Module 8

The Binomial Distribution

• The four requirements of a binomial experiment.

• fixed number of trials (n)

• each trial is independent of the others

• each trial has just two possible outcomes. The outcome of interest is called “success” and the other is “failure”. [the one called “success” is of interest, but it doesn’t have to be a “good” outcome.]

• The probability of success on each trial is the same, and we call it p. The complement of success is failure, and its probability is (1-p).

• Note about the requirement of independence:

Independence occurs in a sample taken with replacement. But if the sample is taken without replacement from a very large population, we can consider that to be “independent” for all practical purposes.

• The guideline is: if the population size (N) is at least 10 times as large as the sample size (n), then we can treat it as independent trials. That is N > 10n.

• Another way to say this: assume independence if the sample size (n) is at most 10% of the population size (N). That is n < 0.10N

Exercises:

1. If a sample of 52 items is taken from a population of 610, can we consider the trials to be independent?

Answer: 10n = 10•52 = 520. The population of 610 is larger than 10n. So yes, we can consider the trials to be independent.

1. From a school with 2,845 students how large a sample can be taken so that we can assume independence?

Answer: 10% of 2845 is 284, so a sample of 284 or less can be taken to assume independence.

• Factorial definition

Example: 5! = 1 • 2 •3 •4 •5, which equals 120.

Example: 7! = 1 • 2 •3 •4 •5• 6 •7, which equals 5040.

Note that the numbers could be multiplied in the other order:

7! = 7• 6 • 5 • 4• 3• 2• 1, which still equals 5040.

The exclamation point is read as “factorial”.

The definition of n factorial is
n! = 1 •2 •3 •… •n or n! = n•(n-1)•(n-2)• … • 2•1

The n must be a positive whole number.

Zero factorial has its own special definition (which doesn’t follow the rule above since the rule wouldn’t make sense). By definition, 0! = 1.

Because factorials come up in some formulas.

In the textbook, we wanted to know the number of ways of getting exactly k successes in n trials.

For example, in n = 5 trials, how many ways can there be 2 successes? We can figure this out “by hand” if we use “s” for success and “f” for “failure (which is not-success)” and then write all the ways that 5 trials could have 2 successes in them. Here’s the list:

ssfff, sfsff, sffsf, sfffs,I am listing these in an organized way

fssff, fsfsf, fsffs, (at least, it looks organized to me).

ffssf, ffsfsYou could list them in another order

fffssthat feels organized to you.

There are 10 items in the list

There are 10 ways to have 2 successes in 5 trials.

The formula for the number of ways to have 2 successes in 5 trials is

Check it out: = = = = 10

Formula for finding Binomial Distribution values (Optional)

The information in this section gives a better understanding of the theory of binomial distributions. Some students enjoy using the formula. But you may instead use the later section on “Calculator Info for Binomial Distributions”. So, this section is optional.

On the TI-83 and 84, to find 7!:

First press 7, then press the MATH button (along the left side), then arrow over to PRB

[PRB is for Probability]. In the menu displayed will be the symbol ! (probably it is #4). Select it and it will appear on the home screen after the 7 you typed earlier.

Press enter, and the calculator will tell you that 7! is 5040

General Formula for the number of ways to have exactly x successes in a list of n trials

is

Another symbol for this is

And another symbol for this is nCx for Combinations of n items taken x at a time

Note: nCx = =

nCx is sometimes read as “n Choose x” … it is the number of ways of choosing, in a list of n items, which x of them will be “successes” while the others are “failures”. It is the number of ways of “combining” x items from a group of n items.

Example: In lists of 7 items, the number of ways to choose 5 of them to be successes (and the rest failures) is 7C5 = =

Binomial Distribution – general formula:

Note: You do not need to use or remember this formula because the calculator has other ways of computing binomial distributions – but you might learn from understanding it.

In general, the number of ways to get x successes (and n - x failures) in n trials is n! / (x!(n−x)!) Therefore, the probability of x successes (and n - x failures) in n trials, where the probability of success in each trial is p (and the probability of failure is (1 - p) is equal to the number of outcomes in which there are x successes out of n trials, times the probability of x successes, times the probability of n - x failures:

P(X=x)= px (1-p)(n-x)

where x may take any value 0, 1, ... , n.

Other ways to write this are:

P(X=x)= px (1-p)(n-x) or

P(X=x)= nCx px (1-p)(n-x)

On the TI-83 and 84, to find nCx, first note that it is called nCr on the calculator.

(of course it doesn’t make any difference if you call it x or r).

Example: to find 5C2 (also called 5 nCr 2):

First press 5, then press the MATH button (along the left side), then arrow over to PRB

In the menu displayed will be nCr (probably it is #3).

Select it and it will appear on the home screen after the 5 you typed earlier.

Then type 2 [so the calculator will have 5 nCr 2].

Press enter, and the calculator will tell you that 5 nCr 2 = 10

That is exactly what we found it to be above when we calculated = 10.

Calculator Information: Binomial Distributions

These are commands for using the TI-83 or TI-84 graphing calculator.

There are two binomial distributions in the DISTR list: binomPdf and binomCdf.

Probability of getting exactly x successes for a Binomial Probability Distribution with

n trials where p is the probability of success on one trial is:

Binompdf( n, p, x) Binomial Probability Distribution function

Details: Display the DISTR menu by pressing 2nd VARS.

Select the option binompdf( which is way down in the list [use the down arrow to get to it].

Note: this is binomPdf, not binomCdf

Some calculators prompt for the inputs. For others you must type them in.

binompdf( will appear on the Home screen.

After it enter the value for n, then a comma, then the value for p, then a comma, then the value for x, then “)”. The numbers must be typed in that order.

Example: In a large population of widgets, 20% of them are defective. If a sample of 7 widgets is taken, what is the probability that exactly 3 of the sample items are defective?
Binompdf( 7, .2, 3) = 0.114688.
The probability of 3 defective items in the sample is about 0.115 or 11.5%.

Cumulative probability of getting x or fewer successes for a Binomial Probability Distribution with n trials where p is the probability of success on one trial is:

Binomcdf( n, p, x) the Binomial Cumulative Distribution function

Details: Display the DISTR menu by pressing 2nd VARS.

Select the option binomcdf( which is way down in the list [use the down arrow to get to it].

Note: this is binomCdf, not binomPdf

Some calculators prompt for the inputs. For others you must type them in.

binomcdf( will appear on the Home screen.

After it enter the value for n, then a comma, then the value for p, then a comma, then the value for x, then “)”. The numbers must be typed in that order.

The result is the cumulative probability of getting 0, 1, 2, …, x successes [that is, the probability of getting 0 through x successes, which is the probability of getting at most x successes.]

Example: Laura has a large collection of stuffed animals. 35% of them are cows.
a) If she randomly selects eight animals to play with, what is the probability she will have at most two cows? This is the probability that she will have 0 or 1 or 2 cows. P(x < 2).
Binomcdf( 8, .35, 2) result is .42781.
So the probability of at most 2 cows in the sample of 8 is 0.428 or 42.8%.
b) If Laura randomly selects 12 animals, what is the probability that she will have at least five cows in the collection? [This is the probability she will have 5, 6, 7, 8, 9, 10, 11, or 12 cows.]
The calculator cannot find this probability directly. Rather, we use the calculator to find the probability of the complement. The complement is getting 0, 1, 2, 3, or 4 cows.
Binomcdf( 12, .35, 4) result is .583345
[this is the probability of getting 0, 1, 2, 3, or 4 cows in the sample of 12]
The probability of getting at least 5 cows is 1 - .583345 = .416655, about .417 = 41.7%

Videos of Calculator commands (TI-83 or 84) for Binomial Distributions

Here are some videos of how to use the calculator commands for the binomial distribution:

(3:26 long)

(6:49 long)

Using the online graphing calculator (optional)

Note: the calculator binomial commands also work on the online graphing calculator at

But for that calculator you simply type in the command in the box at the bottom of the right side. For example, you would type binompdf(7, .2, 3) and then press the “Enter” button.

The calculator would give the value 0.114688

Or type in binomcdf(8, .35, 2), press Enter button, and get .4278137

Binomial Distribution values can be found using EXCEL.
Directions are given in the “Learning Dashboard” from p. 143.
To find probabilities of the type P(X = k) or P(X ≤ k) in Excel, we'll use the following function: =BINOMDIST(k, n, p, cumulative) where:
• k is the number of successes in trials
• n is the number of independent trials
• p is the probability of success in each trial
• cumulative is a logical value that determines the form of the function. If cumulative is TRUE, then BINOMDIST returns the cumulative distribution function, which is the probability that there are at most k successes: P(X ≤ k). If FALSE, it returns the probability mass function, which is the probability that there are exactly k successes: P(X = k).

Mod 8

Mean and Standard Deviation for a Binomial Distribution:

= n•p

Guideline: A number of successes in a Binomial Distribution is considered usual if it is within 2 standard deviations of the mean. So the usual number must be between –2 and +2.

The sensible usual numbers are the whole numbers between those two boundaries.

1. At a certain college 30% of the students are part-time. A random sample of 8 college students from the school is selected.

a) What is the mean number of part-time students in such samples?

b) What is the standard deviation of the number of part-time students?

c) What is the variance of the number of part-time students?

d) What are the usual numbers of part-time students in such samples?

1. Mars Inc. claims that 24% of its M&M plain candies are blue. A sample of 100 M&Ms is randomly selected.

a) Find the mean and standard deviation for the numbers of blue M&Ms in such a group of 100.

b) An actual random sample of 100 M&Ms contained 27 blue. Does it seem that the claimed rate of 24% is wrong? Explain.

1. In a past presidential election, the actual voter turnout was 61%. In a survey, 1002 subjects were asked if they voted in the presidential election.

a) Find the mean and standard deviation for the number of actual voters in groups of 1002.

b) In the survey of 1002 people, 701 said that they voted in the last presidential election (real data). Is this result consistent with the actual voter turnout, or is this result unlikely to occur with an actual voter turnout of 61%?

Explain why or why not.

1a) Mean = n p = 8 ( .3) = 2.41b) = sqrt( np(1-p) ) = sqrt ( 8(.3)(.7) ) = 1.3

1c) variance = standard-deviation squared, but use the value with more decimal places before rounding: var = (1.296)2 = 1.68

1d) usual min = – 2 = 2.4 – 2(1.3) = -0.2 usual max = + 2 = 2.4 + 2(1.3) = 5

Since numbers of students can only be whole numbers, the

Sensible answer is [ 0, 5 ] written in interval notation as wamap wants it.

2a) n = 100, p = prob of blue = 0.24 ; = 24; = 4.3

2b) The usual values of blue would be min of 24 –2(4.3) and max of 24+2(4.3),

which is 15.4 to 32.6. So the sensible whole number range is [ 16, 32 ].

27 is in that usual range, so getting 27 blue is fairly usual when the actual rate is 24%, so there is no evidence that the rate of 24% is wrong. [Note: we are not sure it is right, but we don’t have any evidence it is wrong.]

3a) n=1002; p=.61 since 61% of people actually did vote. = 611.2; = 15.4

3b) the usual range is 580.4 to 642 (which as whole numbers is 581 to 642). 701 people is NOT in that usual range. It could be that a highly unusual event happened. But what is more likely is that the people lied about whether they voted! Some may have wanted to appear to be a good citizen, and so said they voted when they didn’t.

Module 8 – Binomial Distributionp. 1