Example 4.14 from Statistics, 4.3 Binomial Distribution (Optional)

Binomial Distribution (Optional): example 4.14

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

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. This is a binomial experiment since all three characteristics are met. Each person is a trial. Since we sample 200 people, the number of trials is n = 200. For each person, there are two possible outcomes: will develop the disease or not. Since we are measuring the number of people who will develop the disease, a person who will develop the disease is defined as a success and a person who will not develop the disease is defined as a failure. The risk of developing the disease is 1.28 percent. Therefore the probability of a success, $p = 1.28$ percent, $.0128$ , and the probability of a failure, $q = 1 - p = 1 - .0128 = .9872$ . Both p and q remain the same for each person. Therefore, X is a binomial random variable and it can be written as $X ~ B (200, .0128)$ .
  
  We can use a graphing calculator to answer Questions c and d.
2. 1. Mean = np = 200(.0128) = 2.56
  2. Standard Deviation = $\sqrt{n p q} = \sqrt{(200)(0 .128)(.9872)} \approx 1. 5897$
3. Using the TI-83, 83+, 84 calculator with instructions as provided in Example 4.12:
  P(x ≤ 8) = binomcdf(200, .0128, 8) = .9988
4. P(x = 5) = binompdf(200, .0128, 5) = .0707
  P(x = 6) = binompdf(200, .0128, 6) = .0298
  So P(x = 5) > P(x = 6); it is more likely that five people will develop the disease than six.

How did it go?