# II. BINOMIAL DISTRIBUTIONS A. Binomial Experiments 1. A II. BINOMIAL DISTRIBUTIONS A. Binomial Experiments 1. A binomial experiment is a probability experiment that satisfies the following conditions: a. The experiment is repeated for a fixed number of independent trials. b. There are only two possible outcomes of interest for each trial. 1) the outcomes can be classified as a success (S) or as a failure (F). c. The probability of success P(S) is the same for each trial. 1) This is just another way of saying that they are independent. d. The random variable x counts the number of successful trials. 2. Notation for Binomial Experiments. a. n The number of times a trial is repeated.

b. p = P(S) The probability of success in a single trial. c. q = P(F) The probability of failure in a single trial (q = II. BINOMIAL DISTRIBUTIONS Example 1 Page 207 Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q and list the possible values of the random variable x. 1) A certain surgical procedure has an 85% chance of success. A doctor performs the procedure on eight patients. The random variable represents the number of successful surgeries. 2) A jar contains five red marbles, nine blue marbles, and six green marbles. You randomly select three marbles from the jar without replacing them. The

random variable represents the number of red marbles. II. BINOMIAL DISTRIBUTIONS Example 1 Page 207 Solutions: 1) The experiment satisfies the four conditions of a binomial experiment. a) Each surgery represents one trial, and we know that there are eight surgeries. b) Two possible outcomes; the surgery is either a success or a failure. c) The surgeries are independent. Each one has an 85% chance of success. d) n = 8; p = .85; q = .15 (1 - .85) x=

0, 1, 2, 3, 4, 5, 6, 7, or 8. 2) The experiment does NOT satisfy the four conditions of a binomial experiment. a) The number of trials is set, but they are not independent. 1. Because we arent putting the marbles back, the probability of getting a red marble changes every time we pull a marble II. BINOMIAL DISTRIBUTIONS B. Binomial Probability Formula 1. There are several ways to find the probability of success in n trials of binomial experiment. a. One way is to use a tree diagram and the Multiplication Rule. b. Another way is to use the binomial probability formula. 1) Binomial probability formula; . Another way to write this is:

c. A third (and the one you will use) is to allow the calculator to find the probability for you. 1) 2nd VARS, scroll to binompdf and enter the information asked for. II. BINOMIAL DISTRIBUTIONS Example 2 Page 208 Microfracture knee surgery has a 75% chance of success on patients with degenerative knees. The surgery is performed on three patients. Find the probability of the surgery being successful on exactly two patients. Solutions: 1) Tree diagram gives you the following sample space, with the corresponding probabilities: SSS SSF SFS SFF

FSS FSF FFS FFF (we could do this since we only have 8 outcomes). Notice that we are only concerned with the outcomes that involve exactly two successes. SSF, SFS, and FSS are the three that meet this criteria. To find these, we multiply the probabilities of each event and add them together. (.75)(.75)(.25) = .14063; (.75)(.25)(.75) = .14063; (.25)(.75)(.75) = .14063 .140625 + .140625 + .140625 = .421875 II. BINOMIAL DISTRIBUTIONS Example 2 Page 208 2) Use the binomial probability formula.

3) Use the calculator. 2nd VARS A (binompdf); Enter 3 for number of trials, .75 for p, and 2 for xvalue. Highlight Paste and hit Enter twice. The probability is .421875 II. BINOMIAL DISTRIBUTIONS Example 2 Page 208 4) Use the binomial probabilities table. are multiples of 5. one of the other methods. very much in the Look in the column for .75 (p value) Look at the row for n = 3 and x = 2. The probability is .422 Notice that this table only works for values of p that Otherwise, you have to use For that reason, we will not mention the table future.

II. BINOMIAL DISTRIBUTIONS Example 3 Page 209 In a survey, workers in the United States were asked to name their expected sources of retirement income. The results are shown. Seven workers who participated in the survey are randomly selected and asked whether they expect to rely on Social Security for retirement income. Create a binomial probability distribution for the number of workers who respond yes. 401k, IRA, Keough, or other, 47%; Home equity, 26%; Pension, 26%; Social Security, 25%; Savings/CDs, 19%; Stocks/Mutual Funds, 19%; Part-time work, 18%; Annuities or insurance plans, 7%; Inheritance, 7%; Rents/Royalties, 6%. II. BINOMIAL DISTRIBUTIONS Example 3 Page 209 Solution: p = .25, q = .75 (1 - .25), n = 7; possible values of x are 0, 1, 2, 3, 4, 5, 6, and 7

Using the formula, it looks like this: P(0) = OR x P(x) P(1) = OR 0 .13348 P(2) = OR 1 .31146 P(3) = OR P(4) = OR 2 .31146 P(5) = OR 3

.17303 P(6) = OR 4 .05768 P(7) = OR Notice that all the probabilities are between 0 and 1, and they add up 5to very.01154 close to 1 (.99999). This confirms that you have in fact created a probability 6 distribution. .00128 7 .00006 II. BINOMIAL DISTRIBUTIONS Example 5 Page 211 A survey indicates that 41% of women in the United States consider reading their favorite leisure-time activity. You randomly select four U.S. women and ask them if reading is their favorite leisure-time activity.

Find the probability that (1) exactly two of them respond yes, (2) at least two of them respond yes, and (3) fewer than two of them respond yes. Solution: 1. Since we want the probability that exactly two respond yes, we can use the formula, the binompdf function on the calculator or a diagram tree. We could do a diagram tree, since we only have 16 outcomes, but the formula and calculator methods are more effort efficient. We can NOT use the table because p (.41) is not a multiple of 5. Using the formula looks like this: P(2) = OR Using the calculator would go like this: 2nd VARS A, 4, .41, 2 = .35109 II. BINOMIAL DISTRIBUTIONS Example 5 Page 211

2. We want the probability that at least two women respond yes. This means 2, 3, or 4 yeses. Using the formula, we could find each probability and then add them together. P(2) = OR P(3) = OR P(4) = OR Adding these up gives us .542. II. BINOMIAL DISTRIBUTIONS Example 5 Page 211 Another way to do this would be to use the complement rule and save ourselves a little effort. The complement of at least two is less than two, or 0 and 1. P(0) = OR P(1) = OR

Adding these up gives us .45799 Subtracting this from 1 (complement rule) gives us .542, the same answer we got the first time. 1 - .45799 = .54201 So, we can find the answer by finding and adding three probabilities, or by finding and adding two probabilities and then subtracting from 1. II. BINOMIAL DISTRIBUTIONS Example 5 Page 211 Using the calculator, we want to use a new distribution. Because we want the probability of more than one outcome, we want a cumulative probability. We will use the binomcdf function for this. (2 nd VARS B) The calculator will always give the probability of up to the number that you put in for the x-value. Since you want two or more, you need to use 1 for x and subtract from

1. 2nd VARS binomcdf, enter 4 for trials, .41 for p, and 1 for the x-value. You will get .45799, the same answer we got when we did it by hand above. Subtract from 1 to get the answer to the question. 1 - .45799 = .54201 II. BINOMIAL DISTRIBUTIONS Example 5 Page 211 3. We have already found the probability of fewer than two; we used it for the complement rule. The probability of fewer than two is .45799. C. Mean, Variance, and Standard Deviation Although you can use the formula learned in Section 4-1 for mean, variance, and standard deviation of a discrete probability distribution, the properties of a binomial distribution enable you to use much simpler formulas.

Mean: Variance: Standard deviation: II. BINOMIAL DISTRIBUTIONS Example 8 Page 214 In Pittsburgh, Pennsylvania, about 56% of the days in a year are cloudy. Find the mean, variance and standard deviation for the number of cloudy days during the month of June. Interpret the results and determine any unusual values. Solutions: There are 30 days in June. Using n = 30, p = 0.56, and q = 0.44, you can find the mean, variance and standard deviation as follows: On average, the month of June has 16.8 cloudy days. Since values that lie more than 2 standard deviations away from the mean are unusual, we subtract 2.7188 from 16.8 twice and

add it to 16.8 twice to find where our unusual values begin. 16.8 (2*2.7188) = 11.3624 16.8 + (2*2.7188) = 22.2376 So, 11 days or less of clouds would be unusual. So would 23 days or more of clouds. Assignments: Classwork: Pages 216217; #714 All, 1624 Even Homework: Pages 218219; #28 a, d & e and 2932 All