Barbara Illowsky; Susan Dean

There are two types of random variables, discrete random variables and continuous random variables. The values of a discrete random variable are countable, which means the values are obtained by counting. All random variables we discussed in previous examples are discrete random variables. We counted the number of red balls, the number of heads, or the number of female children to get the corresponding random variable values. The values of a continuous random variable are uncountable, which means the values are not obtained by counting. Instead, they are obtained by measuring. For example, let X = temperature of a randomly selected day in June in a city. The value of X can be 68°, 71.5°, 80.6°, or 90.32°. These values are obtained by measuring by a thermometer. Another example of a continuous random variable is the height of a randomly selected high school student. The value of this random variable can be 5'2", 6'1", or 5'8". Those values are obtained by measuring by a ruler.

A discrete probability distribution function has two characteristics:

Each probability is between zero and one, inclusive.
The sum of the probabilities is one.

Example 4.1

A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5.

P(x) = probability that X takes on a value x.

x	P(x)
0	P(x = 0) = $\frac{2}{50}$
1	P(x = 1) = $\frac{11}{50}$
2	P(x = 2) = $\frac{23}{50}$
3	P(x = 3) = $\frac{9}{50}$
4	P(x = 4) = $\frac{4}{50}$
5	P(x = 5) = $\frac{1}{50}$

Table 4.2

X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because we can count the number of values of x and also because of the following two reasons:

Each P(x) is between zero and one, therefore inclusive
The sum of the probabilities is one, that is,

\frac{2}{50} + \frac{11}{50} + \frac{23}{50} + \frac{9}{50} + \frac{4}{50} + \frac{1}{50} = 1

Try It 4.1

A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes on value x. Why is this a discrete probability distribution function (two reasons)?

X	P(x)
0	P(x = 0) = $\frac{4}{50}$
1	P(x = 1) = $\frac{8}{50}$
2	P(x = 2) = $\frac{16}{50}$
3	P(x = 3) = $\frac{14}{50}$
4	P(x = 4) = $\frac{6}{50}$
5	P(x = 5) = $\frac{2}{50}$

Table 4.3

Example 4.2

Suppose Nancy has classes three days a week. She attends classes three days a week 80 percent of the time, two days 15 percent of the time, one day 4 percent of the time, and no days 1 percent of the time. Suppose one week is randomly selected.

Problem

Describe the random variable in words. Let X = the number of days Nancy ________.

Solution

a. Let X = the number of days Nancy attends class per week.

Problem

b. In this example, what are possible values of X?

Solution

b. 0, 1, 2, and 3

Problem

c. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example 4.1. The table should have two columns labeled x and P(x).

Solution

x	P(x)
0	.01
1	.04
2	.15
3	.80

Table 4.4

The sum of the P(x) column is 0.01+0.04+0.15+0.80 = 1.00.

Try It 4.2

Jeremiah has basketball practice two days a week. 90 percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X and what values does it take on?

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable

Problem

Solution

Problem

Solution

Problem

Solution