<meta http-equiv="refresh" content="1; url=/nojavascript/"> Discrete Probability Distributions | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Probability and Statistics - Advanced (Teachers Edition) Go to the latest version.

Introduction

Explore: Probability as Area

In a graph of a probability distribution, area is equal to the probability. This fact becomes especially important when calculating probabilities with a normal distribution in the next chapter or using integrals to find probabilities in later classes. It is helpful to get students accustom to this concept now while working with the more straight forward case of discrete random variables.

Procedure:

Consider the probability experiment of flipping a fair coin four times with the discrete random variable, X = the number of heads in those four flips.

  1. What are the possible values of the discrete random variable?
  2. Calculate the probability of each possible result.
  3. Display the probability distribution for this experiment as a table and as a graph. Check your work by seeing if the two conditions that must be true for all probability distributions are satisfied.
  4. Calculate the area of each rectangular bar in the graph of the probability distribution. What is the sum of the areas?
  5. What is the total area over the random variable values of two and three? What is the probability of getting two or three heads?
  6. What is the relationship between probabilities and area in a probability distribution?

Answers:

  1. \left \{0, 1, 2, 3, 4 \right \}
  2. P(0) = \frac{1}{16}, P(1) = \frac{4}{16}, P(2) = \frac{6}{16}, P(3) = \frac{4}{16}, P(4) = \frac{1}{16}
  3. .
  4. \frac{1}{16} + \frac{4}{16} + \frac{6}{16} + \frac{4}{16} + \frac{1}{16} = \frac{16}{16} = 1
  5. \frac{6}{16} + \frac{4}{16} = \frac{10}{16}, P(2 \ \text{or} \ 3) = P(2) + P(3) - P(2 \ \text{and} \ 3) = \frac{6}{16} + \frac{4}{16} - 0 = \frac{10}{16}
  6. The area is equal to the probability.

Mean and Standard Deviation of Discrete Random Variables

Project: Collecting and Displaying Discrete Data

At this point in the class a project that pulls together many of the important aspects of what the students have been learning in the first few chapters will have substantial benefit. It provides an opportunity to review, and the added benefit of work with real data will improve students experience and intuition.

Objective: To design and perform an experiment that yields values of a discrete random variable and to display the resulting probability distribution as a table and as a bar graph to be analysed.

Procedure:

  1. Design a probability experiment that produces values of a discrete random variable.
  2. Perform the experiment at least 20\;\mathrm{times} and record the results.
  3. Use the collected data to calculate the probability of each value of the random variable.
  4. Make a table and a graph for the probability distribution.
  5. Calculate the mean, variance, and standard deviation for the probability distribution.
  6. Describe the shape of the distribution.

Technology:

Encourage the students to use a spreadsheet to record data, and to make calculations, tables, and graphs. Increasing the amount of data the students must collect will make using technology more rewarding. A quick demonstration of how to use formulas and fill in columns will get students started. If they need additional help, or want to learn more about how to use this tool, additional information and tutorials can be found online.

The Binomial Probability Distribution

Explore: Selection Without Replacement

The characteristics of a binomial experiment require that the probability of success to be constant form trial to trial, and that the trials are independent of each other. These conditions do not hold when selecting from a finite group without replacement, but if the group is large enough, and the number of trials is small enough, maybe we can get “close enough”.

Case One:

In a class of 30 students you have 4 good friends. The instructor is randomly selecting three students to present their project today. Let X = the number of your good friends selected

1. Why isn’t X a binomial random variable?

2. Use classical probability, combinations, and the Multiplicative Rule of Counting to calculate the probability that exactly two of the presenters are your close friends.

3. Use the binomial probability distribution with p = \frac{4}{20} to approximate the probability of having exactly two of your close friends chosen. What do you think of this approximation?

Case Two:

In a school of 1200 students, 3 will be randomly selected to complete a survey about the school lunch program. Let X = the number of those same 4 good friends that are selected for the survey.

4. Use classical probability, combinations, and the Multiplicative Rule of Counting to calculate the probability that exactly two of the four close friends are chosen.

5. Use the binomial probability distribution with p = \frac{4}{700} to approximate the probability of having exactly two of the close friends chosen. What do you think of this approximation?

Answers:

  1. Once a student is selected they are removed from the pool of choices so the probability of success for the next trial changes. The trials are also not independent. If the first trial is a success, the chances of a success on the second trial are lower.
  2. \frac{(_4C_2 *_{26}C_1)}{_{30}C_3} = \frac{156}{4060} \approx 0.0384
  3. _3C_2*(.2)^2(.8)^1 = .096, not a good estimate
  4. \frac{(_4C_2 *_{1196}C_1)}{{1200}C_3} \approx 0.0000250
  5. _3C_2 * \left (\frac{4}{1200} \right )^2 \left (\frac{1196}{1200}\right )^1 \approx 0.0000322, this estimate is much better

The Poisson Probability Distribution

Practice and Extend: Changing Units for the Poisson Distribution The parameter \lambda gives the mean number of events in a certain amount of time, distance, volume, or area. Sometimes though, it is not given in the desired units. For instance, \lambda could be the average number of accidents at a given intersection in a year, but the probability to the calculated is for three accidents in a given month. It is an easy fix; \lambda can just be divided by twelve. Of course, it won’t be month specific, December probably has a higher average than May, and this method will not reflect that difference. In many situations, statisticians calculate the best value that they can, and then consider the inaccuracies.

Exercises:

  1. On average there is one flaw found in every yard of sheetrock produced on a specific machine.
    1. What are the mean and standard deviation of the distribution of flaws per foot?
    2. What is the probability of finding a flaw in the first foot of sheetrock?
  2. A busy executive receives an average of 14 emails an hour.
    1. What is the probability that she will receive more than 150 emails in a ten hour work day?
    2. What is the probability that she will receive more than 200 emails in a ten hour work day?
  3. After being given a free trial, two out of three participants will enroll in a certain telephone service. What is the probability that exactly 70 out of 90 participants will enroll?

Answers:

    1. \mathrm{mean = standard \ deviation} = \frac{1}{3}
    2. p(1) \approx 0.2388
    1. \lambda = 140, P(X > 150) = 1 - P(X \le 150) \approx 1 - 0.8134 = .1866
    2. \lambda = 140, P(X > 200) = 1 - P(X \le 200) \approx 1 - 0.9999992528 \approx 0
  1. \lambda = 60, P(X = 70) \approx 0.02160

The Geometric Probability Distribution

Practice: Identifying Discrete Probability Distributions

Now that students know the basics about the Binomial, Poisson, and Geometric distributions it is time for them to work on identifying which distribution can be used in a given situation.

Exercises:

Identify which, if any, of the discrete probability distributions can be used in the following situations. Give the value(s) of the parameter(s) for the distribution, and find the indicated probability.

  1. In a class of 27, \ 18 students know the answer to question number three on the last exam. If the instructor randomly chooses students, what is the probability that he will have to call on more than two students before he is given the correct answer?
  2. In early August 2009, approximately 60 \% of Americans were in favor of a public option for health care. In a random sample of 10 Americans, what is the probability that less than 4 are in favor of a public plan?
  3. A bowl with four pieces of black liquorish flavored candy and eight pieces of cherry flavored candy is passed around a party. If it is not possible to distinguish between the types of candy when selecting, what is the probability that the first four candies taken are all black liquorish?
  4. A student band sells songs on its website to raise money for their favorite charity. They sell an average of 22 songs each month. What is the probability of more than 30 songs being sold next month?

Answers:

  1. Geometric, p = \frac{2}{3}, P(X > 2) = 1 - P(X \le 2) \approx 1 - 0.8889 = 0.1111
  2. Binomial, p = .6, n = 10, P(X < 4) \approx .0548
  3. None of the distributions learned in this chapter can be used. Each trial does not have the same probability of success and the trials are not independent.
  4. Poisson, \lambda = 22, P(X > 30) = 1 - P(X \le 30) \approx 0.0405

Image Attributions

Files can only be attached to the latest version of None

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.TE.1.Prob-&-Stats-Adv.2.4

Original text