Lesson 4.1 - Binomial Distribution

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 Unit Summary Binomial Experiment and Binomial Random Variable An Activity to Motivate the Binomial Probability Binomial Distribution and How to Evaluate It Mean $$\mu$$ and Standard Deviation $$\sigma$$ of Binomial Random Variable Some Application of Binomial Distribution's Mean and Standard Deviation

An Introduction to Statistical Methods and Data Analysis, chapter 4.8. (You only need to read the material for the Binomial distribution.)

Binomial Experiment and Binomial Random Variable

Binary categorical variable is a variable that has two possible outcomes. For example, gender (male/female), having a tattoo (yes/no) are all examples of a binary categorical varaible.

Binomial experiment satisfies the following four conditions:

1. The experiment consists of n identical trials.
2. Each trial results in one of the two outcomes, called success and failure.
3. The probability of success, denoted π, remains the same from trial to trial.
4. The n trials are independent. That is, the outcome of any trial does not affect the outcome of the others.

An Activity to Motivate the Binomial Probability

An activity to motivate the probability of the binomial distribution:

1. Be creative in thinking of a binary categorical variable which would be "measured" on people. Define the variable and identify the two possible traits. (One possible example is being male [M] or being female [F].)

2. Now, suppose you randomly select three people. You plan to "measure" the people with respect to your binary categorical variable. Using your notation defined above, list the eight possible sequences of outcomes that could occur. (For the gender example, all three people could be male, i.e., MMM, or the first person could be male and the second and third persons could be female, i.e., MFF, and so on. With three people, there are eight possible such outcomes.)

3. Arbitrarily, pick one of the two possible traits for your original variable you defined in #1. Then, for each of the eight outcomes you listed in #2, identify how many times the trait you picked could occur. (For the gender example, if you are interested in the number of males in the sample, then for MMM, there are three males, but for MFF, there is one male.)

4. Assume that each trial has a probability of 0.6 of having the trait that you picked in #3. This probability is different for different problems. Here, we all work on the case π = 0.6 so that we can compare our answers. Then, assuming independence, calculate the probability that if you randomly select three people, none of them will have your trait of interest. (I'd calculate the probability of getting three females, i.e. the probability of getting FFF.)

5. Using the same assumptions as in #4, what is the probability that if you randomly select three people, one (exactly one) of them will have your trait of interest?

6. Using the same assumptions as in #4, what is the probability that if you randomly select three people, two (exactly two) of them will have your trait of interest?

Try to identify a pattern in your calculations in #4, #5, and #6 for P(Y = y).

Binomial Distribution and How to Evaluate It

One can use the formula to find the probability or alternatively, use Minitab to find the probability. In the homework, you may use the one that you are more comfortable with unless specified otherwise.

Using Minitab, find P(y) for n = 20, y = 3, and π = 0.4.

Calc > Probability Distributions > Binomial

Choose Probability since we want to find the probability y = 3. Choose input constant and type in 3 since that is the value you want to evaluate the probability at.

Minitab output:

Probability Density Function

Binomial with n = 20 and p = 0.4

 x P(X = x) 3.00 0.0123

Click on the 'Minitab Movie' icon to view the walk through 'Finding the Probabilities of Binomial Distributions'.

In the following example, we illustrate how to use the formula to compute binomial probabilities. If you don't like to use the formula, you can also just use Minitab to find the probabilities.

Cross-fertilizing a red and a white flower produces red flowers 25% of the time. Now we cross-fertilize five pairs of red and white flowers and produce five offspring.

a. Find the probability that there will be no red flowered plants in the five offspring.

Y = # of red flowered plants in the five offspring. Here, the number of red flowered plants has a binomial distribution with n = 5, π = 0.25.

$P(Y=0)= \frac{5!}{0!(5-0)!} {\pi}^0 {(1-\pi)}^5=1 (0.25)^0 (0.75)^5 =0.237$

b. Find the probability that there will be four or more red flowered plants. Try to figure out your answer first, then click the graphic to compare answers.

Mean μ and Standard Deviation σ of the Binomial Probability Distribution

 $$\mu=n \pi$$ $$\sigma=\sqrt{n \pi (1-\pi)}$$ where π is the probability of success in a given trial, andn is the number of trials in the binomial experiment.

The mean of a distribution is also called the expected value of the distribution.

Of the five cross-fertilized offspring, how many red flowered plants do you expect? Try to figure out your answer first, then click the graphic to compare answers.

Note: Y can only take values 0, 1, 2, ..., n, but the expected value (mean) of Y may be some value other than those that can be assumed by Y.

What is the standard deviation of Y, the number of red flowered plants in the five cross-fertilized offspring? Try to figure out your answer first, then click the graphic to compare answers.

Some Application of Binomial Distribution's Mean and Standard Deviation

A pharmaceutical company claims that a new treatment is successful in reducing fever in more than 60% of the cases. The treatment was tried on 40 randomly selected cases and 11 were successful. Do you doubt the company's claim? (i.e., Can you reject the company's claim?)

Answer: If the claim is valid, then Y (the number of successful cases) has a binomial distribution with n = 40 and π which is greater than 0.6.

We will first consider the boundary case, π = 0.6. Is y = 11 a likely outcome from Y when π = 0.6?

Answer: When π = 0.6, μ = n π = 40 (0.6) = 24

$$\sigma=\sqrt{n \pi (1-\pi)}=\sqrt{40\cdot0.6\cdot0.4}=3.1$$

μ - 3σ = 24 - 3(3.1) = 14.7 , μ + 3σ = 24 + 3(3.1) = 33.3

The observed value, y = 11 is less than 14.7 and not within 3σ of its mean, thus 11 is unlikely to be observed when π = 0.6. Same argument carries through to π greater than 0.6.

Here is Dr. Andrew Wiesner working through this problem:

An alternative approach: Compute the probability of observing a value as small as or smaller than 11, assuming π = 0.6. If the probability is large, do not doubt the claim. If the probability is small, doubt the claim. Using Minitab, we get the following output:

Probability Density Function

Binomial with n = 40 and p = 0.6

 x P(X = x) 0 0.0000000 1 0.0000000 2 0.0000000 3 0.0000000 4 0.0000000 5 0.0000000 6 0.0000000 7 0.0000000 8 0.0000002 9 0.0000013 10 0.0000059 11 0.0000242

We thus obtain that P(X ≤ 11) = 0.0000316. The probability is very small. We, thus doubt the claim. (Note: It is incorrect to just compute the probability at 11 since that is usually very small if sample size is large.)  Here is Dr. Andrew Wiesner again, working through this alternative approach: