1. Chapter 8: The Binomial and Geometric Distributions 8.1 – The Binomial Distributions

2. We frequently encounter random phenomena where there are two outcomes of interest. The coin toss in a football game used to determine which team gets the choice of kicking off or receiving to begin the game. A basketball player shoots a free throw A young couple prepares for their first child

3. Question 1: Gretchen is a 60% free throw shooter. In a season, she shoots an average of 75 free throws. What is the probability that Gretchen  makes exactly 50 out of 75 free throws?  makes more than 50 free throws?  makes no more than 40 free throws? You should be able to answer all of these at the end of this section.

4. All four of these conditions must be met in order to use a binomial distribution. 1. Each observation falls into one of two categories: “success” or “failure” 2. There is a fixed number n of observations 3. The n observations are independent 4. The probability of success, call it p, is the same for each observation.

5. The Binomial Distribution Let X = number of successes X is a binomial random variable B(n, p) represents a binomial distribution where n is the number of observations and p is the probability of success of any one observation X can take on any whole number from 0 to n

6. For each of the following, determine if a binomial distribution applies. 1. Blood type is inherited. If both parents carry genes for O and A blood types, each child has probability 0.25 of getting two O genes and so of having blood type O. A success is the child getting blood type O among 5 children of these parents. Yes. X = number of O-type blood for the 5 children. Each child is independent and the probability of success is 0.25 for each child. X is B(5, 0.25)

7. Binomial or not? 2. Deal 10 cards from a shuffled deck and count the number X of red cards. A “success” is a red card. The observations are not independent. If the first card is black, then the second card is more likely to be red since there are more red cards. This is not binomial.

8. Binomial or not? 3. An engineer chooses an SRS of 10 switches from a shipment of 10,000 switches. Suppose that (unknown to the engineer) 10% of the switches in the shipment are bad. The engineer counts the number X of bad switches in the sample. This is not quite binomial. The 2 nd switch is not independent of the 1 st.

9. Let’s look at example 3 again Assuming that a defective switch is drawn first (p = 0.1), the probability for the second switch being defective changes to p = 0.0999. For practical purposes, this is “close enough” and therefore behaves like a binomial setting even if the condition of independence does not strictly hold. When the population is much larger than the sample, a count of successes in an SRS of size n has approximately the binomial distribution

10. Binomial Formulas You have seen this formula before in Algebra II nCr is the same as

11. Binomial Formulas In some books, they let (1 – p) = q. If you would like to simplify formulas, you must first state how you are doing it

12. Recall that the probability of a child being born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that exactly 2 of them have blood type O? n = 5, p = 0.25 P(X = 2) = (0.25) 2 (1-0.25) 3 = 10(0.25) 2 (0.75) 3 = 0.2637

13. The number X of switches that fail inspection has approximately the binomial distribution with n = 10 and p = 0.1. What is the probability that no more than one switch fails? P(X ≤ 1) = P(X = 0) + P(X = 1) = (0.1) 0 (0.9) 10 + (0.1) 1 (0.9) 9 = 0.7361

14. The mailing list of an agency that markets scuba-diving trips to the Florida Keys contains 70% males and 30% females. The agency calls 30 people chosen at random. a) What is the probability that 20 of the 30 are men? b) What is the probability that the first woman is reached on the fourth call? (That is, the first 4 calls give MMMF.)

15. Let X = the number of men called. X is B(30, 0.7) P(X = 20) = (0.7) 20 (0.3) 10 b) P(MMMF) = (0.7) 3 (0.3) 1 = 0.1416 = 0.1029 a) b) HW: p. 516 #8.1-8.6 and p. 519 #8.7-8.11

16. Using the calculator for calculating binomial probabilities Recall the switch problem… An engineer chooses an SRS of 10 switches from a shipment of 10,000 switches. Suppose that (unknown to the engineer) 10% of the switches in the shipment are bad. The calculator can calculate the probabilities of each value X using the command binompdf(n, p, X)

17. Using the calculator What’s the probability that at most 1 fails? = P(X = 0) + P(X = 1) = binompdf(10, 0.1, 0) + binompdf(10, 0.1, 1) = 0.3487 + 0.3874 = 0.7361 The suffix pdf stands for probability distribution function.

18. Corrine makes 75% of her free throws over the course of a season. In a key game, she shoots 12 free throws and only makes 7 of them. The fans think she failed because she is nervous. Is it unusual for Corinne to perform this poorly? (Assume that each shot is independent of one another) What is the probability of making a basket on at most 7 free throws?

19. Cumulative Binomial Probability In applications we frequently want to find the probability that a random variable takes a range of values. The cumulative binomial probability is useful in these cases. On calculator: binomcdf(n, p, x) B(12, 0.75) P(X ≤ 7) = binomcdf(12, 0.75, 7) = 0.1576

20. Cumulative Binomial Probability It is very important that you understand the difference between binomcdf and binompdf

21. Among employed women, 25% have never been married. Select 10 employed women at random. 1. What is the probability that exactly 2 of the 10 women in your sample have never been married? 2. What is the probability that 2 or fewer have never been married? binompdf(10, 0.25, 2) = 0.2816 binomcdf(10, 0.25, 2) = 0.5256

22. Gretchen is a 60% free throw shooter. In a season, she shoots an average of 75 free throws. What is the probability that Gretchen 1) makes exactly 50 out of 75 free throws? 2) makes more than 50 free throws? 3) makes no more than 40 free throws? P(X = 50) = binompdf(75, 0.6, 50) = 0.0479 P(X > 50) = 1 - binomcdf(75, 0.6, 50) = 0.0963 P(X ≤ 40) = binomcdf(75, 0.6, 40) = 0.1446

23. Caution….. On the AP Exam you are not allowed to just write “calculator talk.” You must write out the formula for each problem, but feel free to use your calculator for the calculations For cumulative, use “+ … +” to show the first two and last part of the formula. HW: pg. 523 #8.13-8.17

24. Binomial Mean and Standard Deviation The binomial distribution is a special case of a probability distribution for a discrete random variable. We could find the mean and standard deviation in the same way as we did before. However, we really don’t want to if our n is large.

25. Binomial Mean and Standard Deviation CAUTION: These formulas are only good for binomial distributions!

26. The count X of bad switches is binomial with n = 10 and p = 0.1. Find the mean and standard deviation. μ = np = 10(0.1) = 1 σ =

27. The Normal Approximation to Binomial Distributions The formula for binomial probabilities becomes awkward as the number of trials n increases. As the number of trials n gets larger, the binomial distribution gets close to a Normal distribution. When n is large, we can use Normal probability calculations to approximate hard- to-calculate binomial probabilities.

28. The Normal Approximation to Binomial Distributions Note: you must verify that np ≥ 10 and n(1 – p) ≥ 10 in order to use Normal approximations

29. a) What are the mean and standard deviation of X? b) Use a Normal approximation to find the probability that the sample will contain between 165 and 195 blacks. Be sure to check that you can safely use the approximation. * One way of checking the effect of undercoverage, nonresponse, and other sources of error in sample surveys is to compare the sample with known facts about the population. About 12% of American adults are black. The number X of blacks in a random sample of 1500 should therefore vary with B(1500, 0.12).

30. B(1500, 0.12) Mean: μ = np = 1500(0.12) = 180 St. Dev.: Verify rule of thumb:np = 180 > 10 n(1-p) = 1320 > 10 Normal Approximation: N(180, 12.59)

31. B(1200, 0.12) N(180, 12.59) Find P(165 ≤ X ≤ 195) HW: pg. 529 #8.19 – 8.23