In this Concept, 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 examples applications of these formulas. In addition, the meaning of expected value will be discussed.

### Watch This

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 (8:52).

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 (4:36).

### Guidance

**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 \begin{align*}\mu\end{align*} (mu), the lowercase \begin{align*}m\end{align*} of the Greek alphabet. A common symbol for standard deviation is \begin{align*}\sigma\end{align*} (sigma), the Greek lowercase \begin{align*}s\end{align*}.

#### Example A

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 \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 \begin{align*}X\end{align*} by its associated probability and then sum over all possible values of \begin{align*}X\end{align*}:

\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 \begin{align*}X\end{align*} is given by the following equation:

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

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

\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 B

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 \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, \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" \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:

\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-year $15,000 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:

\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 \begin{align*}\sigma^2\end{align*}. It is the average of the squared distance of the values of the random variable \begin{align*}X\end{align*} from the mean value, \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:

\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 \begin{align*}\sigma^2\end{align*}, is the standard deviation of a discrete random variable:

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

#### Example C

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 \begin{align*}x\end{align*} successful cures of the five patients is given in the table below:

\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 \begin{align*}\mu\end{align*}.

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

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

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

\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 \begin{align*}\sigma\end{align*}, we first calculate the variance of \begin{align*}X\end{align*}:

\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:

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

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

We can use the mean, or \begin{align*}\mu\end{align*}, and the standard deviation, or \begin{align*}\sigma\end{align*}, to describe \begin{align*}p(x)\end{align*} in the same way we used \begin{align*}\overline{x}\end{align*} and \begin{align*}s\end{align*} to describe the relative frequency distribution. Notice that \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 \begin{align*}\sigma=1.02\end{align*} in this case, measures the spread of the probability distribution \begin{align*}p(x)\end{align*}.

### Guided Practice

*A random variable has the following probability distribution:*

\begin{align*} &\text{Value of X} &&-2 &&0 &&2\\ &\text{Probability} &&.10 &&.80 &&.10 \end{align*}

a) Calculate the mean of X.

b) Calculate the variance of X

c) Calculate the standard deviation of X.

**Solution:**

a) Mean of X is \begin{align*}-2(0.10) + 0(0.80) + 2(0.10) = -0.2 + 0.2 = 0\end{align*}.

b) Variance of \begin{align*}X = (-0.2)^2(0.10)+(0.2)^2(0.10)+0(0.80)=0.4+0.4=0.8.\end{align*}

c) Standard deviation of \begin{align*}X\end{align*} is \begin{align*}\sqrt{0.8}=0.894\end{align*}.

### Explore More

- 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.

a. Find the mean of the distribution.

b. Find the variance.

c. Find the standard deviation.

- An officer at a prison was studying recidivism among the prison inmates. The officer questioned each inmate to find out how many times the inmate had been convicted prior to the inmate’s current conviction. The officer came up with the following table that shows the relative frequencies of X, the number of times previously 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?

- Suppose \begin{align*}X\end{align*} has the following distribution table:

\begin{align*} &\text{Value} &&\text{Probability}\\ &2 &&1/5\\ &3 &&2/5\\ &5 &&2/5 \end{align*}

Find the expected value.

- The possible values for a certain random variable are 1, 2, 3, and 8. Part of its distribution table is given below. Fill in the blank, and find the expected value.

\begin{align*} &\text{Value} &&\text{Probability}\\ &1 &&.3\\ &2 &&.2\\ &3 && \\ &8 &&.4 \end{align*}

- Suppose \begin{align*}n\end{align*} draws are made at random with replacement from a box of numbered balls. Let \begin{align*}S\end{align*} be the sum of the draws. Show that the expected value of \begin{align*}S\end{align*} is equal to \begin{align*}n\end{align*} (the average of the box).
- A die is thrown twice. Let \begin{align*}X_1\end{align*} be the number of spots on the first thrown and \begin{align*}X_2\end{align*} be the number of spots on the second throw. Find \begin{align*}E(X_1\cdot X_2)\end{align*}.
- Find the expected value of the random variable with the following distribution table:

\begin{align*} &\text{Value} &&\text{chance}\\ &-2 &&1/5\\ &0 &&2/5\\ &2 &&1/5\\ &4 &&1/5 \end{align*}

- The possible values for a certain random variable are 1, 3, 4, and 8. Part of its distribution table is given below. Fill in the blank, and find the expected value.

\begin{align*} &\text{Value} &&\text{Chance}\\ &1 &&3/11\\ &2 &&1/11\\ &4 && \\ &7 &&4/11 \end{align*}

- Suppose X represents the number of children in a family. Following is the probability distribution for X for families with particular characteristics;

\begin{align*} &\text{Number of children} &&0 &&1 &&2 &&3\\ &\text{Probability} &&.05 &&.60 &&.30 &&.05 \end{align*}

a. Is this a valid probability distribution? Explain.

b. What is the expected value of X? What does this mean?

- Suppose the probability that you get an A in any class is .4 and the probability that you get a B in the class is 0.6. To construct a grade point average an A is worth 4.0 and a B is worth 3.0.
- Is it possible that you will get a C in this class? Explain.
- What is the expected value of your grade point average?

- Suppose you have to take the bus to school. The probability that you will have to wait for the bus is .25. If you don’t have to wait for the bus the commute takes 20 minutes, but it you have to wait for the bus, the commute takes 30 minutes. What is the expected value of the time it takes you to commute to school?

*Keywords*

Discrete

Expected value

Probability distribution

Random variables

Standard deviation

Variance