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# 4.3: Mean and Standard Deviation of Discrete Random Variables

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Know the definition of the mean, or expected value, of a discrete random variable.
• Know the definition of the standard deviation of a discrete random variable.
• Know the definition of the variance of a discrete random variable.
• Find the expected value of a variable.

## Introduction

In this lesson, you will be presented with the formulas for the mean, variance, and standard deviation of a discrete random variable. You will also be shown many real-world examples of how to use these formulas. In addition, the meaning of expected value will be discussed.

### Characteristics of a Probability Distribution

The most important characteristics of any probability distribution are the mean (or average value) and the standard deviation (a measure of how spread out the values are). The example below illustrates how to calculate the mean and the standard deviation of a random variable. A common symbol for the mean is μ\begin{align*}\mu\end{align*} (mu), the lowercase m\begin{align*}m\end{align*} of the Greek alphabet. A common symbol for standard deviation is σ\begin{align*}\sigma\end{align*} (sigma), the Greek lowercase s\begin{align*}s\end{align*}.

Example: Recall the probability distribution of the 2-coin experiment. Calculate the mean of this distribution.

If we look at the graph of the 2-coin toss experiment (shown below), we can easily reason that the mean value is located right in the middle of the graph, namely, at x=1\begin{align*}x = 1\end{align*}. This is intuitively true. Here is how we can calculate it:

To calculate the population mean, multiply each possible outcome of the random variable X\begin{align*}X\end{align*} by its associated probability and then sum over all possible values of X\begin{align*}X\end{align*}:

μ=(0)(14)+(1)(12)+(2)(14)=0+12+12=1\begin{align*}\mu=(0) \left(\frac{1}{4}\right) + (1) \left(\frac{1}{2}\right) + (2) \left(\frac{1}{4}\right)=0+\frac{1}{2}+\frac{1}{2}=1\end{align*}

### Mean Value or Expected Value

The mean value, or expected value, of a discrete random variable X\begin{align*}X\end{align*} is given by the following equation:

μ=E(x)=xp(x)\begin{align*}\mu=E(x)=\sum_{} xp(x)\end{align*}

This definition is equivalent to the simpler one you have learned before:

μ=1ni=1nxi\begin{align*}\mu = \frac{1}{n} \sum^n_{i=1} x_i\end{align*}

However, the simpler definition would not be usable for many of the probability distributions in statistics.

Example: An insurance company sells life insurance of $15,000 for a premium of$310 per year. Actuarial tables show that the probability of death in the year following the purchase of this policy is 0.1%. What is the expected gain for this type of policy?

There are two simple events here: either the customer will live this year or will die. The probability of death, as given by the problem, is 0.001, and the probability that the customer will live is 10.001=0.999\begin{align*}1-0.001=0.999\end{align*}. The company’s expected gain from this policy in the year after the purchase is the random variable, which can have the values shown in the table below.

Gain, x\begin{align*}x\end{align*} Simple Event Probability
310 Live 0.999 \begin{align*}-\end{align*}14,690 Die 0.001

Figure: Analysis of the possible outcomes of an insurance policy.

Remember, if the customer lives, the company gains $310 as a profit. If the customer dies, the company "gains"$310$15,000=$14,690\begin{align*}\310 - \15,000 = -\14,690\end{align*}, or in other words, it loses $14,690. Therefore, the expected profit can be calculated as follows: μμμ=E(x)=xp(x)=(310)(99.9%)+(31015,000)(0.1%)=(310)(0.999)+(31015,000)(0.001)=309.6914.69=$295=295\begin{align*}\mu &= E(x)=\sum_{} xp(x)\\ \mu &= (310)(99.9 \%)+(310 - 15,000)(0.1 \%)\\ &= (310)(0.999)+(310-15,000)(0.001)\\ &= 309.69 - 14.69=\295\\ \mu &= \295\end{align*} This tells us that if the company were to sell a very large number of the 1-year15,000 policies to many people, it would make, on average, a profit of $295 per sale. Another approach is to calculate the expected payout, not the expected gain: μμ=(0)(99.9%)+(15,000)(0.1%)=0+15=$15\begin{align*}\mu &= (0)(99.9 \%)+(15,000)(0.1 \%)\\ &= 0+15\\ \mu &= \ 15\end{align*}

Since the company charges $310 and expects to pay out$15, the average profit for the company is \$295 per policy.

Sometimes, we are interested in measuring not just the expected value of a random variable, but also the variability and the central tendency of a probability distribution. To do this, we first need to define population variance, or σ2\begin{align*}\sigma^2\end{align*}. It is the average of the squared distance of the values of the random variable X\begin{align*}X\end{align*} from the mean value, μ\begin{align*}\mu\end{align*}. The formal definitions of variance and standard deviation are shown below.

### The Variance

The variance of a discrete random variable is given by the following formula:

σ2=(xμ)2P(x)\begin{align*}\sigma^2 = \sum_{} (x-\mu)^2 P(x)\end{align*}

### The Standard Deviation

The square root of the variance, or, in other words, the square root of σ2\begin{align*}\sigma^2\end{align*}, is the standard deviation of a discrete random variable:

σ=σ2\begin{align*}\sigma=\sqrt{\sigma^2}\end{align*}

Example: A university medical research center finds out that treatment of skin cancer by the use of chemotherapy has a success rate of 70%. Suppose five patients are treated with chemotherapy. The probability distribution of x\begin{align*}x\end{align*} successful cures of the five patients is given in the table below:

xp(x)00.00210.02920.13230.30940.36050.168\begin{align*}& x && 0 && 1 && 2 && 3 && 4 && 5\\ & p(x) && 0.002 && 0.029 && 0.132 && 0.309 && 0.360 && 0.168\end{align*}

Figure: Probability distribution of cancer cures of five patients.

a) Find μ\begin{align*}\mu\end{align*}.

b) Find σ\begin{align*}\sigma\end{align*}.

c) Graph p(x)\begin{align*}p(x)\end{align*} and explain how μ\begin{align*}\mu\end{align*} and σ\begin{align*}\sigma\end{align*} can be used to describe p(x)\begin{align*}p(x)\end{align*}.

a. To find μ\begin{align*}\mu\end{align*}, we use the following formula:

μμμ=E(x)=xp(x)=(0)(0.002)+(1)(0.029)+(2)(0.132)+(3)(0.309)+(4)(0.360)+(5)(0.168)=3.50\begin{align*}\mu &= E(x)=\sum_{} xp(x)\\ \mu &= (0)(0.002)+(1)(0.029)+(2)(0.132)+(3)(0.309)+(4)(0.360)+(5)(0.168)\\ \mu &= 3.50\end{align*}

b. To find σ\begin{align*}\sigma\end{align*}, we first calculate the variance of X\begin{align*}X\end{align*}:

σ2σ2=(xμ)2p(x)=(03.5)2(0.002)+(13.5)2(0.029)+(23.5)2(0.132)+(33.5)2(0.309)+(43.5)2(0.36)+(53.5)2(0.168)=1.05\begin{align*}\sigma^2 &= \sum_{} (x-\mu)^2 p(x)\\ &= (0-3.5)^2(0.002)+(1-3.5)^2 (0.029)+(2-3.5)^2(0.132)\\ & \quad +(3-3.5)^2(0.309)+(4-3.5)^2(0.36)+(5-3.5)^2(0.168)\\ \sigma^2 &= 1.05\end{align*}

Now we calculate the standard deviation:

σ=σ2=1.05=1.02\begin{align*}\sigma = \sqrt{\sigma^2}=\sqrt{1.05}=1.02\end{align*}

c. The graph of p(x)\begin{align*}p(x)\end{align*} is shown below:

We can use the mean, or μ\begin{align*}\mu\end{align*}, and the standard deviation, or σ\begin{align*}\sigma\end{align*}, to describe p(x)\begin{align*}p(x)\end{align*} in the same way we used x¯¯¯\begin{align*}\overline{x}\end{align*} and s\begin{align*}s\end{align*} to describe the relative frequency distribution. Notice that μ=3.5\begin{align*}\mu=3.5\end{align*} is the center of the probability distribution. In other words, if the five cancer patients receive chemotherapy treatment, we expect the number of them who are cured to be near 3.5. The standard deviation, which is σ=1.02\begin{align*}\sigma=1.02\end{align*} in this case, measures the spread of the probability distribution p(x)\begin{align*}p(x)\end{align*}.

## Lesson Summary

The mean value, or expected value, of the discrete random variable X\begin{align*}X\end{align*} is given by μ=E(x)=xp(x)\begin{align*}\mu=E(x)=\sum_{} xp(x)\end{align*}.

The variance of the discrete random variable X\begin{align*}X\end{align*} is given by σ2=(xμ)2p(x)\begin{align*}\sigma^2=\sum_{} (x-\mu)^2 p(x)\end{align*}.

The square root of the variance, or, in other words, the square root of \begin{align*}\sigma^2\end{align*}, is the standard deviation of a discrete random variable: \begin{align*}\sigma=\sqrt{\sigma^2}\end{align*}.

For an example of finding the mean and standard deviation of discrete random variables (5.0)(6.0), see EducatorVids, Statistics: Mean and Standard Deviation of a Discrete Random Variable (2:25).

For a video presentation showing the computation of the variance and standard deviation of a set of data (11.0), see American Public University, Calculating Variance and Standard Deviation (8:51).

For an additional video presentation showing the calculation of the variance and standard deviation of a set of data (11.0), see Calculating Variance and Standard Deviation (4:31).

## Review Questions

1. Consider the following probability distribution: \begin{align*}& x && 0 && 1 && 2 && 3 && 4\\ & p(x) && 0.1 && 0.4 && 0.3 && 0.1 && 0.1\end{align*} Figure: The probability distribution for question 1.
1. Find the mean of the distribution.
2. Find the variance.
3. Find the standard deviation.
2. An officer at a prison questioned each inmate to find out how many times the inmate has been convicted. The officer came up with the following table that shows the relative frequencies of \begin{align*}X\end{align*}, the number of times convicted: \begin{align*}& x && 0 && 1 && 2 && 3 && 4\\ & p(x) && 0.16 && 0.53 && 0.20 && 0.08 && 0.03\end{align*} Figure: The probability distribution for question 2. If we regard the relative frequencies as approximate probabilities, what is the expected value of the number of previous convictions of an inmate?

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