# Chapter 6 the Binomial Probability Distribution and Related Topics Chapter 6 – The Binomial Probability Distribution and Related Topics

Statistical Experiments and Random Variables

• Statistical Experiments – any process by which measurements are obtained.

• A quantitative variable, x, is a random variable if its value is determined by the outcome of a random experiment.

• Random variables can be discrete or continuous.

• Discrete – countable

• Continuous – values fall into an interval of numbers

Random Variables and Their Probability Distributions

A random variable is a variable (function) whose value is a numerical outcome of a random phenomenon. The values of the random variable vary with the outcome of the random phenomenon.

• Discrete random variables – can take on only a countable or finite number of values.

• Continuous random variables – can take on countless values in an interval on the real line

• Probability distributions of random variables

– An assignment of probabilities to the specific values or a range of values for a random variable.

Example - Which measurement involves a discrete random variable?

a). Determine the mass of a randomly-selected penny

b). Assess customer satisfaction rated from 1 (completely satisfied) to 5 (completely dissatisfied).

c). Find the rate of occurrence of a genetic disorder in a given sample of persons.

d). Measure the percentage of light bulbs with lifetimes less than 400 hours.

Remember Discrete means countable.

Continuous Random Variables

If the possible values of a random variable fall in an interval of numbers we say that the random variable is continuous.

A probability distribution is an assignment of probabilities to each distinct value of a discrete random variable or to each interval of values of a continuous random variable.

The probability distribution of a continuous random variable X is given by a probability density curve. The probability of an event is the area under the density curve that is above the values of x that makeup the event.

Note: any single value of a continuous random variable has zero probability of occurring since probability = area under the curve and a single value is a point on the curve.

Discrete Probability Distributions

1) Each value of the random variable has an assigned probability.

2) The sum of all the assigned probabilities must equal 1.

3) Random Variables are denoted by capital letters.

4) X is a discrete random variable if X has a finite (countable) number of possible values x1, x2,…, xk and a probability distribution of possible values.

5) The probability histogram or probability density function is a graph of the possible values of X on the x-axis and the probabilities for each x on the y-axis.

Probability Distribution Features

• Since a probability distribution can be thought of as a relative-frequency distribution for a very large n, we can find the mean and the standard deviation.

• When viewing the distribution in terms of the population, use µ for the mean and σ for the standard deviation.

Example – Dr. Mendoza developed a test to measure boredom tolerance. He administered it to a group of 20,000 adults between the ages of 25 and 35. The possible scores were 0-6, with 6 indicating the highest tolerance for boredom. The test results for this group are shown below:

Score / Number of Students
0 / 1400
1 / 2600
2 / 3600
3 / 6000
4 / 4400
5 / 1600
6 / 400

Means and Standard Deviations for Discrete Probability Distributions Expected Value

The mean of a probability distribution is often called the expected value of the distribution. The terminology reflects the idea that the mean represents a “central point” or “cluster point” for the entire distribution. Of course, the mean or expected value is an average value, and as such, it need not be a point or the sample space.

Example – Compute the expected value and the standard deviation for the given information.

Number of times buyers saw infomercial / 1 / 2 / 3 / 4 / 5+
Percentage of buyers / 27% / 31% / 18% / 9% / 15%

Solution: Make a table

x (number of viewings) / P(x)
1 / 0.27
2 / 0.31
3 / 0.18
4 / 0.09
5 / 0.15
x (number of viewings) / P(x) / xP(x) = μ
1 / 0.27 / 1(0.27) = 0.27
2 / 0.31 / 2(0.31) = 0.62
3 / 0.18 / 3(0.18) = 0.54
4 / 0.09 / 4(0.09) = 0.36
5 / 0.15 / 5(0.15) = 0.75
x (number of viewings) / P(x) / x - μ / /
1 / 0.27 / 1 - 2.54 = -1.54 / / 2.372(0.27) = 0.640
2 / 0.31 / 2 – 2.54 = -0.54 / / 0.292(0.31) = 0.091
3 / 0.18 / 3 – 2.54 = 0.46 / / 0.212(0.18) = 0.038
4 / 0.09 / 4 – 2.54 = 1.46 / / 2.132(0.09) = 0.192
5 / 0.15 / 5 – 2.54 = 2.45 / / 6.052(0.15) = 0.908

Binomial Experiments

The spinner below is spun twice and we are interested in the number of times a 2 is obtained (assume each sector is equally likely).

Think of outcome 2 as a “success” and outcomes 1 and 3 as “failures.” The sample space is

S = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2),

(2, 3), (3, 1), (3, 2), (3, 3)}.

When the outcomes of an experiment are divided into just two categories, success and failure, the associated probabilities are called “binomial.” Repeated trials of the experiment, where the probability of success remains constant throughout all repetitions, are also known as Bernoulli trials.

If x denotes the number of 2s occurring on each pair of spins, then x is an example of a random variable. In S, the number of 2s is 0 in four cases, 1 in four cases, and 2 in one case. Because the table on the next slide includes all the possible values of x and their probabilities it is an example of a probability distribution. In this case, it is a binomial probability distribution.

Probability Distribution for the Number of 2s in Two Spins

X (Number of 2’s) / P(x) = P(X=x)
Case 1: There are no 2’s /
Case 2: There are one 2’s /
Case 3: There are two 2’s /

Binomial Experiments

1) There are a fixed number of trials. This is denoted by n.

2) The n trials are independent and repeated under identical conditions.

3) Each trial has two outcomes:

S = successF = failure

4) For each trial, the probability of success, p, remains the same.

Thus, the probability of failure is 1 – p = q.

5) The central problem is to determine the probability of r successes out of n trials.

Sampling Distribution

Suppose we have a population from which we obtain a random sample (independent samples). In statistical inference we want to make statement about the population based upon the data in the sample. We usually want to make statement about population parameters (example - µ, σ) based on statistic calculated from the sample.

A statistic from a random sample is a random variable. The probability distribution of a statistic is called a sampling distribution.

The sampling distribution depends on the distribution of the population values, the population parameters, and the sample size. We need to study sampling distributions to do statistical inference.

Sampling Distribution of Counts and Proportions- Binomial

Suppose we have a random sample of size n from a population in which we determine if the individual has or has not a certain characteristic. There are only two possible outcomes.

Let x = the number with the characteristic

Let = proportion in the sample unit with the characteristic. tells us something about the proportion in the population with the characteristic. These are called Binomial Experiments.

Example - Which of the following does not involve binomial trials?

a). A recyclable item is placed into a bin meant for paper, plastic, or glass.

b). A randomly-selected student passes a statistics course.

c). A electronic power supply is rejected if it delivers less than 300 milliamps of current.

d). A given concept is classified as “traditional” or “modern”.

Binomial Probability Formula

In general, let

n = the number of repeated trials,

p = the probability of success on any given trial,

q = 1 – p = the probability of failure on any given trial,

and“x” or “r” = the number of successes that occur.

Note that p remains fixed throughout all n trials. This means that all trials are independent. In general, x, successes can be assigned among n repeated trials in nCx different ways.

Example – If you are taking a multiple choice exam with 3 problems and each problem has 4 answer choices A, B, C, D then the probability of success (i.e. picking the correct answer) is 0.25 and the probability of choosing incorrectly is 0.75.

This is a binomial experiment with

Determining Binomial Probabilities

1) Use the Binomial Probability Formula.

2) Use Table 2 of the appendix.

3) Use technology. When n independent repeated trials occur, where

p = probability of success and

q = probability of failure

with p and q (where q = 1 – p) remaining constant throughout all n trials, the probability of exactly x successes is given by

The basic problem of a Binomial experiment is to find the probability of r successes out of n trials.

Example - Find the probability of observing 6 successes in 10 trials if the probability of success is p = 0.4.

a). 0.111b). 0.251c). 0.0002d). 0.022

How to find Binomial Probabilities in Calculator: pg. 221 section 6.2 in text book.

For P(r = #): Binopdf(n,p,r)

binompdf(numtrials,p,x) - binomomial probability distribution function,

numtrials = number of trials,
p = probability of success, and
x is the number of successes.
It is equivalent to P(X=x), where P is the binomial probability distribution function P(x)=C(N,x)px(1-p)N-x

For P(r ≤ #): Binocdf(n,p,r)

binomcdf(numtrials,p,x) - binomomial probability cumulative distribution function,

numtrials = number of trials,
p = probability of success, and
x is the number of successes.
It is equivalent to P(X<x), where P is the binomial probability distribution function P(x)=C(N,x)px(1-p)N-x

Binomial Probabilities

• At times, we will need to calculate other probabilities:

• P(r < k)

• P(r ≤ k)

• P(r > k)

• P(r ≥ k)

where k is a specified value less than or equal to the number of trials, n.

Example - Coin Tossing

Find the probability of obtaining exactly three heads in five tosses of a fair coin.

Example: Rolling a Die

Find the probability of obtaining exactly two 3’s in six rolls of a fair die.

Example: Rolling a Die

Find the probability of obtaining less than two 3’s in six rolls of a fair die.

Example: Baseball Hits

A baseball player has a well-established batting average of .250. In the next series he will bat 10 times. Find the probability that he will get more than two hits.

In many cases we will be interested in the probability of a range of successes. In such cases, we need to use the addition rule for mutually exclusive events.

For example – let , find the probability of 4 or fewer.

Solution:

Using the Binomial Table

1) Locate the number of trials, n.

2) Locate the number of successes, r.

3) Follow that row to the right to the corresponding p column.

Example - Use Table 2 in the Appendix to find the probability of observing 3 successes in 5 trials if

p = 0.7. a). 0.168b). 0.343c). 0.264d). 0.309

Graphing a Binomial Distribution

Example – Jim enjoys playing basketball. He figures that he makes about 50% of the field goals he attempts during a game. Make a histogram showing the probability that Jim will make 0,1,2,3,4,5 or 6 shots out of six attempted field goals.

x / P(x) = P(X=x)
0
1
2
3
4
5
6

Mean and Standard Deviation of a Binomial Distribution

Let X have a Binomial Distribution B(n,p)

mean of X = μ = np

variance of X = = np(1-p)

standard deviation of X =

Rules for means

1. If X has mean μ, then the mean of a+bx is a+bμ
2. If X has mean of and Y has mean then the mean of X+Y is +

Caution: the mean of X*Y is not necessarily *unless X and Y are independent. Also the mean of X/Y /

Rules for Variance

1. if X has variance then a+bx has variance
2. if X and Y are independent random variables with variance and then the

variance of X + Y is also the variance of X–Y is

1. X and Y are independent random variable if any event involving X is independent of any event involving Y

Note: These are population parameters not sample statistics.

Law of Large Numbers

Suppose we have a population with mean μ. Suppose we draw a random sample (independent observations) from the population. Let n = sample size. The sample mean can be made as close to μ as we want by making n large enough.

Example - Calculate the standard deviation of a binomial population with n = 100 and p = 0.3.

a). 21b).9c). 4.5825d). 4.41

Critical Thinking

• Unusual values – For a binomial distribution, it is unusual for the number of successes r to be more than 2.5 standard deviations from the mean:

– This can be used as an indicator to determine whether a specified number of r out of n trials in a binomial experiment is unusual.

Example - A particular binomial experiment has a mean of 14 and a standard deviation of 2. Which of the following is not an unusual number of successes?

a) 16b)17

c) 10d) None of these are unusual.

Example - The probability that someone has a particular disease is .05. If 1000 people are randomly selected. Let X be the number of people with the disease. Find the mean and standard deviation.

The mean of the Binomial Distribution is the expected value of r successes out of n trials.

Example - When Jim shoots field goals in basketball games, the probability that he makes a shot is only 0.5. Out of six throws, what is the expected number of goals Jim will make? For six trials what is the standard deviation of the binomial distribution of the number of successful field goals Jim makes?

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Hannah Province – Mathematics Department Southwest Tennessee Community College