The Binomial Probability Distribution and Related Topics 5 Copyright Cengage Learning. All rights reserved. Section 5.3 Additional Properties of the Binomial Distribution Copyright Cengage Learning. All rights reserved. Focus Points Make histograms for binomial distributions. Compute µ and σ for a binomial distribution. Compute the minimum number of trials n needed to achieve a given probability of success P(r). 3 1
Graphing a Binomial Distribution 4 Graphing a Binomial Distribution Any probability distribution may be represented in graphic form. How should we graph the binomial distribution? Remember, the binomial distribution tells us the probability of r successes out of n trials. Therefore, we ll place values of r along the horizontal axis and values of P(r) on the vertical axis. The binomial distribution is a discrete probability distribution because r can assume only whole-number values such as 0, 1, 2, 3,... Therefore, a histogram is an appropriate graph of a binomial distribution. 5 Graphing a Binomial Distribution Procedure: 6 2
Example 7 Graph of a Binomial Distribution A waiter at the Green Spot Restaurant has learned from long experience that the probability that a lone diner will leave a tip is only 0.7. During one lunch hour, the waiter serves six people who are dining by themselves. Make a graph of the binomial probability distribution that shows the probabilities that 0, 1, 2, 3, 4, 5, or all 6 lone diners leave tips. 7 Example 7 Solution This is a binomial experiment with n = 6 trials. Success is achieved when the lone diner leaves a tip, so the probability of success is 0.7 and that of failure is 0.3: n = 6 p = 0.7 q = 0.3 8 Example 7 Solution We want to make a histogram showing the probability of r successes when r = 0, 1, 2, 3, 4, 5, or 6. It is easier to make the histogram if we first make a table of r values and the corresponding P(r) values (Table 5-12). Binomial Distribution for n = 6 and p = 0.70 Table 5-12 9 3
Example 7 Solution We ll use Table 3 of Appendix II to find the P(r) values for n = 6 and p = 0.70. To construct the histogram, we ll put r values on the horizontal axis and P(r) values on the vertical axis. Our bars will be one unit wide and will be centered over the appropriate r value. 10 Example 7 Solution The height of the bar over a particular r value tells the probability of that r (see Figure 5-3). Graph of the Binomial Distribution for n = 6 and p = 0.7 Figure 5-3 11 Example 7 Solution The probability of a particular value of r is given not only by the height of the bar over that r value but also by the area of the bar. Each bar is only one unit wide, so its area (area = height times width) equals its height. Since the area of each bar represents the probability of the r value under it, the sum of the areas of the bars must be 1. In this example, the sum turns out to be 1.001. It is not exactly equal to 1 because of rounding error. 12 4
Mean and Standard Deviation of a Binomial Distribution 13 Mean and Standard Deviation of a Binomial Distribution Two other features that help describe the graph of any distribution are the balance point of the distribution and the spread of the distribution about that balance point. The balance point is the mean µ of the distribution, and the measure of spread that is most commonly used is the standard deviation σ. The mean µ is the expected value of the number of successes. 14 Mean and Standard Deviation of a Binomial Distribution For the binomial distribution, we can use two special formulas to compute the mean µ and the standard deviation σ. 15 5
Mean and Standard Deviation of a Binomial Distribution Procedure: 16 Example 8 Compute µ and σ Let s compute the mean and standard deviation for the distribution of Example 7 that describes that probabilities of lone diners leaving tips at the Green Spot Restaurant. Solution: In Example 7, n = 6 p = 0.7 q = 0.3 For the binomial distribution, µ = np = 6(0.7) = 4.2 17 Example 8 Solution The balance point of the distribution is at µ = 4.2. The standard deviation is given by 18 6
Quota Problems: Minimum Number of Trials for a Given Probability 19 Quota Problems: Minimum Number of Trials for a Given Probability In applications, you do not want to confuse the expected value of r with certain probabilities associated with r. Quotas occur in many aspects of everyday life. The manager of a sales team gives every member of the team a weekly sales quota. In some districts, police have a monthly quota for the number of traffic tickets issued. Nonprofit organizations have recruitment quotas for donations or new volunteers. 20 Quota Problems: Minimum Number of Trials for a Given Probability The basic ideas used to compute quotas also can be used in medical science (how frequently checkups should occur), quality control (how many production flaws should be expected), or risk management (how many bad loans a bank should expect in a certain investment group). To have adequate power, a satellite must have a quota of three working solar cells. Such problems come from many different sources, but they all have one thing in common: They are solved using the binomial probability distribution. 21 7
Quota Problems: Minimum Number of Trials for a Given Probability To solve quota problems, it is often helpful to use equivalent formulas for expressing binomial probabilities. These formulas involve the complement rule and the fact that binomial events are independent. Equivalent probabilities will be used in Example 9. 22 Quota Problems: Minimum Number of Trials for a Given Probability Procedure: 23 Example 9 Quota Junk bonds can be profitable as well as risky. Why are investors willing to consider junk bonds? Suppose you can buy junk bonds at a tremendous discount. You try to choose good companies with a good product. The company should have done well but for some reason did not. 24 8
Example 9 Quota Suppose you consider only companies with a 35% estimated risk of default, and your financial investment goal requires four bonds to be good bonds in the sense that they will not default before a certain date. Remember, junk bonds that do not default are usually very profitable because they carry a very high rate of return. The other bonds in your investment group can default (or not) without harming your investment plan. 25 Example 9 Quota Suppose you want to be 95% certain of meeting your goal (quota) of at least four good bonds. How many junk bond issues should you buy to meet this goal? Solution: Since the probability of default is 35%, the probability of a good bond is 65%. Let success S be represented by a good bond. 26 Example 9 Solution Let n be the number of bonds purchased, and let r be the number of good bonds in this group. We want P(r 4) 0.95 This is equivalent to 1 P(0) P(1) P(2) P(3) 0.95 Since the probability of success is p = P(S) = 0.65, we need to look in the binomial table under p = 0.65 and different values of n to find the smallest value of n that will satisfy the preceding relation. 27 9
Example 9 Solution Table 3 of Appendix II shows that if n = 10 when p = 0.65, then, 1 P(0) P(1) P(2) P(3) = 1 0 0 0.004 0.021 = 0.975 The probability 0.975 satisfies the condition of being greater than or equal to 0.95. We see that 10 is the smallest value of n for which the condition P(r 4) 0.95 is satisfied. 28 Example 9 Solution Under the given conditions (a good discount on price, no more than 35% chance of default, and a fundamentally good company), you can be 95% sure of meeting your investment goal with n = 10 (carefully selected) junk bond issues. In this example, we see that by carefully selecting junk bonds, there is a high probability of getting some good bonds that will produce a real profit. What do you do with the other bonds that aren t so good? Perhaps the quote from Liar s Poker will suggest what is sometimes attempted. 29 10