Chapter 4: Discrete Random Variables

4.7 Using Technology for the Binomial Distribution

Learning Objectives

By the end of this section, the student should be able to:

  • use technology to solve problems involving the Binomial Distribution

This is another section to show technology tools that can be used to help with the statistics concepts taught in this chapter. This section will cover tools to help solve problems involving the Binomial Distribution.

Graphing Calculator

Similar to all the other technology sections, the main tool introduced will be the TI-83, 83+, 84, or 84+ Graphing Calculator. The graphing calculator can help to find the probabilities of binomial experiments, without having to use the Binomial Probability Function ([latex]P(X = x) = \frac{n!}{x!(n-x)!} p^x q^{(n-x)}[/latex]).

 

Using the TI-83, 83+, 84, 84+ Calculator for Binomial Distributions

  1. Access [DISTR] by pressing 2nd key, vars key. (Note: DISTR stands for "Distributions")
  2. The functions for the Binomial Distribution are:
    • <binompdf()> corresponds to P(X = x)
    • <binomcdf()> corresponds to P(X ≤ x)

The syntax for them is as follows:

  • binompdf(n, p, value) to calculate P(X = value). If "value" is left out of the function, the result is the list of all probabilities for x: 0, 1, . . . , n.
  • binomcdf(n, p, value) to calculate P(X ≤ value). If "value" is left out of the function, the result is the list of cumulative probabilities.

Legend

  • A blank calculator button represents a button press
  • [ ] represents yellow command or green letter behind a key
  • < > represents items on the screen

 

Let's use these calculator functions with some of the same examples we did in 4.3 Binomial Distribution.

Example

It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. Twenty adult workers are randomly selected. Let [latex]X =[/latex] the number of workers who have a high school diploma but do not pursue any further education.

Then [latex]X[/latex] takes on the values [latex]0, 1, 2, \ldots, 20[/latex] where [latex]n = 20, p = 0.41[/latex], and [latex]q = 1-0.41 = 0.59[/latex]. Finally, [latex]X \sim B(20,0.41).[/latex]

Find the probability that:

  1. Exactly 12 of them have a high school diploma but do not pursue any further education.
  2. At most 12 of them have a high school diploma but do not pursue any further education.
  3. More than 12 of them have a high school diploma but do not pursue any further education.

Using the binomial distribution functions, we can do the following in our graphing calculator:

  1. If you want to find exactly 12, the probability statement would be [latex]P(x = 12)[/latex]. We use the PDF function: [latex]\text{binompdf}[/latex]. [latex]P(x = 12) = \text{binompdf}(20,0.41,12) = 0.0417[/latex]
  2. To find at most 12, you would want to find [latex]P(X \le 12)[/latex]. We use the CDF function. On a calculator, that's [latex]\text{binomcdf}(20,0.41,12) = 0.9738[/latex]
  3. If you want to find more than 12, we are finding [latex]P(x > 12)[/latex]. We would have to use [latex]1 - \text{binomcdf}(20,0.41,12) = 0.0262[/latex].

 

Note

To compute binomial probabilities on a graphing calculator, the syntax is as follows:

  • To calculate [latex]P(X = x)[/latex]: binompdf(n, p, x).
  • To calculate [latex]P(X \leq x)[/latex]: binomcdf(n, p, x).
  • If you wanted to instead find [latex]P(X > x)[/latex], use 1 - binomcdf(n, p, x).

 

Here's some more examples, with solution, for you to try, from the problems covered in the Section.

Your Turn!

About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find:

  1. The probability that exactly 14 of them participate in a community volunteer program outside of school.
  2. The probability that at most 14 of them participate in a community volunteer program outside of school.
  3. The probability that more than 14 of them participate in a community volunteer program outside of school.

Solution

  1. [latex]P(x = 14) = \text{binompdf}(30,0.32,14) = 0.0359[/latex]
  2. [latex]P(x \le 14) = \text{binomcdf}(30,0.32,14) = 0.9695[/latex]
  3. [latex]P(x > 14) = 1 – P(x \le 14) = 1 – \text{binomcdf}(30,0.32,14) = 1 – 0.9695 = 0.0305[/latex]

Your Turn!

In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let [latex]X =[/latex] the number of pages that feature signature artists.

What is the probability distribution? Find the following probabilities:

  1. the probability that two pages feature signature artists
  2. the probability that at most six pages feature signature artists
  3. the probability that more than three pages feature signature artists.

Solution

[latex]X \sim B \left(100,\frac{8}{560}\right)[/latex]

  1. [latex]P(x = 2) = \text{binompdf}(100,\frac{8}{560},2) = 0.2466[/latex]
  2. [latex]P(x \le 6) = \text{binomcdf}(100,\frac{8}{560},6) = 0.9994[/latex]
  3. [latex]P(x > 3) = 1 – P(x \le 3) = 1 – \text{binomcdf}(100,\frac{8}{560},3) = 1 – 0.9443 = 0.0557[/latex]

Your Turn!

According to a Gallup poll, 60% of American adults prefer saving over spending. Let [latex]X =[/latex] the number of American adults out of a random sample of 50 who prefer saving to spending.

What is the probability distribution for [latex]X[/latex]? Use your calculator to find the following probabilities:

  1. the probability that 25 adults in the sample prefer saving over spending
  2. the probability that at most 20 adults prefer saving
  3. the probability that more than 30 adults prefer saving

Solution

[latex]X \sim B(50,0.6)[/latex]

  1. [latex]P(x = 25) = \text{binompdf}(50, 0.6, 25) = 0.0405[/latex]
  2. [latex]P(x \le 20) = \text{binomcdf}(50, 0.6, 20) = 0.0034[/latex]
  3. [latex]P(x > 30) = 1 - \text{binomcdf}(50, 0.6, 30) = 1 – 0.5535 = 0.446[/latex]

Your Turn!

During the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let [latex]X =[/latex] the number of shots that scored points.

What is the probability distribution for [latex]X[/latex]?

  1. Use your calculator to find the probability that DeAndre scored with 60 of these shots.
  2. Find the probability that DeAndre scored with more than 50 of these shots.

Solution

[latex]X \sim B(80, 0.613)[/latex]

  1. [latex]P(x = 60) = \text{binompdf}(80,0.613,60)=0.0036[/latex]
  2. [latex]P(x > 50) = 1 – P(x \le 50) = 1 – \text{binomcdf}(80, 0.613, 50) = 1 – 0.6282 = 0.3718[/latex]

Helpful Videos for the Graphing Calculator

Below are links to helpful videos for using the graphing calculator for the concepts covered on this page:

 

Additional Technology Tools

In addition to the graphing calculator, there are some additional technology tools that can be used for the concepts covered on this page. Below are links to helpful videos for those tools:

 

Microsoft Excel

In Microsoft Excel, both binomial probabilities are computed using BINOM.DIST(x, n, p, True/False), where False computes [latex]P(X=x)[/latex] and is equivalent to the graphing calculator binompdf(), and True computes [latex]P(X \leq x)[/latex] and is equivalent to binomcdf(). Note the order of the values are different than what is goes in the calculator. Let's see some Examples with using Excel:

Your Turn!

It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. Twenty adult workers are randomly selected. Let [latex]X =[/latex] the number of workers who have a high school diploma but do not pursue any further education.

Then [latex]X[/latex] takes on the values [latex]0, 1, 2, \ldots, 20[/latex] where [latex]n = 20, p = 0.41[/latex], and [latex]q = 1-0.41 = 0.59[/latex]. Finally, [latex]X \sim B(20,0.41).[/latex]

Find the probability that at most 12 of them have a high school diploma but do not pursue any further education.

Solution

In Excel, we enter BINOM.DIST(12, 20, 0.41, TRUE) and we get about 0.9738.

Your Turn!

About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that exactly 14 of them participate in a community volunteer program outside of school.

Solution

In Excel, we enter BINOM.DIST(14, 30, 0.32, FALSE) and would get 0.0359.

 

Section Practice

A student takes a ten-question true-false quiz but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct.

Solution

[latex]X =[/latex] number of questions answered correctly
[latex]X \sim B(10, 0.5)[/latex]
We are interested in AT LEAST 70% of ten questions correct. 70% of ten is seven. We want to find the probability that X is greater than or equal to seven. The event "at least seven" is the complement of "less than or equal to six".
Using your calculator's distribution menu: 1 – binomcdf(10, .5, 6) gives 0.171875
The probability of getting at least 70% of the ten questions correct when randomly guessing is approximately 0.172.

A student takes a 32-question multiple-choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75% of the questions correctly.

Solution
  • [latex]X =[/latex]number of questions answered correctly
  • [latex]X \sim B \left(\text{32, }\frac{\text{1}}{\text{3}}\right)[/latex]
  • We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find [latex]P(x > 24)[/latex]. The event "more than 24" is the complement of "less than or equal to 24."
  • Using your calculator's distribution menu: 1 – binomcdf[latex]\left(\text{32, }\frac{\text{1}}{\text{3}},\text{ 24}\right)[/latex]
  • [latex]P(x > 24) = 0[/latex]
  • The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero.

According to The World Bank, only 9% of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let [latex]X =[/latex] the number of people who have access to electricity.

  1. What is the probability distribution for [latex]X[/latex]?
  2. Using the formulas, calculate (i) the mean and (ii) standard deviation of [latex]X[/latex].
  3. Use your calculator to find the probability that 15 people in the sample have access to electricity.
  4. Find the probability that at most ten people in the sample have access to electricity.
  5. Find the probability that more than 25 people in the sample have access to electricity.
Solution
  1. [latex]X \sim B(150,0.09)[/latex]
  2. (i) [latex]\text{Mean} = np = 150(0.09) = 13.5[/latex]
    (ii) [latex]\text{Standard Deviation} =npq=150(0.09)(0.91)\approx{3.5050}[/latex]
  3. [latex]P(x = 15) = \text{binompdf} (150, 0.09, 15) = 0.0988[/latex]
  4. [latex]P(x \le 10) = \text{binomcdf} (150, 0.09, 10) = 0.1987[/latex]
  5. [latex]P(x > 25) = 1 – P(x \le 25) = 1 – \text{binomcdf}(150, 0.09, 25) = 1 – 0.9991 = 0.0009[/latex]

The literacy rate for a nation measures the proportion of people age 15 and over that can read and write. The literacy rate in Afghanistan is 28.1%. Suppose you choose 15 people in Afghanistan at random.

Let [latex]X =[/latex] the number of people who are literate.

  1. Sketch a graph of the probability distribution of [latex]X[/latex].
  2. Using the formulas, calculate the (i) mean and (ii) standard deviation of [latex]X[/latex].
  3. Find the probability that more than five people in the sample are literate.
  4. Is it more likely that three people or four people are literate?
Solution
  1. [latex]X \sim B(15, 0.281)[/latex]
    Histogram of Binomial Distribution with 15 trials and success probability 0.281
    Figure 1. Graph of [latex]X \sim B(15, 0.281)[/latex]
  2. (i) [latex]\text{Mean} = \mu = np = 15(0.281) = 4.215[/latex]
    (ii) [latex]\text{Standard Deviation} = \sigma = \sqrt{npq} = \sqrt{15\left(0.281\right)\left(0.719\right)} = 1.7409[/latex]
  3. [latex]P(x > 5) = 1 – P(x \le 5) = 1 – \text{binomcdf} (15, 0.281, 5) = 1 – 0.7754 = 0.2246[/latex]
  4. [latex]P(x = 3) = \text{binompdf} (15, 0.281, 3) = 0.1927[/latex] [latex]P(x = 4) = \text{binompdf} (15, 0.281, 4) = 0.2259[/latex] It is more likely that four people are literate than three people are.

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