as cumulative probabilities, we’re ready to use software to find the area under the curve!

Finding Area Under the Curve: R

We will use statistical software called R to find areas under the curve. R is an incredibly powerful statistical programming language, but we’re going to keep it simple. $$P(Z < z)$$ is found using the command ‘pnorm(z).’ To find $$P(Z<1)$$, I would type pnorm(1). That entry and R output look like this:

pnorm(1)
## [1] 0.8413447

so $$P(Z < 1) = 0.8413447$$. Since we are only going to use R for a few simple commands, we will run it completely online at the website rdrr.io/snippets (bookmark this website!)

For now, you can run R right here in the course notes! This is exactly what you will see on the rdrr.io website. Type in your command and click the green “Run” button. Try finding $$P(Z < 2)$$.

Make sure you are able to run the command and get $$P(Z<2)=0.9772499$$. (If it prints out “Sorry, something went wrong. All I know is:” just press the “Run” button again.)

We can also find a z-score given a specified area/probability. The notation $$z_{\alpha}$$ (z-alpha) is the z-score corresponding to a right-tail area of $$\alpha$$. That is, $P(Z>z_{\alpha}) = \alpha$ We can find $$z_{\alpha}$$ using the command qnorm(p, lower.tail=FALSE). To find $$P(Z>z_{\alpha}) = 0.1$$, I would type

qnorm(0.1, lower.tail=FALSE)
## [1] 1.281552

so if $$P(Z>z_{\alpha}) = 0.1$$, then $$z_{\alpha}=1.281552$$. (If you wanted to consider $$P(Z < z) = p$$, you would replace “FALSE” with “TRUE.”)

A quick note about R: R will print very large numbers and numbers close to 0 using scientific notation. However, R’s scientific notation may not look the way you’re used to! Check out the R output for $$P(Z < -5)$$:

pnorm(-5)
## [1] 2.866516e-07

When you see e-07, that means $$\times10^{-7}$$… so $$P(Z < -5) = 2.8665 \times 10^{-7} \approx 0.00000029$$.

Finding Area Under the Curve: Applets

Another option for finding probabilities and z-scores associated with the normal curve is to use an online applet. The Rossman and Chance Normal Probability Calculator is my preferred applet. It’s relatively straightforward to use and would be difficult to demonstrate in these course notes! We will demonstrate this applet in class. I recommend you bookmark any websites you use to find probabilities!

You can also find the area under a normal distribution using a Normal Distribution Table. These are outdated and not used anywhere but the statistics classroom. As a result, I do not teach them. However, if you wish to use the table instead of R, there is a short tutorial here.

## 4.6 Working with Normally Distributed Variables

### 4.6.1 Normal Distribution Probabilities

Using z-scores and area under the standard normal curve, we can find probabilities for any normal distribution problem!

Determining Normal Distribution Probabilities

1. Sketch the normal curve for the variable.
2. Shade the region of interest and mark its delimiting x-value(s).
3. Find the z-score(s) for the value(s).
4. Use the pnorm command in R to find the associated area.

Example: Find the proportion of SAT-takers who score between 1150 and 1300. Assume that SAT scores are approximately normally distributed with mean $$\mu=1100$$ and standard deviation $$\sigma = 200$$.

First, let’s figure out what we want to calculate. Using area under the curve concepts, the proportion of test-takers who score between 1150 and 1300 will be $$P(1150 < X < 1300)$$.

1. Sketch:

1. Calculate z-scores: $x = 1150 \rightarrow z = \frac{1150-1100}{200} = 0.25$ and $x=1300 \rightarrow z = \frac{1300-1100}{200} = 1.$
2. Use R with pnorm to find $$P(Z < 0.25)$$ and $$P(Z < 1)$$:
pnorm(0.25)
## [1] 0.5987063
pnorm(1)
## [1] 0.8413447

Note that $P(1150 < X < 1300) = P\left(\frac{1150-1100}{200} < Z < \frac{1300-1100}{200}\right) = P(0.25 < Z < 1)$ and, using cumulative probability concepts, $P(0.25 < Z < 1) = P(Z < 1) - P(Z < 0.25).$ Using R, we found $$P(Z < 0.25) \approx 0.5987$$ and $$P(Z < 1) \approx 0.8413$$, so $P(Z < 1) - P(Z < 0.25) \approx 0.8413 - 0.5987 = 0.2426.$ That is, approximately 26.26% of test-takers score between 1150 and 1300 on the SAT.

### 4.6.2 Empirical Rule for Variables

For any (approximately) normally distributed variable,

1. Approximately 68% of all possible observations lie within one standard deviation of the mean: $$\mu \pm \sigma.$$
2. Approximately 95% of all possible observations lie within two standard deviations of the mean: $$\mu \pm 2\sigma.$$
3. Approximately 99.7% of all possible observations lie within three standard deviations of the mean: $$\mu \pm 3\sigma.$$

Given some data, you can check if approximately 68% of the data falls within $$\bar{x}\pm s$$, 95% within $$\bar{x}\pm 2s$$, and 99.7% within $$\bar{x}\pm 3s$$ to examine whether the data follow the empirical rule.

Note that a z-score tells us how many standard deviations an observation is from the mean. A positive z-score $$z>0$$ is above the mean; a negative z-score $$z<0$$ is below the mean.

Example: $$z=-0.23$$ is 0.23 standard deviations below the mean.

### 4.6.3 Percentiles

We can also find the observation associated with a percentage/proportion.

The $$w$$th percentile $$p_w$$ is the observation that is higher than w% of all observations $P(X < p_w) = w$

Finding a Percentile

1. Sketch the normal curve for the variable.
2. Shade the region of interest and label the area.
3. Use the applet to determine the z-score for the area.
4. Find the x-value using $$z$$, $$\mu$$, and $$\sigma$$.

Note that if $$z = \frac{x-\mu}{\sigma}$$, then $$x = \mu + z\sigma$$.

Example: Find the 90th percentile for SAT scores.

From the previous example, we know that SAT scores are approximately Normal($$\mu=1100$$, $$\sigma=200$$). 1. Sketch the normal curve.

1. Shade the region of interest and label the area.

1. Use R with qnorm to determine the z-score for the area:
qnorm(0.9)
## [1] 1.281552

Find the x-value using $$z\approx 1.2816$$, $$\mu=1100$$, and $$\sigma=200$$: $x = 1100 + 1.2816(200) = 1356.32$ so 90% of SAT test-takers score below 1356.32.