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4.4 The Binomial Distribution

LEARNING OBJECTIVES

  • Recognize the binomial probability distribution and apply it appropriately.

A binomial experiment has the following characteristics:

  1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter [latex]n[/latex] denotes the number of trials.
  2. There are only two possible outcomes, called “success” and “failure,” for each trial. The letter [latex]p[/latex] denotes the probability of a success on any one trial, and [latex]1-p[/latex] denotes the probability of a failure on one trial.
  3. The [latex]n[/latex] trials are independent and are repeated using identical conditions.
  4. For each individual trial, the probability of a success, [latex]p[/latex], and the probability of a failure, [latex]1-p[/latex], remain the same. Because the [latex]n[/latex] trials are independent, the outcome of one trial does not affect the outcome of another trial.

For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with probability [latex]p=0.6[/latex]. Then, [latex]1-p=0.4[/latex]. This means that for every true-false statistics question Joe answers, his probability of success [latex]p=0.6[/latex] and his probability of failure [latex]1-p=0.4[/latex] remain the same.

The outcomes of a binomial experiment fit a binomial probability distribution. The random variable [latex]X[/latex] is the number of successes obtained in the [latex]n[/latex] independent trials. The mean of a binomial probability distribution is [latex]\displaystyle{\mu=n \times p}[/latex], and the standard deviation is [latex]\displaystyle{\sigma=\sqrt{n \times p \times (1-p)}}[/latex]

Any experiment with the characteristics of a binomial experiment and where [latex]n=1[/latex] is called a Bernoulli Trial (named after Jacob Bernoulli, who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.

EXAMPLE

At ABC College, the withdrawal rate from an elementary physics course is [latex]30\%[/latex] for any given term. This implies that, for any given term, [latex]70\%[/latex] of the students stay in the class for the entire term. A “success” could be defined as an individual who withdrew from the course. The random variable [latex]X[/latex] is the number of students who withdraw from the randomly selected elementary physics class.

TRY IT

The health board is concerned about the amount of fruit available in school lunches. [latex]48\%[/latex] of schools in the state offer fruit in their lunches every day. This implies that [latex]52\%[/latex] do not. What would a “success” be in this case?

 

Click to see Solution

 

A success would be a school that offers fruit in their lunch every day.

EXAMPLE

Suppose a game has only two outcomes: win or lose. The probability of winning any game is [latex]55\%[/latex], and the probability of losing is [latex]45\%[/latex]. Each game played is independent. Suppose someone plays the game [latex]20[/latex] times. Is this a binomial experiment?

Solution

  • There are [latex]20[/latex] trials (games).
  • There are only two possible outcomes in any trial (game):  win (success) or lose (failure). The probability of success (winning) is [latex]p=0.55[/latex]. The probability of a failure is [latex]1-p=0.45[/latex].
  • Each trial (game) is independent. The outcome of any game does not affect the outcome of any other game.
  • The probability of success and the probability of failure remain the same from game to game.

EXAMPLE

Approximately [latex]70\%[/latex] of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of [latex]50[/latex] students, what is the probability that at least [latex]40[/latex] will do their homework on time? Students are selected randomly.

  1. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is [latex]0.70[/latex] for each trial.
  2. If we are interested in the number of students who do their homework on time, then how do we define [latex]X[/latex]?
  3. What values does [latex]x[/latex] take on?
  4. What is a “failure,” in words?
  5. What is the probability of “failuere”?
  6. The words “at least” translate to what kind of inequality for the probability question [latex]\displaystyle{P(x....40)}[/latex].

Solution

  1. failure
  2. [latex]X[/latex] is the number of statistics students who do their homework on time.
  3. [latex]0, 1, 2, …, 50[/latex].
  4. Failure is defined as a student who does not complete his or her homework on time.
  5. [latex]1-p=0.30[/latex]
  6. “At least” means greater than or equal to ([latex]\geq[/latex]). The probability question is [latex]\displaystyle{P(x\geq40)}[/latex].

TRY IT

Sixty-five percent of people pass the state driver’s exam on the first try. A group of [latex]50[/latex] individuals who have taken the driver’s exam is randomly selected. Why this is a binomial problem?

 

Click to see Solution
  • There are only two outcomes on any exam (pass or fail).
  • There is a fixed number of trials ([latex]n=50[/latex]).
  • The probability of pass ([latex]65\%[/latex]) is the same for each trial.
  • The trials are independent. (The fact that any one person passes or fails the exam does not affect whether or not any other person passes or fails.)

EXAMPLE

The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students?

Solution

The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson, and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is [latex]\displaystyle{\frac{6}{16}}[/latex], and the probability of a student on the second draw is [latex]\displaystyle{\frac{5}{15}}[/latex]. The probability of drawing a student’s name changes for each of the trials and, therefore, violates the condition of independence.

TRY IT

A high school lacrosse team is selecting a captain. The names of all the seniors are put into a hat and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). We want to see if the captains all play the same position. State whether or not this is binomial and state why.

 

Click to see Solution

 

This is not binomial because the names are not replaced after each draw, which means the probability changes for each time a name is drawn. This violates the condition of independence.

Calculating Binomial Probabilities

CALCULATING BINOMIAL PROBABILITIES IN EXCEL

To calculate probabilities associated with binomial random variables in Excel, use the binom.dist(x,n,p,logic operator) function.

  • For x, enter the number of successes.
  • For n, enter the number of trials.
  • For p, enter the probability of success.
  • For the logic operator, enter false to find the probability of exactly x successes and enter true the find the probability of at most (less than or equal to) x successes.

The output from the binom.dist function is:

  • The probability of getting exactly x success in n trials with a probability of success p when the logic operator is false.
  • The probability of at most x successes in n trials with a probability of success p when the logic operator is true.

Visit the Microsoft page for more information about the binom.dist function.

NOTE

Because we can only enter false or true into the logic operator, the binom.dist function can only directly calculate the probability of getting exactly x successes in n trials or getting at most x success in n trials. In order to calculate other binomial probabilities, such as fewer than x successes, more than x successes or at least x successes, we need to manipulate how we use the binom.dist function by changing what we enter into the binom.dist function, using the complement rule, or both.

EXAMPLE

It has been stated that about [latex]41\%[/latex] of adult workers have a high school diploma but do not pursue any further education. Suppose [latex]20[/latex] adult workers are randomly selected.

  1. How many adult workers in the sample are expected to have a high school diploma but do not pursue any further education?
  2. What is the probability that exactly [latex]8[/latex] of the workers in the sample have a high school diploma but do not pursue further education?
  3. What is the probability that at most [latex]12[/latex] of the workers in the sample have a high school diploma but do not pursue further education?

Solution

Let [latex]X[/latex] be the number of workers in the sample who have a high school diploma but do not pursue further education. The number of trials is [latex]n=20[/latex], and the probability of success is [latex]p=0.41[/latex].

  1. [latex]\displaystyle{\mu=n \times p=20 \times 0.41=8.2}[/latex].  On average, in any sample of [latex]20[/latex] workers, [latex]8.2[/latex] have a high school diploma but do not pursue further education.
  2. We want to find [latex]\displaystyle{P(x=8)}[/latex].
    Function binom.dist
    Field 1 8
    Field 2 20
    Field 3 0.41
    Field 4 false
    Answer 0.1790

    The probability that exactly [latex]8[/latex] of the workers in the sample have a high school diploma but do not pursue further education is [latex]17.9\%[/latex].

  3. We want to find [latex]\displaystyle{P(x \leq 12)}[/latex].
    Function binom.dist
    Field 1 12
    Field 2 20
    Field 3 0.41
    Field 4 true
    Answer 0.9738

    The probability that at most [latex]12[/latex] of the workers in the sample have a high school diploma but do not pursue further education is [latex]97.38\%[/latex].

TRY IT

About [latex]32\%[/latex] of students participate in a community volunteer program outside of school. Suppose [latex]30[/latex] students are selected at random.

  1. What is the expected number of students in the sample that participate in a community volunteer program?
  2. What is the probability that exactly [latex]10[/latex] of the students in the sample participate in a community volunteer program?
  3. What is the probability that at most [latex]14[/latex] of the students in the sample participate in a community volunteer program?
Click to see Solution
  1. [latex]\displaystyle{\mu=n \times p=30 \times 0.32=9.6}[/latex]
  2. Function binom.dist
    Field 1 10
    Field 2 30
    Field 3 0.32
    Field 4 false
    Answer 0.1512
  3. Function binom.dist
    Field 1 14
    Field 2 30
    Field 3 0.32
    Field 4 true
    Answer 0.9695

EXAMPLE

In the 2013 Jerry’s Artarama art supplies catalogue, there are [latex]560[/latex] pages and [latex]1.5\%[/latex] of the pages feature signature artists. Suppose [latex]100[/latex] pages are randomly selected from the catalogue.

  1. What is the probability that fewer than [latex]3[/latex] of the pages in the sample feature signature artists?
  2. What is the probability that more than [latex]5[/latex] of the pages in the sample feature signature artists?
  3. What is the probability that at least [latex]4[/latex] of the pages in the sample feature signature artists?
  4. What is the probability that between [latex]2[/latex] and [latex]6[/latex] of the pages in the sample feature signature artists?

Solution

  1. We want to find [latex]\displaystyle{P(x \lt 3)}[/latex]. We cannot find this probability directly in Excel because the binom.dist function can only calculate [latex]=[/latex] or [latex]\leq[/latex] probabilities.  Because [latex]x[/latex] must be an integer (it is the number of pages), [latex]x \lt 3[/latex] is the same as [latex]x \leq 2[/latex] (of course, in general, this is not true). So [latex]\displaystyle{P(x \lt 3)=P(x \leq 2)}[/latex] and [latex]\displaystyle{P(x \leq 2)}[/latex] is a probability we can calculate with the binom.dist function.
    Function binom.dist
    Field 1 2
    Field 2 100
    Field 3 0.015
    Field 4 true
    Answer 0.8098
  2.  We want to find [latex]\displaystyle{P(x \gt 5)}[/latex].  We cannot find this probability directly in Excel because the binom.dist function can only calculate [latex]=[/latex] or [latex]\leq[/latex] probabilities. The complement of [latex]\gt[/latex] is [latex]\leq[/latex], so [latex]\displaystyle{P(x \gt 5)=1-P(x \leq 5)}[/latex] and [latex]\displaystyle{P(x \leq 5)}[/latex] is a probability we can calculate with the binom.dist function.
    Function 1-binom.dist
    Field 1 5
    Field 2 100
    Field 3 0.015
    Field 4 true
    Answer 0.0177
  3. We want to find [latex]\displaystyle{P(x \geq 4)}[/latex]. We cannot find this probability directly in Excel because the binom.dist function can only calculate [latex]=[/latex] or [latex]\leq[/latex] probabilities.  The complement of [latex]\geq[/latex] is [latex]\lt[/latex], so [latex]\displaystyle{P(x \geq 4)=1-P(x \lt 4)}[/latex].  Because [latex]x[/latex] must be an integer (it is the number of pages), [latex]x \lt 4[/latex] is the same as [latex]x \leq 3[/latex]. So [latex]\displaystyle{P(x \geq 4)=1-P(x \lt 4)=1-P(x \leq 3)}[/latex] and [latex]\displaystyle{P(x \leq 3)}[/latex] is a probability we can calculate with the binom.dist function.
    Function 1-binom.dist
    Field 1 3
    Field 2 100
    Field 3 0.015
    Field 4 true
    Answer 0.0642
  4.  We want to find [latex]\displaystyle{P(2 \leq x \leq 6)}[/latex].  We cannot find this probability directly in Excel because the binom.dist function can only calculate [latex]=[/latex] or [latex]\leq[/latex] probabilities.  But, [latex]\displaystyle{P(2 \leq x \leq 6)=P(x \leq 6)-P(x \leq 1)}[/latex].  So we can calculate [latex]\displaystyle{P(2 \leq x \leq 6)}[/latex] as the difference of two binom.dist functions.
Function binom.dist -binom.dist
Field 1 6 1
Field 2 100 100
Field 3 0.015 0.015
Field 4 true true
Answer 0.4426

TRY IT

According to a Gallup poll, [latex]60\%[/latex] of American adults prefer saving over spending. Suppose [latex]50[/latex] American adults are selected at random.

  1. What is the probability that at least [latex]35[/latex] adults in the sample prefer saving over spending?
  2. What is the probability that fewer than [latex]20[/latex] adults in the sample prefer saving over spending?
  3. What is the probability between [latex]15[/latex] and [latex]25[/latex] adults in the sample prefer saving over spending?
  4. What is the probability that more than [latex]30[/latex] adults prefer saving over spending?
Click to see Solution
  1. Function 1-binom.dist
    Field 1 34
    Field 2 50
    Field 3 0.6
    Field 4 true
    Answer 0.0955
  2. Function binom.dist
    Field 1 19
    Field 2 50
    Field 3 0.6
    Field 4 true
    Answer 0.0014
  3. Function binom.dist -binom.dist
    Field 1 25 14
    Field 2 50 50
    Field 3 0.6 0.6
    Field 4 true true
    Answer 0.0978
  4. Function 1-binom.dist
    Field 1 30
    Field 2 50
    Field 3 0.6
    Field 4 true
    Answer 0.4465

TRY IT

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 [latex]61.3\%[/latex] of his shots. Suppose we take a random sample of [latex]80[/latex] shots made by DeAndre during the 2013 season.

  1. What is the expected number of shots that scored points in a sample of [latex]80[/latex] of DeAndre’s shots?
  2. What is the probability that DeAndre scored on [latex]60[/latex] of the [latex]80[/latex] shots?
  3. What is the probability that DeAndre scored on more than [latex]50[/latex] of the [latex]80[/latex] shots?
  4. What is the probability that DeAndre scored between [latex]65[/latex] and [latex]75[/latex] of the [latex]80[/latex] shots?
Click to see Solution
  1. [latex]\mu=n \times p=80 \times 0.613=49.04[/latex]
  2. Function binom.dist
    Field 1 60
    Field 2 80
    Field 3 0.613
    Field 4 false
    Answer 0.0036
  3. Function 1-binom.dist
    Field 1 50
    Field 2 80
    Field 3 0.613
    Field 4 true
    Answer 0.3718
  4. Function binom.dist -binom.dist
    Field 1 75 64
    Field 2 80 80
    Field 3 0.613 0.613
    Field 4 true true
    Answer 0.0001

Video: “Binomial Probability in Excel – Word problems” by Joshua Emmanuel [7:00] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.

Exercises

  1. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about [latex]4\%[/latex]. Suppose a sample of [latex]25[/latex] patient calls is taken.
    1. On average, for every [latex]25[/latex] patient calls, how many do you expect to have the flu?
    2. Calculate the standard deviation for this probability distribution.
    3. Find the probability that exactly [latex]5[/latex] of the [latex]25[/latex] patients who called have the flu.
    4. Find the probability that at most [latex]3[/latex] of the [latex]25[/latex] patients who called have the flu.
    5. Find the probability that between [latex]3[/latex] and [latex]9[/latex] of the patients who called have the flu.
    6. Find the probability that at least [latex]6[/latex] of the [latex]25[/latex] patients who called have the flu.
    7. Find the probability that fewer than [latex]5[/latex] of the [latex]25[/latex] patients who called have the flu.
    8. Find the probability that more than [latex]4[/latex] of the [latex]25[/latex] patients who called have the flu.
    Click to see Answer
    1. [latex]1[/latex]
    2. [latex]0.98[/latex]
    3. [latex]0.0024[/latex]
    4. [latex]0.9835[/latex]
    5. [latex]0.0765[/latex]
    6. [latex]0.0004[/latex]
    7. [latex]0.9972[/latex]
    8. [latex]0.0165[/latex]

     

  2. The probability that the San Jose Sharks will win any given game is [latex]0.3694[/latex] based on a 13-year win history of [latex]382[/latex] wins out of [latex]1,034[/latex] games played (as of a certain date). An upcoming monthly schedule contains [latex]12[/latex] games.
    1. What is the expected number of wins for that upcoming month?
    2. What is the probability that the San Jose Sharks win exactly [latex]6[/latex] games in that upcoming month?
    3. What is the probability that the San Jose Sharks win at least [latex]5[/latex] games in that upcoming month?
    4. What is the probability that the San Jose Sharks win between [latex]3[/latex] and [latex]9[/latex] games in that upcoming month?
    5. What is the probability that the San Jose Sharks win fewer than [latex]6[/latex] games in that upcoming month?
    6. What is the probability that the San Jose Sharks win more than [latex]8[/latex] games in that upcoming month?
    7. What is the probability that the San Jose Sharks win at most [latex]7[/latex] games in that upcoming month?
    Click to see Answer
    1. [latex]4.4328[/latex]
    2. [latex]0.1476[/latex]
    3. [latex]0.4734[/latex]
    4. [latex]0.7024[/latex]
    5. [latex]0.7427[/latex]
    6. [latex]0.0084[/latex]
    7. [latex]0.9644[/latex]

     

  3. 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 [latex]70\%[/latex] of the questions correct.
    Click to see Answer

    [latex]0.1719[/latex]

     

  4. More than [latex]96\%[/latex] of the very largest colleges and universities (more than [latex]15,000[/latex] total enrollments) have some online offerings. Suppose you randomly select [latex]13[/latex] such institutions.
    1. On average, how many schools would you expect to offer such courses?
    2. Find the probability that at most [latex]10[/latex] offer such courses.
    3. Find the probability that more than [latex]9[/latex] offer such courses.
    4. Find the probability that exactly [latex]11[/latex] offer such courses.
    5. Find the probability that between [latex]7[/latex] and [latex]12[/latex] offer such courses.
    6. Is it more likely that [latex]12[/latex] or that [latex]13[/latex] will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence.
    Click to see Answer
    1. [latex]12.48[/latex]
    2. [latex]0.0135[/latex]
    3. [latex]0.9986[/latex]
    4. [latex]0.0797[/latex]
    5. [latex]0.4118[/latex]
    6. [latex]13[/latex] is more likely because the probability that [latex]13[/latex] offer such courses is [latex]0.5882[/latex] and the probability that [latex]12[/latex] offer such courses is [latex]0.3186[/latex].

     

  5. Suppose that about [latex]85\%[/latex] of graduating students attend their graduation. A group of [latex]22[/latex] graduating students is randomly chosen.
    1. How many are expected to attend their graduation?
    2. Find the probability that between [latex]16[/latex] and [latex]18[/latex] attend graduation.
    3. Find the probability that fewer than [latex]15[/latex] attend graduation.
    4. Find the probability that at least [latex]20[/latex] attend graduation.
    5. Would you be surprised if all [latex]22[/latex] attended graduation? Justify your answer numerically.
    Click to see Answer
    1. [latex]18.7[/latex]
    2. [latex]0.3880[/latex]
    3. [latex]0.0114[/latex]
    4. [latex]0.3382[/latex]
    5. Yes, because the probability that all [latex]22[/latex] attend is only [latex]0.028[/latex].

     

  6. At The Fencing Center, [latex]60\%[/latex] of the fencers use the foil as their main weapon. We randomly survey [latex]25[/latex] fencers at The Fencing Center.
    1. How many of the [latex]25[/latex] fencers are expected to use the foil as their main weapon?
    2. Find the probability that [latex]6[/latex] of the [latex]25[/latex] fencers use the foil as their main weapon.
    3. Find the probability that at least [latex]15[/latex] of the [latex]25[/latex] fencers use the foil as their main weapon.
    4. Find the probability that at most [latex]10[/latex] of the [latex]25[/latex] fencers use the foil as their main weapon.
    5. Find the probability that between [latex]12[/latex] and [latex]17[/latex] of the [latex]25[/latex] fencers use the foil as their main weapon.
    6. Find the probability that fewer than [latex]13[/latex] of the [latex]25[/latex] fencers use the foil as their main weapon.
    7. Find the probability that more than [latex]20[/latex] of the [latex]25[/latex] fencers use the foil as their main weapon.
    8. Would you be surprised if all [latex]25[/latex] fencers in the sample use the foil as their main weapon? Justify your answer numerically.
    Click to see Answer
    1. [latex]15[/latex]
    2. [latex]0.0002[/latex]
    3. [latex]0.4246[/latex]
    4. [latex]0.0344[/latex]
    5. [latex]0.7686[/latex]
    6. [latex]0.1538[/latex]
    7. [latex]0.0095[/latex]
    8. Yes, because the probability that all [latex]25[/latex] use the foil is only [latex]0.0000028[/latex].

     

  7. Approximately [latex]8\%[/latex] of students at a local high school participate in after-school sports all four years of high school. A sample of [latex]60[/latex] seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school.
    1. How many seniors in the sample are expected to have participated in after-school sports all four years of high school?
    2. What is the probability that at least [latex]10[/latex] of the seniors in the sample have participated in after-school sports all four years of high school?
    3. What is the probability that between [latex]7[/latex] and [latex]15[/latex] of the seniors in the sample have participated in after-school sports all four years of high school?
    4. What is the probability that less than [latex]6[/latex] of the seniors in the sample have participated in after-school sports all four years of high school?
    5. Would you be surprised if none of the seniors in the sample participated in after-school sports all four years of high school? Justify your answer numerically.
    6. Is it more likely that [latex]4[/latex] or that [latex]5[/latex] of the seniors in the sample participated in after-school sports all four years of high school? Justify your answer numerically.
    Click to see Answer
    1. [latex]4.8[/latex]
    2. [latex]0.02[/latex]
    3. [latex]0.202[/latex]
    4. [latex]0.6526[/latex]
    5. Yes, because the probability that [latex]0[/latex] of the [latex]60[/latex] seniors in the sample participated in after-school sports all four years of high school is only [latex]0.0067[/latex].
    6. [latex]4[/latex] is more likely because the probability that [latex]4[/latex] of the seniors in the sample participated in after-school sports all four years of high school is [latex]0.1873[/latex] and the probability that [latex]5[/latex] of the seniors in the sample participated in after-school sports all four years of high school is [latex]0.1824[/latex].

     

  8. It has been estimated that only about [latex]30\%[/latex] of British Columbia residents have adequate earthquake supplies. Suppose [latex]11[/latex] BC residents are randomly surveyed about their earthquake supplies.
    1. What is the probability that at least [latex]8[/latex] of the residents in the sample have adequate earthquake supplies?
    2. What is the probability that between [latex]5[/latex] and [latex]9[/latex] of the residents in the sample have adequate earthquake supplies?
    3. What is the probability that at most [latex]3[/latex] of the residents in the sample have adequate earthquake supplies?
    4. Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why?
    5. How many residents do you expect will have adequate earthquake supplies?
    Click to see Answer
    1. [latex]0.0043[/latex]
    2. [latex]0.2103[/latex]
    3. [latex]0.5696[/latex]
    4. None is more likely because the probability that [latex]0[/latex] residents in the sample have adequate earthquake supplies is [latex]0.0198[/latex], and the probability that [latex]11[/latex] residents in the sample have adequate earthquake supplies is [latex]0.0000018[/latex].
    5. [latex]3.3[/latex]

     

“4.4 The Binomial Distribution” and “4.6 Exercises” from Introduction to Statistics by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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Introduction to Statistics - Second Edition Copyright © 2025 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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