Binomial Distribution

Example

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

Example

Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times.

[reveal-answer q=”328186″]Show Solution[/reveal-answer]
[hidden-answer a=”328186″]Here, if you define X as the number of wins, then X takes on the values 0, 1, 2, 3, …, 20. The probability of a success is [latex]p=0.55[/latex]. The probability of a failure is [latex]q=0.45[/latex]. The number of trials is [latex]n=20[/latex]. The probability question can be stated mathematically as [latex]P(x=15)[/latex].

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Example

A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more than ten heads? Let [latex]X=[/latex] the number of heads in 15 flips of the fair coin. X takes on the values 0, 1, 2, 3, …, 15. Since the coin is fair, [latex]p=0.5[/latex] and [latex]q=0.5[/latex]. The number of trials is [latex]n=15[/latex]. State the probability question mathematically.

[reveal-answer q=”995613″]Show Solution[/reveal-answer]
[hidden-answer a=”995613″][latex]P(x>10)[/latex][/hidden-answer]

Example

Approximately 70% 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 50 students, what is the probability that at least 40 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 0.70 for each trial.
2. If we are interested in the number of students who do their homework on time, then how do we define X?
3. What values does x take on?
4. What is a “failure,” in words?
5. If [latex]p+q=1[/latex], then what is q?
6. The words “at least” translate as what kind of inequality for the probability question P(x ____40).

[reveal-answer q=”872394″]Show Solution[/reveal-answer]
[hidden-answer a=”872394″]

1. failure
2. [latex]X=[/latex] the number of statistics students who do their homework on time
3. 0, 1, 2, …, 50
4. Failure is defined as a student who does not complete his or her homework on time. The probability of a success is [latex]p=0.70[/latex]. The number of trials is [latex]n=50[/latex].
5. [latex]q=0.30[/latex]
6. greater than or equal to (≥)The probability question is [latex]P(x\geq40)[/latex].

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Example

It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?

Let [latex]X=[/latex] the number of workers who have a high school diploma but do not pursue any further education.

X takes on the values 0, 1, 2, …, 20 where [latex]n=20[/latex], [latex]p=0.41[/latex], and [latex]q=1–0.41=0.59[/latex]. [latex]X\sim{B}(20,0.41)[/latex]

Find [latex]P(x\leq12)[/latex]. [latex]P(x\leq12)=0.9738[/latex]. (calculator or computer)

[reveal-answer q=”156496″]Show Solution[/reveal-answer]
[hidden-answer a=”156496″]

• Go into 2nd DISTR. The syntax for the instructions are as follows:
• To calculate ([latex]x=[/latex] value): binompdf(n, p, number) if “number” is left out, the result is the binomial probability table.
• To calculate P([latex]x\leq[/latex] value): binomcdf(n, p, number) if “number” is left out, the result is the cumulative binomial probability table.
• For this problem: After you are in 2nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,0.41,12). The result is [latex]P(x\leq12)=0.9738[/latex].

Note

If you want to find [latex]P(x=12)[/latex], use the pdf (binompdf). If you want to find [latex]P(x>12)[/latex], use [latex]1-\text{binomcdf}(20,0.41,12)[/latex].

The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.

The graph of [latex]X\sim{B}(20,0.41)[/latex] is as follows:

The y-axis contains the probability of x, where [latex]X=[/latex] the number of workers who have only a high school diploma.

The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, [latex]\mu=np=(20)(0.41)=8.2[/latex].

The formula for the variance is [latex]\sigma^{2}=npq[/latex]. The standard deviation is [latex]\sqrt{{{n}{p}{q}}}[/latex].

[latex]\sigma=\sqrt{(20)(0.41)(0.59)}={2.20}[/latex]

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Example

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.

1. What values does x take on?
2. 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.
3. Using the formulas, calculate the (i) mean and (ii) standard deviation.

[reveal-answer q=”81853″]Show Solution[/reveal-answer]
[hidden-answer a=”81853″]

1. [latex]x=0,1,2,3,4,5,6,7,8[/latex]
2. [latex]{X}\sim{B}{({100},\frac{{8}}{{560}})}[/latex]
   1. [latex]{P}{({x}={2})}=\text{binompdf}{({100},\frac{{8}}{{560}},{2})}={0.2466}[/latex]
   2. [latex]{P}{({x}\leq{6})}=\text{binomcdf}{({100},\frac{{8}}{{560}},{6})}={0.9994}[/latex]
   3. [latex]{P}{({x}{>}{3})}={1}-{P}{({x}\leq{3})}={1}-\text{binomcdf}{({100},\frac{{8}}{{560}},{3})}={1}-{0.9443}={0.0557}[/latex]
   1. [latex]\text{Mean}={n}{p}={({100})}{(\frac{{8}}{{560}})}=\frac{{800}}{{560}}\approx{1.4286}[/latex]
   2. [latex]\text{Standard Deviation}=\sqrt{{{n}{p}{q}}}=\sqrt{{{({100})}{({8560})}{({552560})}}}\approx{1.1867}[/latex]

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Example

The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let [latex]X=[/latex] the number of people who will develop pancreatic cancer.

1. What is the probability distribution for X?
2. Using the formulas, calculate the (i) mean and (ii) standard deviation of X.
3. Use your calculator to find the probability that at most eight people develop pancreatic cancer.
4. Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically.

[reveal-answer q=”586098″]Show Solution[/reveal-answer]
[hidden-answer a=”586098″]

1. [latex]X\sim{B}(200,0.0128)[/latex]
   1. Mean[latex]=np=200(0.0128)=2.56[/latex]
   2. [latex]\text{Standard Deviation}=\sqrt{{{n}{p}{q}}}=\sqrt{{{({200})}{({0.0128})}{({0.9872})}}}\approx{1.5897}[/latex]
2. Using the TI-83, 83+, 84 calculator:
   [latex]P(x\leq8)=\text{binomcdf}(200,0.0128,8)=0.9988[/latex]
3. [latex]P(x=5)=\text{binompdf}(200,0.0128,5)=0.0707[/latex]
   [latex]P(x=6)=\text{binompdf}(200,0.0128,6)=0.0298[/latex] So [latex]P(x=5)>P(x=6)[/latex]; it is more likely that five people will develop cancer than six.

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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?

[reveal-answer q=”52898″]Show Solution[/reveal-answer]
[hidden-answer a=”52898″]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]\frac{6}{16}[/latex], when the first draw selects a staff member. The probability of drawing a student’s name changes for each of the trials and, therefore, violates the condition of independence.[/hidden-answer]