1. The Binomial Probability Distribution and Related Topics 5

2. Section 5.3 Additional Properties of the Binomial Distribution

3. 3 Focus Points Make histograms for binomial distributions. Compute  and  for a binomial distribution. Compute the minimum number of trials n needed to achieve a given probability of success P (r).

4. 4 1. Graphing a Binomial Distribution

5. 5 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.

6. 6 1. Graphing a Binomial Distribution Procedure:

7. 7 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. n = 6 p = 0.7 q = 0.3

8. 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). cont’d Binomial Distribution for n = 6 and p = 0.70 Table 5-12

9. 9 Example 7 – Solution The height of the bar over a particular r value tells the probability of that r (see Figure 5-3). cont’d Graph of the Binomial Distribution for n = 6 and p = 0.7 Figure 5-3

10. 10 2. Mean and Standard Deviation of a Binomial Distribution

11. 11 2. 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  of the distribution, and the measure of spread that is most commonly used is the standard deviation . The mean  is the expected value of the number of successes. For the binomial distribution, we can use two special formulas to compute the mean  and the standard deviation .

12. 12 2. Mean and Standard Deviation of a Binomial Distribution Procedure:

13. 13 Example 8 – Compute  and  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,  = np = 6(0.7) = 4.2

14. 14 The balance point of the distribution is at  = 4.2. The standard deviation is given by cont’d Example 8 – Solution

15. 15 3. Quota Problems: Minimum Number of Trials for a Given Probability

16. 16 Guided Exercise 8: Find the minimum value of n for a give P(r) A satellite is powered by three solar cells. The probability that any one of these cells will fail is 0.15, and the cells operate or fail independently. Part I: In this part, we want to find the least number of cells the satellite should have so that the expected value of the number of working cells is no smaller than 3. In this situation, n represents the number of cells, r is the number of successful or working cells, p is the probability that a cell will work, q is the probability that a cell will fail, and m is the expected value that should be no smaller than 3.

17. 17 Guided Exercise 8: Find the minimum value of n for a give P(r) a) What is the value of q? of p? n=? q=0.15, p=0.85 b) The expected value μ for the number of working cells is given by μ=np. The expected value of the number of working cells should be no smaller than 3, so 3 ≤ n(0.85), n ≥ 3/0.85 = 3.53 c) Since n is between 3 and 4, should you round it to 3 or to 4 to make sure that m is at least 3? n = 4.

18. 18 Guided Exercise 8: Find the minimum value of n for a give P(r) Part II: In this part, we want to find the smallest number of cells the satellite should have to be 97% sure that there will be adequate power—that is, that at least three cells work. a) The letter r has been used to denote the number of successes. In this case, r represents the number of working cells. We are trying to find the number n of cells necessary to ensure that (choose the correct statement) (i) P(r ≥ 3) = 0.97 or (ii) P(r ≤ 3) = 0.97

19. 19 Guided Exercise 8: Find the minimum value of n for a give P(r) b) We need to find a value for n such that P(r ≥ 3) = 0.97 Try n = 4. Then r ≥ 3 means r = 3 or 4, so P(r ≥ 3) = P(3) + P(4) Use Table 3 (Appendix II) with n = 4 and p = 0.85 to find values of P(3) and P(4). Then compute P(r ≥ 3) for n = 4. Will n = 4 guarantee that P(r ≥ 3) is at least 0.97? P(3) = 0.368 P(4) = 0.522 P(r ≥ 3) = 0.890 n = 4 is not sufficient.

20. 20 Guided Exercise 8: Find the minimum value of n for a give P(r) c) Now try n = 5 cells. For n = 5, P(r ≥ 3) = P(3) + P(4) + P(5) since r can be 3, 4, or 5. Are n = 5 cells adequate? [Be sure to find new values of P(3) and P(4), since we now have n = 5.] P( r ≥ 3) = P(3) + P(4) + P(5) = 0.138 + 0.392 + 0.444 = 0.974 Thus, n = 5 cells are required.

21. 21 3. Quota Problems: Minimum Number of Trials for a Given Probability Quotas occur in many aspects of everyday life. For examples: 1. The manager of a sales team gives every member of the team a weekly sales quota. 2. In some districts, police have a monthly quota for the number of traffic tickets issued. 3. Nonprofit organizations have recruitment quotas for donations or new volunteers.

22. 22 3. 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.

23. 23 3. Quota Problems: Minimum Number of Trials for a Given Probability To solve quota problems, it is often helpful to use equivalent formulas (defined in the next page) 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.

24. 24 3. Quota Problems: Minimum Number of Trials for a Given Probability Procedure:

25. 25 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.

26. 26 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. cont’d

27. 27 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. cont’d

28. 28 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. cont’d

29. 29 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. cont’d

30. 30 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. cont’d

31. 31 Answers to some even hw: 2. The expected value is higher for the first distribution. 4. a) p=0.5; μ=5; Yes. b) right c) Left 8. n=8; p=0.01 b) μ=0.08 b) 0.997 c) 0.281 10. n=5; p=0.85 b) μ=4.25; σ≈0.798; expected number is about 4. 14. n=22 gives P(r≥1) at least 0.90 [correction: P(0)≤0.10] 16. p=0.65. a) P(r≥1)=1-P(0)≥98% or P(0)≤2%: n=4 b) 2.6 20. p=0.55 a) P(r≥1)=1-P(0)≥99% or P(0)≤1%: n=6 b) 4.95, or about 5 22. n=6; p=0.45 a) 0.008 b) 0.0277 c) 0.836 d) μ=2.7 σ=1.219 e) 10 professors