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Discrete Random Variables: Teacher's Guide

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

Summary: This module is the complementary teacher's guide for the "Discrete Random Variables" chapter of the Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.

This chapter introduces expected value (long term average) and four of the common discrete random variables (binomial, geometric, hypergeometric, and Poisson). The authors cover expected value and two of the discrete random variables (binomial and Poisson). Depending on your background, you may want to cover the binomial (usually required) together with none or some of the other discrete random variables

Random Variables

Explain random variable (assigns numerical values to the outcomes of a statistical experiment). Upper case letters denote random variables. Example: Let XX = the number of cars in your household. (The phrase "the number of" tells you that XX takes on discrete values.) XX takes on the values 0, 1, 2, 3, ...

The Probability Distribution Function

A probability distribution function (pdf) is best shown with an example: A controversial drug is given to two patients. Let X = the number of patients cured.

  • P(a cure)=56P(a cure)=56
  • P(no cure)=16P(no cure)=16

A pdf is easiest to understand in a table.

Table 1: Each probability is between 0 and 1.
X X P(X)P(X) or P(X=x)P(X=x)
0 P ( X = 0) = ( 1 6 ) ( 1 6 ) = ( 1 36 ) P(X=0)=( 1 6 )( 1 6 )=( 1 36 )
1 P ( X = 1) = 2 ( 1 6 ) ( 5 6 ) = ( 10 36 ) P(X=1)=2( 1 6 )( 5 6 )=( 10 36 )
2 P ( X = 2) = ( 5 6 ) ( 5 6 ) = ( 25 36 ) P(X=2)=( 5 6 )( 5 6 )=( 25 36 )

The previous example can be used as an example of expected value or long term average ( μμ). Make a third column labeled (x)(P(x))(x)(P(x)). Calculate the three values and add them. The result, (0)( 1 36 )+(1)(1036)+(2)(2536)=6036=1.67(0)( 1 36 )+(1)(1036)+(2)(2536)=6036=1.67, is the expected number of patients who are cured if the drug is administered many times to two patients.

The binomial is a special discrete pdf or pattern. A binomial experiment consists of counting the number of successes in one or more Bernoulli trials. (A Bernoulli trial has only two possible outcomes, success or failure. In every Bernoulli trial, the probability of a success (or failure) remains the same.)

Example 1

Problem 1

John comes to his stat class and discovers he must take a true-false quiz . There are 20 questions on the quiz. John has not attended class recently and must guess randomly at the questions. Let XX = the number of questions John answers correctly out of 20 questions. XX takes on the values 0, 1, 2, 3, ..., 20. P(correct answer: a success)P(correct answer: a success) =0.5=0.5. John's guessing at the answers is a binomial experiment.

Notation: XX ~ B(20,0.5)B(20,0.5) where the number of trials, nn , is 20 and the probability of a success, pp, on any trial is 0.5.

Students can find the mean (μ = npμ=np ), and the standard deviation (σ=σ= square root of npqnpq) either by hand or with technology. ( qq is the probability of a failure.) Have students help you fill in the blanks and answer the questions:

  1. σ=σ=
  2. Draw the graph. (horizontal axis is the number of successes; vertical is the probability of 0 successes, 1 success, 2 successes, ..., 20 successes. Draw vertical lines or boxes.
    • What is the probability that John gets 15 questions correct? P(X=15)P(X=15)
    • More than 15 questions correct? P(X>15)P(X>15)
    • At least 15 questions correct? (P(X=15)+P(X>15))(P(X=15)+P(X>15))

A geometric experiment takes place when at least one Bernoulli trial is performed and all are failures except the last one which is the only success. Example: Liz likes to play darts. The probability that she hits the bull's eye (success) on any throw is 85%. (Liz is good!) Liz throws darts at the bull's eye until she hits it. Let XX = the number of times Liz throws the dart at the bull's eye until she hits it. Have students help you fill in the blanks:

Fill in the blanks.

  • XX~ _______ ( XX ~ G(p)G(p) where p=p= probability of a success= 0.85)
  • Draw the graph. (Number of throws until the first success versus probability)
  • 4. What is the probability that Liz hits the bull's eye for the first time on the third throw? That it takes more than three throws for Liz to hit the bull's eye for the first time? That it takes at least three throws?
  • XX takes on the values _______.
  • μ=μ= _______. In words, μμ is _______

The Geometric Equation


Hypergeometric Distribution

The hypergeometric distribution is characterized by choosing a sample without replacement from two distinct groups. One of the two groups is what is of interest in the sample. Some lotteries are based on the hypergeometric distribution. click to edit note

Example 2

Suppose a shipment of 20 tape recorders contains 5 defectives. An inspector randomly chooses 8 of the tape recorders to inspect. He is interested in the number of defectives in the sample of 8. Have the class answer questions similar to those for the binomial and the geometric.


XX ~ H(r,b,n)H(r,b,n) where rr = size of the group of interest, bb = size of the other group, and nn = size of the sample.

Poisson Distribution

The Poisson distribution is concerned with the number of times an event takes place in a certain interval. It is used in the field of reliability. The Poisson approximates the binomial when n is "large" (say, more than 100) and p is "small" (say, less than 0.1).

Example 3

Suppose the average number of accidents that occur in a week at a particularly busy intersection is one. The interval is one week. The average is one accident. Let XX = the number of accidents that occur in a one week period at the intersection. Have the students help fill in the blanks and answer the questions:

  1. X~X~ _______ (X ~ P ( μX~P(μ where μ=oneaccident)μ=oneaccident))
  2. What values does XX take on?
  3. What is the probability that at most one accident occurs in a week?

The Poisson Distribution Formula

The parameter for the Poisson is the mean, μμ size 12{μ} {}. Some books and calculators use the Greek letter, λλ size 12{λ} {} (lambda) as the mean. The equation for the Poisson is:

P(X=x)=μxeμx! where x=0,1,2,3,...P(X=x)=μxeμx! where x=0,1,2,3,...

Assign Practice

Have the students complete the portion of the practice that is appropriate for what you have covered in class. Expected Value, Binomial, and Poisson are dealt with Practice 1, Practice 2, and Practice 3. Practice 4 is based on the Geometric Distribution, while Practice 5 is focused on reviewing the Hypergeometric Distribution.

Calculator Instructions

If you are using the TI-83/TI-84 series, there are probability functions for the binomial, Poisson, and geometric. Each has a pdf and a cdf (for example binompdf and binomcdf).These functions are located in 2nd DISTR. If you use, say, binompdf(n,p) , you will get the table of probabilities for 0, 1, 2, ..., nn. If you use binompdf(n,p) , you will get the probability of xx. If you use binomcdf(n, p, x), you will get the cumulative probability (P(X=0)+P(X=1)+P(X=2)+...+P(X=n))(P(X=0)+P(X=1)+P(X=2)+...+P(X=n)).

Assign Homework

Assign Homework. Suggested homework: 1 - 17 odds, 23, 33 - 37 (Binomial and Poisson).

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