1. 5.1 Sampling Distributions for Counts and Proportions

2. Binomial Experiments Consider the familiar discrete probability distribution of a fair 6-sided die. Now suppose we are interested in obtaining a 3 or a 4 over 100 rolls. Here we say that n=100 and p=1/3. It seems like over 100 rolls, roughly 100(1/3)≈33 would turn up a 3 or 4. But what if we want the probability of exactly seven 3s? Or more than ten 4s? This is an example of a binomial experiment, which we denote B(100,1/3).

3. What is it? We define a binomial experiment, B(n,p), to satisfy the following properties: 1.The experiment consists of a fixed number n of trials. (100 rolls of a die) 2.Each trial has exactly two outcomes, success (S) or failure (F). We write P(S)=p and P(F)=q. Then p+q=1 so q=1-p and p=1-q. Each trial is identical and hence p and q do not change from one trial to another. (S= obtaining a 3 or 4. F= obtaining 1,2,5,6) 3.The trials are independent of one another; that is, the outcome of one trial has no bearing on the outcome of any other trial. This is referred to as a Bernoulli trial. Then the random variable x, called the binomial random variable, is the number of successes in the n trials. The probability distribution of x is a binomial probability distribution.

4. Suppose we roll a fair die 20 times. What is the probability of obtaining exactly four 3s? Not quite obvious… Table C (pp. T-6 to T-10 in the book) gives a binomial probability distribution for various values of n, k, and various probabilities. Now we may answer this question.

5. 40% of voters are opposed to a proposed development. If 15 voters are randomly selected, find the probability that 1.7 are opposed to the development. 2.Fewer than 4 are opposed to the development. 3.More than 2 are opposed to the development.

6. Practice Problem A person has a 5% chance of winning a free ticket in a state lottery. If she plays the game 12 times, what is the probability she will win more than 1 free ticket?

7. Expected value Recall that for a discrete probability distribution, the expected value was the mean. One can show that for a binomial distribution, It can also be shown that

8. Thus, we can now compute the mean, variance, and standard deviation for both the voting example and the lottery example. Let’s do that.