Chapter 4 Discrete Probability Distributions 2012 Pearson Education

Chapter Outline • 4. 1 Probability Distributions • 4. 2 Binomial Distributions • 4.

Section 4. 1 Probability Distributions © 2012 Pearson Education, Inc. All rights reserved. 3

Section 4. 1 Objectives • Distinguish between discrete random variables and continuous random variables • Construct a discrete probability distribution and its graph • Determine if a distribution is a probability distribution • Find the mean, variance, and standard deviation of a discrete probability distribution • Find the expected value of a discrete probability distribution © 2012 Pearson Education, Inc. All rights reserved. 4 of 63

Random Variables Random Variable • Represents a numerical value associated with each outcome of a probability distribution. • Denoted by x • Examples § x = Number of sales calls a salesperson makes in one day. § x = Hours spent on sales calls in one day. © 2012 Pearson Education, Inc. All rights reserved. 5 of 63

Random Variables Discrete Random Variable • Has a finite or countable number of possible outcomes that can be listed. • Example § x = Number of sales calls a salesperson makes in one day. x 0 1 © 2012 Pearson Education, Inc. All rights reserved. 2 3 4 5 6 of 63

Random Variables Continuous Random Variable • Has an uncountable number of possible outcomes, represented by an interval on the number line. • Example § x = Hours spent on sales calls in one day. x 0 1 2 © 2012 Pearson Education, Inc. All rights reserved. 3 … 24 7 of 63

Example: Random Variables Decide whether the random variable x is discrete or continuous. 1. xx = The number of Fortune 500 companies that lost money in the previous year. Solution: Discrete random variable (The number of companies that lost money in the previous year can be counted. ) {0, 1, 2, 3, …, 500} © 2012 Pearson Education, Inc. All rights reserved. 8 of 63

Example: Random Variables Decide whether the random variable x is discrete or continuous. 2. xx = The volume of gasoline in a 21 -gallon tank. Solution: Continuous random variable (The amount of gasoline in the tank can be any volume between 0 gallons and 21 gallons. ) © 2012 Pearson Education, Inc. All rights reserved. 9 of 63

Example • Discrete or Continuous? • X represents the number of shares of stock in the Dow Jones Industrial Average that have price increases on a given day • X represents the volume of bottled water in a 32 ounce container • The length of time it takes to take a test • The number of home runs hit during a baseball game Hint: Discrete Random Variables are usually counted data, while Continuous Random Variables are usually measured data

Discrete Probability Distributions Discrete probability distribution • Lists each possible value the random variable can assume, together with its probability. • Must satisfy the following conditions: In Words In Symbols 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 the probabilities is 1. ΣP (x) = 1 © 2012 Pearson Education, Inc. All rights reserved. 11 of 63

Constructing a Discrete Probability Distribution Let x be a discrete random variable with possible outcomes x 1, x 2, … , xn. 1. Make a frequency distribution for the possible outcomes. 2. Find the sum of the frequencies. 3. Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies. 4. Check that each probability is between 0 and 1, inclusive, and that the sum of all probabilities is 1. © 2012 Pearson Education, Inc. All rights reserved. 12 of 63

Example: Constructing a Discrete Probability Distribution 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 Score, x Frequency, f neither trait. Construct a 1 24 probability distribution for the 2 33 random variable x. Then graph the 3 42 distribution using a histogram. 4 30 5 © 2012 Pearson Education, Inc. All rights reserved. 21 13 of 63

Solution: Constructing a Discrete Probability Distribution • Divide the frequency of each score by the total number of individuals in the study to find the probability for each value of the random variable. • Discrete probability distribution: x P(x) 1 0. 16 2 0. 22 © 2012 Pearson Education, Inc. All rights reserved. 3 0. 28 We read this as: If X occurs, the probability of X is… 4 0. 20 5 0. 14 14 of 63

Solution: Constructing a Discrete Probability Distribution x 1 2 3 4 5 P(x) 0. 16 0. 22 0. 28 0. 20 0. 14 This is a valid discrete probability distribution since 1. Each probability is between 0 and 1, inclusive, 0 ≤ P(x) ≤ 1. 2. The sum of the probabilities equals 1, ΣP(x) = 0. 16 + 0. 22 + 0. 28 + 0. 20 + 0. 14 = 1. © 2012 Pearson Education, Inc. All rights reserved. 15 of 63

Solution: Constructing a Discrete Probability Distribution • Histogram Passive-Aggressive Traits Probability, P(x) 0, 3 0, 25 0, 2 0, 15 0, 1 0, 05 0 1 2 3 4 5 Score, x Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome. © 2012 Pearson Education, Inc. All rights reserved. 16 of 63

Guidelines • What is the probability of having a 1 or a 2 on the survey? • Just add the probabilities: . 16 +. 22 =. 38 • Verifying probability distributions: § Each probability should be between 0 and 1 § The sum of the probabilities = 1

Example • Verify that this is a probability distribution • What is the probability that a randomly selected person involved in a traffic accident is in the 16 -34 age group • ∑ P(ages) = 1 -check • . 30 +. 27 =. 57 Series 1 • 57% probability That a randomly selected person is in the 16 -34 age group 0, 35 0, 3 Series 1 0, 25 0, 2 0, 15 0, 1 0, 05 0 16 -24 25 -34 35 -44 Series 1 45 -54 55 -64

Mean of a discrete probability distribution • μ = Σx. P(x) • Each value of x is multiplied by its corresponding probability and the products are added. © 2012 Pearson Education, Inc. All rights reserved. 19 of 63

Example: Finding the Mean The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean score. Solution: What does this mean? It means that for this group of 50 people, they average a little towards the passive side (remember 1 is passive, 5 is aggressive) x 1 2 3 P(x) 0. 16 0. 22 0. 28 x. P(x) 1(0. 16) = 0. 16 2(0. 22) = 0. 44 3(0. 28) = 0. 84 4 5 0. 20 0. 14 4(0. 20) = 0. 80 5(0. 14) = 0. 70 © 2012 Pearson Education, Inc. All rights reserved. μ = Σx. P(x) = 2. 94 20 of 63

Variance and Standard Deviation Variance of a discrete probability distribution • σ2 = Σ(x – μ)2 P(x) Standard deviation of a discrete probability distribution • © 2012 Pearson Education, Inc. All rights reserved. 21 of 63

Example: Finding the Variance and Standard Deviation The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the variance and standard deviation. ( μ = 2. 94) x 1 2 3 P(x) 0. 16 0. 22 0. 28 4 5 0. 20 0. 14 © 2012 Pearson Education, Inc. All rights reserved. 22 of 63

Solution: Finding the Variance and Standard Deviation Recall μ = 2. 94 x P(x) x–μ (x – μ)2 P(x) 1 0. 16 1 – 2. 94 = – 1. 94 (– 1. 94)2 ≈ 3. 764(0. 16) ≈ 0. 602 2 0. 22 2 – 2. 94 = – 0. 94 (– 0. 94)2 ≈ 0. 884(0. 22) ≈ 0. 194 3 0. 28 3 – 2. 94 = 0. 06 (0. 06)2 ≈ 0. 004(0. 28) ≈ 0. 001 4 0. 20 4 – 2. 94 = 1. 06 (1. 06)2 ≈ 1. 124(0. 20) ≈ 0. 225 5 0. 14 5 – 2. 94 = 2. 06 (2. 06)2 ≈ 4. 244(0. 14) ≈ 0. 594 Variance: σ2 = Σ(x – μ)2 P(x) = 1. 616 Standard Deviation: © 2012 Pearson Education, Inc. All rights reserved. 23 of 63

Blaise Pascal • (June 19, 1623, at Clermont, France – August 19, 1662) was a French mathematician, physicist, and religious philosopher. He was a child prodigy who was educated by his father, a civil servant. Pascal's earliest work was in the natural and applied sciences where he made important contributions to the construction of mechanical calculators, the study of fluids, and clarified the concepts of pressure and vacuum by generalizing the work of Evangelista Torricelli. Pascal also wrote in defense of the scientific method. • Pascal was a mathematician of the first order. He helped create two major new areas of research. He wrote a significant treatise on the subject of projective geometry at the age of sixteen, and later corresponded with Pierre de Fermat on probability theory, strongly influencing the development of modern economics and social science. • Pascal is also generally known as the father of Expected Value

Expected Value • A mathematical way to use probability to determine what to expect in various situations over the long run • For example used to determine premiums on insurance companies, payouts on gambling games, risk/benefit analysis in business propositions • The standard way to find expected value is to multiply each possible outcome by its probability and then add these products • We use E as the symbol for Expected Value • Expected Value is also used in Gaming Theory –such as the Prisoner’s Dilemma

Prisoner’s Dilemma • The dilemma arises when the person analyzing this assumes that both prisoners only care about minimizing their own jail terms. • Each prisoner has two options: to cooperate with his accomplice and stay quiet, or to break the code and squeal on his accomplice in return for a lighter sentence. • The outcome of each choice depends on the choice of the accomplice, but each prisoner must choose without knowing what his accomplice has chosen. Larson/Farber 5 th ed 26

Expected Value Expected value of a discrete random variable • Equal to the mean

Example • Find the expected value for the outcome of one roll of the dice • E = 1 x 1/6 + 2 x 1/6 + 3 x 1/6 + 4 x 1/6 + 5 x 1/6 + 6 x 1/6 =each number 1 through 6 is the side of a die, and each side has a probability of 1/6 of being the side which lands up • = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3. 5 • The expected value -the most likely number to show if I roll the dice- is 3. 5 • While this value cannot occur in 1 roll of the dice, it should emerge as the average over a long period of tests • This idea comes up a lot – “over a long period of time”

Example • Find the expected value for the number of girls for a family with 3 children –in other words, how many girls would you expect to see in a family with 3 children? • No Girls: Boy/Boy 1 way • 1 Girl: GBB/BGB/BBG 3 ways • 2 Girls: GGB/BGG/GBG 3 ways • 3 Girls: GGG 1 way • 8 ways total to have 3 children • E = 0 x 1/8 + 1 x 3/8 + 2 x 3/8 +3 x 1/8 = 12/8 = 3/2 = 1. 5 • Again, this makes sense. If you had 3 children, then you would expect half to be boys, and half to be girls

Example • An automobile insurance company has determined the probabilities for various claim amounts for drivers age 16 -21, as shown below: Amount of Claim Probability $0 0. 70 $2, 000 0. 15 $4, 000 0. 08 $6, 000 0. 05 $8, 000 0. 01 $10, 000 0. 01 • Calculate the expected value and describe what this means in practical terms • How much should the company charge for premiums?

Example • E = $0 x. 70 + $2, 000 x. 15 + $4, 000 x. 08 + $6, 000 x. 05 + $8, 000 x. 01 + $10, 000 x. 01 • = $0 + $300 + $320 + $300 + $80 + $100 = $1100 • The expected value of paying out is $1100. In the long run, the average cost of a claim is $1100 • Therefore, the company needs to charge premiums of at least $1100 Amount of Claim Probability $0 0. 70 $2, 000 0. 15 $4, 000 0. 08 $6, 000 0. 05 $8, 000 0. 01 $10, 000 0. 01

Expected Value for Payoffs • • But what does this mean? A game is played using 1 die. If a 1, 2, or 3 is rolled, the player wins nothing If a 4 or 5 is rolled, the player wins $3 If a 6 is rolled, the player wins $9 If there is a charge of $1 to play the game, what is the game’s expected value? Expected Outcome Gain/Loss Probability Value 1, 2, or 3 ($1) 3/6 -3/6 4, 5 $2 2/6 4/6 6 $8 1/6 8/6 Total: 9/6 This chart is how you should calculate expected values. It is easy and well laid out

Example • E = (-$1) x (3/6) + ($2) x (2/6) + ($8) x (1/6) • = (-$3 + $4 + $8)/6 = $9/6 = $1. 50 • In the long run, a player can expect to win an average of $1. 50 for each game played • This doesn’t mean a player will win $1. 50 for a single game • Therefore (for example), if a player played 1000 games, he or she could expect to win about $1, 500 • Obviously you may not win if you play just one game. But “over time” you would expect to win…

Casino Games • Unlike the dice game, casino games are set up for a player to lose in the long run • For example, Roullette: § You bet $1 on a single number § You can win $35, and you keep your $1 bet § There are 38 slots –or 1 winning slot and 37 losing • What is the expected value for betting $1 on the number 20

Example Expected Outcome Gain/Loss Probability Value Ball on 20 $35 1/38 35/38 Ball not on 20 ($1) 37/38 -2/38 • E = $35(1/38) + -$1(37/38) (you have a 1/38 probability of winning $35 and you lose $1 37/38 times) • = ($35 -$37)/38 = -$2/38 = -$. 05 • The expected value is about minus 5 cents • In the long run, a player will lose about 5 cents per game • If a player played 2000 games, they would lose about $100

Example: Finding an Expected Value At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. You buy one ticket for 2 dollars. What is the expected value of your gain? © 2012 Pearson Education, Inc. All rights reserved. 36 of 63

Solution: Finding an Expected Value • To find the gain for each prize, subtract the price of the ticket from the prize: § Your gain for the $500 prize is $500 – $2 = $498 § Your gain for the $250 prize is $250 – $2 = $248 § Your gain for the $150 prize is $150 – $2 = $148 § Your gain for the $75 prize is $75 – $2 = $73 • If you do not win a prize, your gain is $0 – $2 = –$2 © 2012 Pearson Education, Inc. All rights reserved. 37 of 63

Solution: Finding an Expected Value • Probability distribution for the possible gains (outcomes) Gain, x $498 $248 $148 $73 –$2 P(x) You can expect to lose an average of $1. 35 for each ticket you buy. © 2012 Pearson Education, Inc. All rights reserved. 38 of 63

Doing it with my chart Outcome Gain/Loss Probability Expected Value Win $500 $498 1/1500 498*(1/1500) 0. 33 Win $250 $248 1/1500 248*(1/1500) 0. 17 Win $150 $148 1/1500 148*(1/1500) 0. 10 Win $75 $73 1/1500 73*(1/1500) 0. 05 Lose $2 -$2 1496/1500 -2*(1496/1500) Sum: -1. 99 -1. 35 Note: Gain/Lost is “List 1” and Probability is “List 2” • Enter Lists into stats • Go back into stats, then calculate, then 1 -Var Stats • Hit ENTER, and BEFORE you hit enter again, type L 1, L 2 (use the 2 nd button for each) 39

Section 4. 1 Summary • Distinguished between discrete random variables and continuous random variables • Constructed a discrete probability distribution and its graph • Determined if a distribution is a probability distribution • Found the mean, variance, and standard deviation of a discrete probability distribution • Found the expected value of a discrete probability distribution © 2012 Pearson Education, Inc. All rights reserved. 40 of 63

Assignment • Page 197 -198 5 -25 odd Larson/Farber 5 th ed 41

Assignment • Page 197 -198 5 -25 odd • Page 198 199 • 27 -32 , 35 -46 § Find the mean, variance, and standard deviation of each chart § Interpret the mean to the average reader Larson/Farber 5 th ed 42

Chapter 4 Quiz 1 (30 points, 5 points each) • There are 38 slots on a roulette wheel. 18 red, 18 black, and 2 green. • A normal bet is 1 dollar, and if you win you receive 36 dollars back (total) 1. What is the expected value if you bet on #20 red? (use a chart) 2. Using question # 1, if you bet 100 times, how much money would you expect to win or lose? 3. What is the expected value if you bet just on red? (use a chart) 4. Using question # 3, if you bet 149 times, how much money would you expect to win or lose? 5. What is the expected value if you bet on green? (use a chart) 6. Using question # 5, if you bet 520 times, how much money would you expect to win or lose? This Chart Larson/Farber 5 th ed Expected Outcome Gain/Loss Probability Value Total: 43

Chapter 4 Quiz 2 (10 points, 5 points each) • • • 1. 2. • You pay 10 dollars to play a game. 5000 people play the game. There are 5 grand prize winners. They win 500 dollars (they receive 500) There are 10 1 st prize winners. They win 50 dollars (they receive 50) There are 15 2 nd prize winners. They win 25 dollars (they receive 25) There are 20 3 rd prize winners. They win 10 dollars. (they receive 10) Everyone else, obviously, is not a winner. What is the expected value of this game? If you played this game 100 times, how much money would you expect to win or lose? Create a chart to calculate this answer This Chart Larson/Farber 5 th ed Expected Outcome Gain/Loss Probability Value Total: 44

Section 4. 2 Binomial Distributions © 2012 Pearson Education, Inc. All rights reserved. 45

Section 4. 2 Objectives • Determine if a probability experiment is a binomial experiment • Find binomial probabilities using the binomial probability formula • Find binomial probabilities using technology, formulas, and a binomial probability table • Graph a binomial distribution • Find the mean, variance, and standard deviation of a binomial probability distribution © 2012 Pearson Education, Inc. All rights reserved. 46 of 63

Binomial Experiments 1. The experiment is repeated for a fixed number of trials, where each trial is independent of other trials. 2. There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success (S) or as a failure (F). 3. The probability of a success P(S) is the same for each trial. 4. The random variable x counts the number of successful trials. © 2012 Pearson Education, Inc. All rights reserved. 47 of 63

Notation for Binomial Experiments Symbol n p = P(S) q = P(F) x Description The number of times a trial is repeated The probability of success in a single trial The probability of failure in a single trial (q = 1 – p) The random variable represents a count of the number of successes in n trials: x = 0, 1, 2, 3, … , n. © 2012 Pearson Education, Inc. All rights reserved. 48 of 63

Example • • From a standard deck of cards, pick a card Note whether it is a club or not and replace Repeat 5 times, so n=5 The outcome is classified as S = clubs, and F = not clubs Probability of success: p=p(S)= ¼ Probability of failure: q=p(F) = ¾ The random variable x represents the number of clubs selected during your test. So x can equal 0, 1, 2, 3, 4, or 5 • Note that X is a discrete random variable –we can list (count) all possible outcomes for x

Example: Binomial Experiments 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. © 2012 Pearson Education, Inc. All rights reserved. 50 of 63

Solution: Binomial Experiments Binomial Experiment 1. Each surgery represents a trial. There are eight surgeries, and each one is independent of the others. 2. There are only two possible outcomes of interest for each surgery: a success (S) or a failure (F). 3. The probability of a success, P(S), is 0. 85 for each surgery. 4. The random variable x counts the number of successful surgeries. © 2012 Pearson Education, Inc. All rights reserved. 51 of 63

Solution: Binomial Experiments Binomial Experiment • n = 8 (number of trials) • p = 0. 85 (probability of success) • q = 1 – p = 1 – 0. 85 = 0. 15 (probability of failure) • x = 0, 1, 2, 3, 4, 5, 6, 7, 8 (number of successful surgeries) © 2012 Pearson Education, Inc. All rights reserved. 52 of 63

Example: Binomial Experiments 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. 2. A jar contains five red marbles, nine blue marbles, and six green marbles. You randomly select three marbles from the jar, without replacement. The random variable represents the number of red marbles. © 2012 Pearson Education, Inc. All rights reserved. 53 of 63

Solution: Binomial Experiments Not a Binomial Experiment • The probability of selecting a red marble on the first trial is 5/20. • Because the marble is not replaced, the probability of success (red) for subsequent trials is no longer 5/20. • The trials are not independent and the probability of a success is not the same for each trial. © 2012 Pearson Education, Inc. All rights reserved. 54 of 63

Binomial Probability Formula • The probability of exactly x successes in n trials is • • This is really a mixture of a combination problem (n. Cx) times a probability of success times a probability of failure we have “n choose x” times probability of success^number of successes x probability of failure^number of failures n = number of trials p = probability of success q = 1 – p probability of failure x = number of successes in n trials © 2012 Pearson Education, Inc. All rights reserved. 55 of 63

$Example: Finding Binomial Probabilities Microfracture knee surgery has a 75% chance of success on$

Example: Finding Binomial Probabilities 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. © 2012 Pearson Education, Inc. All rights reserved. 56 of 63

Solution: Finding Binomial Probabilities Method 1: Draw a tree diagram and use the Multiplication

Solution: Finding Binomial Probabilities Method 2: Binomial Probability Formula © 2012 Pearson Education, Inc.

Solution: Finding Binomial Probabilities • Using a Calculator: • Binomial Distribution: § 2 nd Vars Binomial. PDF(N, P, S) –this solves for 1 point on the curve § N = number of trials § P = probability of success § S = number of successes • Therefore we type in: Binomial. PDF(3, . 75, 2) § =. 421875 Larson/Farber 5 th ed 59

Binomial Probability Distribution • List the possible values of x with the corresponding probability of each. • Example: Binomial probability distribution for Microfracture knee surgery: n = 3, p = x 0 1 2 3 P(x) 0. 016 0. 141 0. 422 § Use the binomial probability formula to find probabilities. § Or: Use the calculator, and calculate each of these successes, then create a distribution table © 2012 Pearson Education, Inc. All rights reserved. 60 of 63

Example • A six sided die is rolled 3 times. Find the probability of rolling exactly 1 six • We could use the tree diagram, and the multiplication rule • Or we could use the calculator: • 2 nd vars (dist) binompdf then enter (in order): § N trials, (3), then probability of success (1/6), then number of successes (1) § So you would have binompdf(3, 1/6, 1) (3 rolls, 1/6 probability of getting a 6, 1 success) • =. 34722222

Example: Constructing a Binomial Distribution In a survey, U. S. adults were asked to give reasons why they liked texting on their cellular phones. Seven adults who participated in the survey are randomly selected and asked whether they like texting because it is quicker than calling. Create a binomial probability distribution for the number of adults who respond yes. Hint: What percent is success? © 2012 Pearson Education, Inc. All rights reserved. 62 of 63

Solution: Constructing a Binomial Distribution • 56% of adults like texting because it is quicker than Remember: This is for a chart with every possible calling. probability calculated • n = 7, p = 0. 56, q = 0. 44, x = 0, 1, 2, 3, 4, 5, 6, 7 P(0) = 7 C 0(0. 56)0(0. 44)7 = 1(0. 56)0(0. 44)7 ≈ 0. 0032 P(1) = 7 C 1(0. 56)1(0. 44)6 = 7(0. 56)1(0. 44)6 ≈ 0. 0284 P(2) = 7 C 2(0. 56)2(0. 44)5 = 21(0. 56)2(0. 44)5 ≈ 0. 1086 P(3) = 7 C 3(0. 56)3(0. 44)4 = 35(0. 56)3(0. 44)4 ≈ 0. 2304 P(4) = 7 C 4(0. 56)4(0. 44)3 = 35(0. 56)4(0. 44)3 ≈ 0. 2932 P(5) = 7 C 5(0. 56)5(0. 44)2 = 21(0. 56)5(0. 44)2 ≈ 0. 2239 P(6) = 7 C 6(0. 56)6(0. 44)1 = 7(0. 56)6(0. 44)1 ≈ 0. 0950 P(7) = 7 C 7(0. 56)7(0. 44)0 = 1(0. 56)7(0. 44)0 ≈ 0. 0173 © 2012 Pearson Education, Inc. All rights reserved. 63 of 63

Solution: Constructing a Binomial Distribution x 0 1 2 P(x) 0. 0032 0. 0284 0. 1086 3 4 5 6 7 0. 2304 0. 2932 0. 2239 0. 0950 0. 0173 All of the probabilities are between 0 and 1 and the sum of the probabilities is 1. © 2012 Pearson Education, Inc. All rights reserved. 64 of 63

Example: Finding Binomial Probabilities Using Technology The results of a recent survey indicate that 67% of U. S. adults consider air conditioning a necessity. If you randomly select 100 adults, what is the probability that exactly 75 adults consider air conditioning a necessity? Use a technology tool to find the probability. (Source: Opinion Research Corporation) Solution: • Binomial with n = 100, p = 0. 67, x = 75 • Solve on the calculator © 2012 Pearson Education, Inc. All rights reserved. 65 of 63

Solution: Finding Binomial Probabilities Using Technology From the displays, you can see that the probability that exactly 75 adults consider air conditioning a necessity is about 0. 02. © 2012 Pearson Education, Inc. All rights reserved. 66 of 63

Chapter 4 Quiz 2 (5 points each) • A game is played using 1 die. If a 1, 3, or 5 is rolled, the player wins nothing. If a 2 or 4 is rolled, the player gets $5 back. If a 6 is rolled, the player gets $8 back. There is a charge of $1 to play the game. Outcome 1, 3, or 5 Gain/Loss Probability Expected Value 1/6 7/6 $4 1. 2. 3. 4. Make this chart, and fill in all missing numbers What is the game’s expected value? What is the mean of this distribution chart? Over time, the person running this game will either make or lose money. How much money will they expect to make or lose total, if 500 games were played? Larson/Farber 5 th ed 67

Example: Finding Binomial Probabilities A survey indicates that 41% of women in the U. S. 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 at least two of them respond yes. Solution: • n = 4, p = 0. 41, q = 0. 59 • At least two means two or more. • Find the sum of P(2), P(3), and P(4). • You can solve each on the calculator and add © 2012 Pearson Education, Inc. All rights reserved. 68 of 63

Solution: Finding Binomial Probabilities P(2) = 4 C 2(0. 41)2(0. 59)2 = 6(0. 41)2(0. 59)2 ≈ 0. 351094 P(3) = 4 C 3(0. 41)3(0. 59)1 = 4(0. 41)3(0. 59)1 ≈ 0. 162654 P(4) = 4 C 4(0. 41)4(0. 59)0 = 1(0. 41)4(0. 59)0 ≈ 0. 028258 P(x ≥ 2) = P(2) + P(3) + P(4) ≈ 0. 351094 + 0. 162654 + 0. 028258 ≈ 0. 542 © 2012 Pearson Education, Inc. All rights reserved. 69 of 63

Solution Using the Calculator • There is a way to calculate cumulative probabilities • In this case, we choose 2 nd Vars binomcdf( § This stands for binomial cumulative (binompdf stands for binomial point –when you want a specific value at one point on the curve) • This solves “up to” problems § In other words –We want to know the probability up to and including a certain number of successes • There’s only one problem: This particular problem isn’t an “up to” problem, but a “more than and including problem” Larson/Farber 5 th ed 70

Solving With a Calculator (Continued) • Enter 2 nd Vars Binomcdf(N, P, S) § N = number of trials (4 in this example) § P = probability (. 41 in this example) § S = number of successes (in this case just 1) • Enter Binomcdf(4, . 41, 1) § We get. 457 • This is not the solution -what now? • This is the COMPLEMENT of our solution –we were searching for 2 or more successes. We calculated up to and including 1 success • To find our solution, we simply take 1 -. 457 and we get. 542, which is our solution Larson/Farber 5 th ed 71

Example: Finding Binomial Probabilities Using a Table About ten percent of workers (16 years and over) in the United States commute to their jobs by carpooling. You randomly select eight workers. What is the probability that exactly four of them carpool to work? Use a table to find the probability. (Source: American Community Survey) Solution: • Binomial with n = 8, p = 0. 10, x = 4 • Solve this on the calculator © 2012 Pearson Education, Inc. All rights reserved. 72 of 63

Solution: Finding Binomial Probabilities Using a Table • A portion of Table 2 is shown The probability that exactly four of the eight workers carpool to work is 0. 005. © 2012 Pearson Education, Inc. All rights reserved. 73 of 63

Example: Graphing a Binomial Distribution Sixty percent of households in the U. S. own a video game console. You randomly select six households and ask each if they own a video game console. Construct a probability distribution for the random variable x. Then graph the distribution. (Source: Deloitte, LLP) Solution: • n = 6, p = 0. 6, q = 0. 4 • Find the probability for each value of x • Solve each solution on the calculator and then graph by hand © 2012 Pearson Education, Inc. All rights reserved. 74 of 63

Solution: Graphing a Binomial Distribution x 0 1 2 3 4 5 6 P(x)

Mean, Variance, and Standard Deviation • Mean: μ = np • Variance: σ2 =

Example: Finding the Mean, Variance, and Standard Deviation 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. (Source: National Climatic Data Center) Solution: n = 30, p = 0. 56, q = 0. 44 Mean: μ = np = 30∙ 0. 56 = 16. 8 Variance: σ2 = npq = 30∙ 0. 56∙ 0. 44 ≈ 7. 4 Standard Deviation: © 2012 Pearson Education, Inc. All rights reserved. 77 of 63

Solution: Finding the Mean, Variance, and Standard Deviation μ = 16. 8 σ2 ≈ 7. 4 σ ≈ 2. 7 • On average, there are 16. 8 cloudy days during the month of June. • The standard deviation is about 2. 7 days. • Values that are more than two standard deviations from the mean are considered unusual. § 16. 8 – 2(2. 7) =11. 4, a June with 11 cloudy days or fewer would be unusual. § 16. 8 + 2(2. 7) = 22. 2, a June with 23 cloudy days or more would also be unusual. © 2012 Pearson Education, Inc. All rights reserved. 78 of 63

Section 4. 2 Summary • Determined if a probability experiment is a binomial experiment • Found binomial probabilities using the binomial probability formula • Found binomial probabilities using technology and a binomial table • Graphed a binomial distribution • Found the mean, variance, and standard deviation of a binomial probability distribution © 2012 Pearson Education, Inc. All rights reserved. 79 of 63

Assignment • Page 211 -214 3 -6, 9 -26, 27, 29 Larson/Farber 5 th ed 80

Chapter 4 Quiz 3 • Do problem 30 on page 214 • Do parts a, b and c (skip part d) • 30 points –ten points for each Larson/Farber 5 th ed 81

Section 4. 3 More Discrete Probability Distributions © 2012 Pearson Education, Inc. All rights

Section 4. 3 Objectives • Find probabilities using the geometric distribution • Find probabilities

Geometric Distribution • A discrete probability distribution. • Satisfies the following conditions § A trial is repeated until a success occurs. § The repeated trials are independent of each other. § The probability of success p is constant for each trial. § x represents the number of the trial in which the first success occurs. • The probability that the first success will occur on trial x is P(x) = p(q)x – 1, where q = 1 – p. © 2012 Pearson Education, Inc. All rights reserved. 84 of 63

Example: Geometric Distribution Basketball player Le. Bron James makes a free throw shot about 74% of the time. Find the probability that the first free throw shot Le. Bron makes occurs on the third or fourth attempt. Solution: • P(shot made on third or fourth attempt) = P(3) + P(4) • Geometric with p = 0. 74, q = 0. 26, x = 3 © 2012 Pearson Education, Inc. All rights reserved. 85 of 63

Solution: Geometric Distribution • P(3) = 0. 74(0. 26)3– 1 = 0. 050024 • P(4) = 0. 74(0. 26)4– 1 ≈ 0. 013006 P (shot made on third or fourth attempt) = P(3) + P(4) ≈ 0. 050024 + 0. 013006 ≈ 0. 063 © 2012 Pearson Education, Inc. All rights reserved. 86 of 63

Geometric Probability on the Calculator • 2 nd Vars Geometpdf (p, x) –this solves for one specific probability, or point on the curve • P = Probability of Success • X = number of trials until success • If you want to solve a cumulative probability: • 2 nd Vars Geometcdf (p, x) –this solves for cumulative probability – “up to” • P = Probability of Success • X = number of trials you want to solve “up to” Larson/Farber 5 th ed 87

Geometric Probability on the Calculator Basketball player Le. Bron James makes a free throw shot about 74% of the time. Find the probability that the first free throw shot Le. Bron makes occurs on the third or fourth attempt. On the calculator: • 2 nd Vars Geometpdf (p, x) • 2 nd Vars Geometpdf (. 74, 3) =. 0500 • 2 nd Vars Geometpdf (. 74, 4) =. 013 • Add them and get. 063 Larson/Farber 5 th ed 88

Example • From previous experience you know that the probability of making a sale on any given phone call is. 23 • Find the probability that the sale will occur on the fourth or fifth call • Find the probability for the 4 th call, then find the probability for the 5 th call, then add them • P(x) = P*(q) x-1 • p(4) =. 23*. 77^3 =. 105 • p(5) =. 23*. 77^4 =. 08 • So p(sale on 4 or 5 call) =. 105 +. 08 =. 186

Example • Find the probability that you will make a sale before the 4 th call • You will need to find the p(1) + p(2) + p(3) Then sum them • 2 nd VARS geometpdf(p, x) –do this 3 times with X as 1, then 2, then 3 (remember the probability of success is. 23) • You should get. 543467 • You can also do a cumulative probability: Geometcdf(. 23, 3) and you will get the same answer • Notice: This type of distribution is discrete (x number of trials is easily counted) but we could go on forever –where x is any large number. Therefore, the sum of the probabilities will approach 1, but won’t ever be exactly 1 • Therefore, you can NOT calculate the complement to this

Poisson Distribution • So far, in binomial experiments we have focused on the probability of success for any given number of trials • What if we want to know the probability that a specific event occurs a specific number of times within a given amount of time or space? • (In other words, we may not be focused on success this time, just that something occurred) • For example, what is the probability that an employee will take 15 days sick leave within a year? • To calculate these types of problems, we use the Poisson Distribution

Poisson Distribution • Siméon-Denis Poisson (21 June 1781 – 25 April 1840), was a French mathematician, geometer, and physicist. • He was first published at the age of 18, and was considered a giant in his field until he passed away • One of his most famous theories is called “The Poisson Distribution”

Poisson Distribution Poisson distribution • A discrete probability distribution. • Satisfies the following conditions § The experiment consists of counting the number of times x an event occurs in a given interval. The interval can be an interval of time, area, or volume. § The probability of the event occurring is the same for each interval. § The number of occurrences in one interval is independent of the number of occurrences in other intervals. © 2012 Pearson Education, Inc. All rights reserved. 93 of 63

Poisson Distribution Poisson distribution • Conditions continued: § The probability of exactly x occurrences in an interval is where e is an irrational number ≈ 2. 71828 and μ is the mean number of occurrences per interval unit. © 2012 Pearson Education, Inc. All rights reserved. 94 of 63

The Constant e (FYI) • The mathematical constant e is the unique real number such that the value of the derivative (slope of the tangent line) of the function f(x) = ex at the point x = 0 is exactly 1. • The number e is of eminent importance in mathematics, alongside 0, 1, π and i. Besides being abstract objects, all five of these numbers play important and recurring roles across mathematics, and, coincidentally are the five constants appearing in one formulation of Euler's identity. • The number e is irrational; it is not a ratio of integers. Furthermore, it is transcendental; it is not a root of any non-zero polynomial with rational coefficients. The numerical value of e truncated to 20 decimal places is • 2. 7182845904523536….

Example: Poisson Distribution The mean number of accidents per month at a certain intersection is 3. What is the probability that in any given month four accidents will occur at this intersection? Solution: • Poisson with x = 4, μ = 3 © 2012 Pearson Education, Inc. All rights reserved. 96 of 63

On a Calculator • The mean number of accidents per month at a certain intersection is 3. What is the probability that in any given month four accidents will occur at this intersection? • On a calculator: • 2 nd VARS • Poissonpdf(μ, x) • Where μ is the mean and x is the number of occurrences • Thus: poissonpdf(3, 4) =. 168 Larson/Farber 5 th ed 97

Cumulative Poisson • What is the probability that more than 4 accidents will occur in any given month at the intersection? • 2 ways: • Use Poissonpdf(μ, x) and solve for each probability • Thus we solve P(0), P(1), P(2), P(3) and P(4) • Add all these probabilities =. 81526 • Subtract this answer from 1. 1847 • This new answer will be all the probabilities more than P(4) • 2 nd way: • Use Poissoncdf(μ, x) and solve for up to and including 4 accidents • Poissoncdf(3, 4) =. 81526 • Subract from 1 =. 1847 Larson/Farber 5 th ed 98

Assignment • Page 222 1 -24 (skip 9 and 10) Larson/Farber 5 th ed 99

Section 4. 3 Summary • Found probabilities using the geometric distribution • Found probabilities

Chapter 4 Quiz 4 (on your own) • Do problem 25 on page 224 1. Part A) Solve for a binomial point distribution. § On your paper, write what you used for: n, p, s (5 pts) § Show the probability of success (5 pts) 2. Part B) Solve for a Poisson point distribution. You must calculate the mean to solve this problem. § On your paper, write your equation and what you calculated for the mean (5 pts) § Show the probability of success (5 pts) Larson/Farber 5 th ed 101