PROBABILITY AND STATISTICS CHAPTER 4 NOTES DISCRETE PROBABILITY DISTRIBUTIONS

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1 PROBABILITY AND STATISTICS CHAPTER 4 NOTES DISCRETE PROBABILITY DISTRIBUTIONS I. INTRODUCTION TO RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS A. Random Variables 1. A random variable x represents a value associated with each of a probability experiment. a. The word indicates that x is determined by chance. There are two types of random variables: and. 1) A random variable is if it has a finite or number of possible outcomes that can be. 2) A random variable is if it has an infinite or number of possible outcomes represented by an on the. Example 1 Page 195 Decide whether the random variable x is discrete or continuous. Explain your reasoning. 1) x represents the number of stocks in the Dow Jones Industrial Average that have share price increases on a given day. 2) x represents the volume of water in a 32-ounce container. 1) The number of stocks whose share price increases can be. So, x is a random variable. 2) The amount of water in the container can be any volume between 0 ounces and 32 ounces, including decimals. We the volume of water. So, x is a variable. B. Discrete Probability Distributions. 1. Each value of a discrete random variable can be assigned a probability. By listing each value of the random variable with its corresponding probability, you are forming a probability distribution. a. A probability distribution lists each possible value the random variable can assume, together with its probability. A probability distribution must satisfy the following conditions: 1) The probability of each value of the discrete random variable is between 0 and 1, inclusive. 0 P(x) 1 2) The sum of all probabilities is exactly 1. P(x) = 1 2. How to Construct a Discrete Probability Distribution a. Make a frequency distribution for the outcomes. b. Find the of the frequencies c. Find the of each possible outcome by dividing its by the sum of the. d. Check that each probability is between and, inclusive, and that the sum is. Example 2 Page 196 An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were given a score from 1 to 5, where 1 was extremely passive and 5 extremely aggressive. A score of 3 indicated neither trait. The results are shown in the table. Construct a probability distribution for the random variable x. the frequency of each score by the of individuals in the study to find the probability for each value of the random variable. Score, x Frequency, P(x) x P(x)

2 Example 3 Page 197 Verify that the distribution in the table is a probability distribution. Days of Rain (x) Probability P(x) If the distribution is a probability distribution, then (1) each probability is between and, inclusive, and (2) the sum of the probabilities equals. 1) Each probability between 0 and 1 2) P(x) = = 1 Because conditions are met, the distribution a probability distribution. Example 4 Page 197 Decide whether each distribution is a probability distribution. 1) 2) x P(x) x P(x) ) Each probability is between 0 and1 but the sum of the probabilities is 1.07, which is greater than 1. So, it is not a probability distribution. 2) The sum of the probabilities is equal to 1, but P(3) and P(4) are not between 0 and 1. So, it is not a probability distribution. Probabilities can never be negative or greater than 1. C. Mean, Variance, and Standard Deviation. 1. You can measure the of a probability distribution with its mean and measure with its variance and standard deviation. The mean of a discrete random variable is defined as follows: a. The mean of a discrete random variable is given by μ = xp(x). 1) Each value of x is by its corresponding probability and the products are added. Example 5 Page 198 The probability distribution for the personality inventory test for passive-aggressive traits discussed in Example 2 is given. Find the mean score. What can you conclude? x P(x) This is different from what the book tells you to do; it uses the calculator to do the math. 1) Enter the values of x into L1 using the STAT Edit menu. 2) Enter the values for P(x) into L2. 3) Use the arrow keys to highlight the L3 at the top of the third column. Type L1*L2 here, and these values will automatically populate L3 for you. 4) Press 2 nd and Mode (Quit) to go back to the main screen. 5) Press 2 nd STAT, arrow to MATH, and select 5 (sum). Designate L3 as the list that you want the sum of by pressing 2 nd and 3. Press Enter You will see that the total of L3 is. That is your mean. 2. Although the mean of the random variable of probability distribution describes a typical outcome, it gives no information about how the outcomes vary. a. To study the variation of the outcomes, you can use the variance and standard deviation of the random variable of a probability distribution. 1) The of a discrete random variable is σ 2 = (x μ) 2 P(x) 2) The of a discrete random variable is the square root of the variance, so σ = (x μ) 2 P(x), or σ = σ 2 5 4

3 Example 6 Page 199 We are still using the probability distribution for the personality inventory test for passive-aggressive traits discussed in Example 2. Find the variance and the standard deviation. Again, different from the book; we are using the calculator. We are keeping the 3 lists we already created solving Example 5, and will add 3 more columns (lists) to finish the process. 1) In order to find the variance, we need to know what each x - μ is, so we highlight L4 and type L1 (2 nd STAT MATH 5 2 nd 3). You could also just manually type L1 2.94, if you wanted to. This takes each value in L1 and subtracts the total of L3 (the mean) from it. Notice that some of these values are negative, and that they all add up to zero. This does us no good. 2. Highlight L5 and type 2 nd 4 and press the x 2 key. This squares L4, making them positive and useful. 3. Highlight L6 and type L2*L5. This multiplies the values for P(x) and the values for (x μ) Exit from the STAT menu, and press 2 nd STAT, arrow to MATH, press 5 and 2 nd 6 to get the sum of list 6. This is your variance. In this case, the variance is. 5. To find the standard deviation, simply take the of the. Hit 2 nd and press the x 2 key, then press 2 nd and the negative sign key. This takes the square root of the last answer, which was the variance. This is your standard deviation. In this case, the standard deviation is D. Expected Value 1. The mean of a random variable represents what you would expect to happen for thousands of trials. It is also called the. a. Remember that the expected value can be negative or greater than 1; it is not a probability, but the answer that we expect to get. Example 7 Page 200 At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. You buy one ticket. What is the expected value of your gain? Create a table using the values for x, and the probabilities of achieving each of those values. Gain, (x) Probability, P(x) In the calculator, L1 is the Gain, and L2 is the P(x) You can actually type the fraction into the list; the calculator will convert to decimal. This prevents having to enter rounded values too many times Highlight L3 and type in L1*L2. Find the total of L3 2 nd STAT, MATH, 5, 2 nd 3. The total of L3 is That means that you can expect to lose an average of $1.35 for each $2 ticket that you buy.

4 II. BINOMIAL DISTRIBUTIONS A. Binomial Experiments 1. A binomial experiment is a experiment that satisfies the following conditions: a. The experiment is for a number of independent trials. b. There are only possible outcomes of interest for each trial. 1) the outcomes can be classified as a ( ) or as a ( ). c. The probability of success P(S) is for each trial. 1) This is just another way of saying that they are. d. The random variable 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 = 1 p) d. x The random variable represents a count of the number of successes in n trials. 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. 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 = ; p = ; q = ( ) 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. 1. Because we aren t putting the marbles back, the probability of getting a red marble every time we pull a marble out. 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. This only works when there are not a lot of possible outcomes, and so is not recommended. b. The second way is to use the binomial probability formula. 1) Binomial probability formula; P(x) = n! (n x)!x! px q n x. Another way to write this is: P(x) = n C x p x q n x c) The third way is to use the calculator. (This is probably the one that most of you will use). d) One last way to find binomial probabilities is to use a binomial probability table. There is a copy of the binomial probability table in your book it is Table 2 in Appendix B. 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. 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 successes. SSF, SFS, and FSS are the three that meet this criteria. To find these, we the probabilities of each event and add them together. (.75)(.75)(.25) =.14063; (.75)(.25)(.75) =.14063; (.25)(.75)(.75) = =

5 2) Use the binomial probability formula. P(2 successful surgeries) = 3! (3 2)!2! p2 q 3 2 = 3!. (1)!2! = 3) Use the calculator. 2 nd VARS A (binompdf); Enter 3 for number of trials,.75 for p, and 2 for x-value. Highlight Paste and hit Enter twice. The probability is. 4) Use the binomial probabilities table. Look in the column for.75 (p value) and the row for n = 3 and x = 2. The probability is. Notice that this table only works for values of p that are multiples of 5. Otherwise, you have to use one of the other methods. For that reason, we will not mention the table very much in the future. 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%. The probability of someone expecting to rely on Social Security is. p =.25, q =.75 (1 -.25), n = 7, and the possible values of x are 0, 1, 2, 3, 4, 5, 6, and 7 Using the formula, it looks like this: P(0) = (7 0)!0! (.250 )( ) OR P(0) = 7 C 0 (.25 0 )( ) = x P(x) P(1) = (7 1)!1! (.251 )( ) OR P(1) = 7 C 1 (.25 1 )( ) = 0 P(2) = (7 2)!2! (.252 )( ) OR P(2) = 7 C 2 (.25 2 )( ) = 1 P(3) = (7 3)!3! (.253 )( ) OR P(3) = 7 C 3 (.25 3 )( ) = 2 P(4) = (7 4)! (.254 )( ) OR P(4) = 7 C 4 (.25 4 )( ) = 4 P(5) = (7 5)!5! (.255 )( ) OR P(5) = 7 C 5 (.25 5 )( ) = 5 P(6) = (7 6)!6! (.256 )( ) OR P(6) = 7 C 6 (.25 6 )( ) = 6 P(7) = (7 7)! (.257 )( ) OR P(7) = 7 C 7 (.25 7 )( ) = 7 Notice that all the probabilities are between 0 and 1, and they add up to very close to 1 (.99999). This confirms that you have in fact created a probability distribution. 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. 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 is not a multiple of 5. Using the formula looks like this: P(2) = (4 2)!2! (.412 )( ) = OR P(2) = 4 C 2 (.41 2 )( ) =

6 Using the calculator would go like this: 2 nd VARS A, 4,.41, 2 =. Either way, the probability that exactly two women respond yes is around. 2. We want the probability that at least two women respond yes. This means yeses. Using the formula, we could find each probability and then them together. P(2) = (4 2)!2! (.412 )( ) OR P(2) = 4 C 2 (.41 2 )( ) =. P(3) = (4 3)!3! (.413 )( ) OR P(3) = 4 C 3 (.41 3 )( ) =. P(4) = (4 4)! (.414 )( ) OR P(4) = 4 C 4 (.41 4 )( ) =. Adding these up gives us. 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, or and. P(0) = (4 0)!0! (.410 )( ) OR P(0) = 4 C 0 (.41 0 )( ) =. P(1) = (4 1)!1! (.411 )( ) OR P(1) = 4 C 1 (.41 1 )( ) =. Adding these up gives us. Subtracting this from (complement rule) gives us, the same answer we got the first time. 1 - =. So, we can find the answer by finding and adding three probabilities, or by finding and adding two probabilities and then subtracting from 1. Using the calculator, we want to use a new distribution. Because we want the probability of more than one outcome, we want a probability. We will use the binomcdf function for this. (2 nd VARS B) The calculator will 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. 2 nd 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 = We have already found the probability of fewer than two; we used it for the complement rule. The probability of fewer than two is. 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: μ = np Variance: σ 2 = npq Standard deviation: σ = npq 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. 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: μ = np = ( )( ) = σ 2 = npq = ( )( )( ) = σ = npq = ( )()() On average, the month of June has 16.8 days cloudy days. Since values that lie more than standard deviations away from the mean are unusual, we subtract from 16.8 twice and add it to 16.8 twice to find where our unusual values begin.

7 16.8 (2*2.7188) = (2*2.7188) = So, days or less of clouds would be unusual. So would days or more of clouds. III. More Discrete Probability Distributions A. The Geometric Distribution Many actions in life are until a occurs. An accounting major will take the CPA exam as many times as it takes to pass it. You may have to call someone several times before they answer, but you keep trying until they do. Situations such as these can be represented by a geometric distribution. 1. A geometric distribution is a discrete probability distribution of a random variable x that satisfies the following conditions: a. A trial is repeated until a success occurs. b. The repeated trials are of each other. c. The probability of success p is for each trial. 2. The probability that the first success will occur on trial number x is found using the formula: P(s) = p(q) x 1, where q 1 = p 3. You can also find this probability with the calculator using the geometpdf function (2 nd VARS E) Example 1 Page 222 From experience, you know that the probability that you will make a sale on any given telephone call is Find the probability that your first sale on any given day will occur on your fourth or fifth sales call. Since this is an or question, we simply the probability of a sale on the fourth call to the probability of a sale on the fifth call. Using the formula, you have P(4) =.23( ) = and P(5) =.23(. 77) 4 = Adding these, we get =. Using the calculator, you access the geometpdf function at 2 nd VARS E Input.23 for p and 4 for x. You get.105 for x = 4. Repeat for 5; 2 nd VARS E,.23, 5 and get =. These are the same answers you got doing it by hand. Try it Yourself 1 Page 223 Find the probability that your first sale will occur before your fourth sales call. Since we want success before the 4 th call, we want it to occur on the 1 st, 2 nd, or 3 rd call. We can use either the formula or the calculator to find these probabilities and add them up. P(1) =.23( ) =.23; P(2) =.23( ) =.1771; P(3) =.23( ) = =. We can also use the geometcdf function (2 nd VARS F) on the calculator. Since we want the probability of up to 3 calls, enter 3 for the x-value. 2 nd VARS F,.23 and 3 gives us the probability of, which is exactly what we got doing it by adding 3 probabilities together. B. Poisson Distribution In a binomial experiment you are interested in finding the probability of a specific number of successes in a given number of trials. Suppose instead that you want to know the probability that a specific number of occurrences takes place within a given unit of time or space. For instance, to determine the probability that an employee will take 15 sick days within a year, you can use the Poisson distribution. 1. The Poisson distribution is a discrete probability distribution of a random variable x that satisfies the following conditions: a. The experiment consists of the number of times, x, an event occurs in a given interval. The interval can be an interval of time, area, or volume. b. The probability of the event occurring is for each interval.

8 c. The number of occurrences in one interval is of the number of occurrences in other intervals. 2. The formula for finding the probability of exactly x occurrences in an interval is: P(x) = μx e μ, where e is an irrational number approximately equal to and μ is the mean x! number of occurrences per interval unit. 3. The calculator will also help us with the Poisson distribution (2 nd VARS C for pdf and D for cdf) Example 2 Page 223 The mean number of accidents per month at a certain intersection is three. What is the probability that in any given month four accidents will occur at this intersection? Solution Using x = and μ =, the probability that 4 accidents will occur in any given month at the intersection is: P(4) = 34 e 3 = ; you can enter the e by pressing 2 nd and LN On the calculator, 2 nd VARS C gets you into the distribution. Enter for the mean and for the x-value. You get, the same answer as above. Try it Yourself Page 224 What is the probability that more than four accidents will occur in any given month at the intersection? By hand, we need to find the probabilities for 1, 2, 3, and 4 accidents and add them up. We then subtract that from 1 to get the probability of more than 4 accidents occurring. P(0) =.04979; P(1) =.14936; P(2) =.22404; P(3) =.22404; P(4) = ( ) = Using the calculator, we use the Poissoncdf function to find the probability of up to 4 accidents occurring and then subtract that from 1. 2 nd VARS D, 3 and 4 =. 1 - =, the same answer we got doing it by hand.