3.3. Expected (mean) value and variance for discrete random variables

3.3.1. Expected value

Once we have constructed the probability distribution for a random variable, we often want to compute the mean or expected value of the random variable. The mean of discrete random variable X, denoted either

or , is actually the mean of its probability distribution. The mean

(or expected) value of a discrete random variable is the value that we expect to observe per repetition, on average, if we perform an experiment a large number of times. For example, we may expect a house salesperson to sell on average, 3.50 houses per month. It does not mean that every month this salesperson will sell exactly 3.50 houses. (Actually he (or she) can not sell exactly 3.50 houses). This simply means that if we observe for many months, this salesperson will sell a different number of houses different months. However, the average of all sold houses in these months will be 3.50.

Definition:

The mean (or expected value) of discrete random variable X is defined as

Here the sum extends over all distinct values x of X.

In order to compute the expected value of a discrete random variable we must multiply each value of the random variable by the corresponding value of its probability function. We then add the resulting terms.

Example:

Sales show that five is the maximum number of cars sold on a given day at car selling company. Table 3.4 shows probability distribution of cars sold per day. Find the expected number of cars sold.

Table 3.4

x	P(x)
0 1 2 3 4 5	0.18 0.39 0.24 0.14 0.04 0.01

Solution:

To find the expected number (or mean) of cars sold, we multiply each value of x by its probability and add these results.

x	P(x)
0 1 2 3 4 5	0.18 0.39 0.24 0.14 0.04 0.01	0.00 0.39 0.48 0.42 0.16 0.05

In fact, it is impossible for company to sell exactly 1.50 cars in any given day. But we examine selling cars at this company for many days into the future, and see that, the expected value of 1.50 cars provides a good estimate of the mean or average daily sales volume. The expected value can be important to the managers from both planning and decision making points of view.

For example, suppose that this company will be open 40 days during next

2-month. How many cars should the owner expect to be sold during this time?

While we can not specify the exact value of 1.50 cars, it provides an expected sale of cars for the next 2-month period.

3.3.2. Variance and standard deviation of discrete random variable

While the expected value gives us an idea of the average or central value for the random variable, often we would also like to measure the dispersion or variability of the possible values of the random variable. The variance of discrete random variable X, denoted by , measures the spread of its probability distribution. In defining the variance of a random variable, a weighted average of the squares of its possible discrepancies about the means is formed; the weight associated with is the probability that the random variable takes the value x. The variance can be viewed as the average value that will be taken by the function over a very large number of repeated trials.

The mathematical expression for the variance of a discrete random variable is

.

The standard deviation, , is the positive square root of the variance.

In some particular cases, an alternative but equivalent (sometimes called shortcut formula) formula for the variance can be used:

Example:

Find the variance for the example in previous topic, for the number of cars sold per day at a car selling company.

Solution:

Let us apply .

The calculations are shown in the table 3.5:

Table3.5

x			P(x)
0 1 2 3 4 5	0-1.50=-1.50 1-1.50=-0.50 2-1.50=0.50 3-1.50=1.50 4-1.50=2.50 5-1.50=3.50	2.25 0.25 0.25 2.25 6.25 12.25	0.18 0.39 0.24 0.14 0.04 0.01	2.25·0.18=0.4050 0.25·0.39=0.0975 0.25·0.24=0.0600 2.25·0.14=0.3150 6.25· 0.04=0.2500 12.25· 0.01=0.1225

We see that the variance for the number of cars sold per day is 1.25.

The standard deviation of the number of cars sold per day is .

Remark:

For the purpose of easier managerial interpretation the standard deviation may be preferred over the variance because it is measured in the same units as the random variable.

Example:

The following table gives the probability distribution of X.

x	0 1 2 3 4 5
P(x)	0.02 0.20 0.30 0.30 0.10 0.08

Compute the standard deviation of x.

Solution:

Let us apply equivalent (shortcut formula) formula for the variance

The following table shows all the calculations required for the computation of the standard deviation of x.

x	P(x)	x P(x)
0 1 2 3 4 5	0.02 0.20 0.30 0.30 0.10 0.08	0.00 0.20 0.60 0.90 0.40 0.40	0 1 4 9 16 25	0.00 0.20 1.20 2.70 1.60 2.00

We perform the following steps to compute the standard deviation by shortcut formula:

Step1: Compute the mean of discrete random variable:

Step2: Compute the value of .

Step3: Substitute the values of and in the shortcut formula for the variance

=

Step4: Take positive square root of variance.

.

Example:

A farmer will earn a profit of $30 thousand in case of heavy rain next year, $60 thousand in case of a moderate rain, and $15 thousand in case of little rain. A meteorologist forecasts that the probability is 0.35 for heavy rain, 0.40 for moderate rain, and 0.25 for little rain next year. Let X be the random variable that represents next year’s profit in thousands of dollars for this farmer. Write the probability distribution of x. Find the mean and standard deviation of x. Give a brief interpretation of the values of the mean and standard deviation.

Solution: Table 3.6

x	P(x)
30 60 15	0.35 0.40 0.25

The table3.6 lists the probability distribution of x

The table 3.7 shows all calculations needed for the computation of the mean and standard deviation.

Table 3.7

x	P(x)	x P(x)
30 60 15	0.35 0.40 0.25	10.5 24 3.75	900 3600 225	315 1440 56.25

The mean of x is $38.25 thousand. The standard deviation is $18.660.

Thus, it is expected that a farmer will earn an average of $38.25 thousand profits in next year with a standard deviation of $18.660 thousand.