Elementary Statistics Eighth Edition Chapter 4 Discrete Probability Distributions Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 1 Slide - 2 Chapter Outline 4.1 Probability Distributions 4.2 Binomial Distributions Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 3 Section 4.1 Probability Distributions Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 4 Section 4.1 Objectives 1. How to distinguish between discrete random variables and continuous random variables 2. How to construct a discrete probability distribution and its graph and how to determine if a distribution is a probability distribution 3. How to find the mean, variance, and standard deviation of a discrete probability distribution 4. How to find the expected value of a discrete probability distribution Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 5 Random Variables (1 of 3) 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. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 6 Random Variables (2 of 3) 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. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 7 Random Variables (3 of 3) 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. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 8 Example: Discrete and Continuous Variables (1 of 2) Determine whether each random variable x is discrete or continuous. Explain your reasoning. 1. Let x represent 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.) Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 9 Example: Discrete and Continuous Variables (2 of 2) Determine whether each random variable x is discrete or continuous. Explain your reasoning. 2. Let x represent 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.) Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 10 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) 2. The sum of all the probabilities is 1. P( x) = 1 0 is less than or equal to P of x is less than or equal to 1 the sum of P of x = 1 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 11 Constructing a Discrete Probability Distribution Let x be a discrete random variable with possible outcomes x1 x2 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 the probabilities is 1. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 12 Example: Constructing and Graphing a Discrete Probability Distribution An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Each individual was given a whole number score from 1 to 5, where 1 is extremely passive and 5 is extremely aggressive. A score of 3 indicated neither trait. The results are shown. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram. Score, x Frequency, f 1 24 2 33 3 42 4 30 5 21 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 13 Solution: Constructing and Graphing a Discrete Probability Distribution (1 of 3) • 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. 24 P(1) = = 0.16 150 33 P(2) = = 0.22 150 30 P(4) = = 0.20 150 P(5) = 42 P(3) = = 0.28 150 21 = 0.14 150 • Discrete probability distribution: x P( x) P of x 1 0.16 2 0.22 3 0.28 4 0.20 5 0.14 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 14 Solution: Constructing and Graphing a Discrete Probability Distribution (2 of 3) x P of x P( x) 1 0.16 2 0.22 3 0.28 4 0.20 5 0.14 This is a valid discrete probability distribution since 1. Each probability is between 0 and 1, inclusive, 0 P ( x ) 2. The sum of the probabilities equals 1, P ( x ) = 016 + 022 + 028 + 020 + 014 = Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 15 Solution: Constructing and Graphing a Discrete Probability Distribution (3 of 3) x P( x) 1 2 3 4 5 0.16 0.22 0.28 0.20 0.14 P of x Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome. Also, the probability of an event corresponds to the sum of the areas of the outcomes included in the event. Passive-Aggressive Traits You can see that the distribution is approximately symmetric. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 16 Example: Verifying a Probability Distribution Verify that the distribution for the three-day forecast and the number of days of rain is a probability distribution. Days of Rain, x Probability, P ( x ) P of x 0 0.216 1 0.432 2 0.288 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved 3 0.064 Slide - 17 Solution: Verifying a Probability Distribution Solution If the distribution is a probability distribution, then (1) each probability is between 0 and 1, inclusive, and (2) the sum of all the probabilities equals 1. 1. Each probability is between 0 and 1. 2. P ( x ) = 0216 + 0432 + 0288 + 0064 = Days of Rain, x Probability, P ( x ) P of x 0 1 2 3 0.216 0.432 0.288 0.064 Because both conditions are met, the distribution is a probability distribution. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 18 Example: Identifying Probability Distributions (1 of 2) Determine whether each distribution is a probability distribution. Explain your reasoning. 1. x 5 6 7 8 0.28 0.21 0.43 0.15 P of x P( x) Solution • Each probability is between 0 and 1, but the sum of all the probabilities is 1.07, which is greater than 1. • The sum of all the probabilities in a probability distribution always equals 1. So, this distribution is not a probability distribution. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 19 Example: Identifying Probability Distributions (2 of 2) Determine whether each distribution is a probability distribution. Explain your reasoning. 2. x P( x) P of x 1 2 3 4 1 2 1 4 5 4 −1 half 1 fourth 5 fourths negative 1 Solution • The sum of all the probabilities is equal to 1, but P ( 3) and P ( 4 ) are not between 0 and 1. • Probabilities can never be negative or greater than 1. So, this distribution is not a probability distribution. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 20 Mean Mean of a discrete probability distribution • = xP ( x ) • Each value of x is multiplied by its corresponding probability and the products are added. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 21 Example: Finding the Mean The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean score. Solution: x 1 2 3 4 5 P( x) xP ( x ) 0.16 0.22 0.28 0.20 0.14 1(0.16) = 0.16 P of x x P of x 1 left parenthesis 0.16 right parenthesis, = 0.16 2(0.22) = 0.44 2 left parenthesis 0.22 right parenthesis, = 0.44 3(0.28) = 0.84 3 left parenthesis 0.28 right parenthesis, = 0.84 4(0.20) = 0.80 4 left parenthesis 0.20 right parenthesis, = 0.80 5(0.14) = 0.70 5 left parenthesis 0.14 right parenthesis, = 0.70 = xP ( x ) = 294 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 22 Solution: Finding the Mean The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean score. Solution: = xP ( x ) = 294 Recall that a score of 3 represents an individual who exhibits neither passive nor aggressive traits and the mean is slightly less than 3. So, the mean personality trait is neither extremely passive nor extremely aggressive, but is slightly closer to passive. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 23 Variance and Standard Deviation Variance of a discrete probability distribution • 2 = ( x − ) P( x) 2 Standard deviation of a discrete probability distribution • = = ( x − ) P( x) 2 2 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 24 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. Score, x Probability, P ( x ) P of x 1 2 0.16 0.22 3 0.28 4 5 0.20 0.14 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 25 Solution: Finding the Variance and Standard Deviation = 2.94 Recall ( x – )2 x– x P( x) 1 0.16 1 – 2.94 = –1.94 ( –1.94 ) 2 2 0.22 2 – 2.94 = –0.94 ( –0.94 ) 2 3 0.28 3 – 2.94 = 0.06 ( 0.06 ) 4 0.20 4 – 2.94 = 1.06 (1.06 ) 5 0.14 5 – 2.94 = 2.06 ( 2.06 ) P of x x minus mu 1 minus 2.94 = negative 1.94 2 minus 2.94 = negative 0.94 3 minus 2.94 = 0.06 4 negative 2.94 = 1.06 5 minus 2.94 = 2.06 ( x – )2 P ( x ) left parenthesis x minus mu right parenthesis squared left parenthesis x minus mu right parenthesis squared P of x = 3.7636 3.7636 ( 0.16 ) = 0.60218 = 0.8836 0.8836 ( 0.22 ) = 0.19439 left parenthesis negative 1.94 right parenthesis squared = 3.7636 3.7636 left parenthesis 0.16 right parenthesis, = 0.60218 left parenthesis negative 0.94 right parenthesis squared = 0.8836 2 = 0.0036 0.0036 ( 0.28 ) = 0.00101 = 1.1236 1.1236 ( 0.20 ) = 0.22472 = 4.2436 4.2436 ( 0.14 ) = 0.59410 left parenthesis 0.06 right parenthesis squared = 0.0036 2 0.0036 left parenthesis 0.28 right parenthesis = 0.00101 left parenthesis 1.06 right parenthesis squared = 1.1236 2 1.1236 left parenthesis 0.20 right parenthesis, = 0.22472 left parenthesis 2.06 right parenthesis squared = 4.2436 Variance: Standard Deviation: 0.8836 left parenthesis 0.22 right parenthesis = 0.19439 4.2436 left parenthesis 0.14 right parenthesis, = 0.59410 = ( x − ) P ( x ) = 16164 2 = 2 = 16164 1.27 13 Most of the data values differ from the mean by no more than 1.3. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 26 Expected Value Expected value of a discrete random variable • Equal to the mean of the random variable. • E ( x ) = = xP ( x ) Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 27 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. Find the expected value and interpret its meaning. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 28 Solution: Finding an Expected Value (1 of 2) • 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 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 29 Solution: Finding an Expected Value (2 of 2) • Probability distribution for the possible gains (outcomes) Gain, x $498 $248 $148 P( x) 1 1500 1 1500 1 1500 P of x start fraction 1 over 1500 end fraction start fraction 1 over 1500 end fraction start fraction 1 over 1500 end fraction $73 1 1500 start fraction 1 over 1500 end fraction −$2 negative $2 1496 1500 start fraction 1496 over 1500 end fraction E ( x ) = xP ( x ) 1 1 1 1 1496 + $248 + $148 + $73 + ( −$2 ) 1500 1500 1500 1500 1500 = −$1.35 You can expect to lose an average of $1.35 for each ticket you buy. = $498 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 30 Section 4.2 Binomial Distributions Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 31 Section 4.2 Objectives 1. How to determine whether a probability experiment is a binomial experiment 2. How to find binomial probabilities using the binomial probability formula 3. How to find binomial probabilities using technology, formulas, and a binomial probability table 4. How to construct and graph a binomial distribution 5. How to find the mean, variance, and standard deviation of a binomial probability distribution Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 32 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. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 33 Notation for Binomial Experiments Symbol Description n The number of times a trial is repeated p The probability of success in a single trial q The probability of failure in a single trial (q = 1 − p) left parenthesis q = 1 minus p right parenthesis x The random variable represents a count of the number of successes in n trials: x = 0, 1, 2, 3, , n. x = 0, 1, 2, 3, ellipsis, n. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 34 Example: Identifying and Understanding Binomial Experiments (1 of 2) Decide whether each 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. If it is not, explain why. 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. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 35 Solution: Identifying and Understanding Binomial Experiments (1 of 3) 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. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 36 Solution: Identifying and Understanding Binomial Experiments (2 of 3) 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) Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 37 Example: Identifying and Understanding Binomial Experiments (2 of 2) Decide whether each 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. If it is not, explain why. 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. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 38 Solution: Identifying and Understanding Binomial Experiments (3 of 3) 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. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 39 Binomial Probability Formula Binomial Probability Formula • The probability of exactly x successes in n trials is P( x ) = n C x p q x n− x n = p x q n− x ( n − x ) x • n = number of trials • p = probability of success • q = 1 − p probability of failure • x = number of successes in n trials • Note: number of failures is n−x Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 40 Example: Finding a Binomial Probability Rotator cuff surgery has a 90% chance of success. The surgery is performed on three patients. Find the probability of the surgery being successful on exactly two patients. (Source: The Orthopedic Center of St. Louis) Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 41 Solution: Finding a Binomial Probability (1 of 2) Method 1: Draw a tree diagram and use the Multiplication Rule. 81 3 1000 = 0.243 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 42 Solution: Finding a Binomial Probability (2 of 2) Method 2: Use the binomial probability formula. 9 1 n = 3, p = , q = , and x = 2. 10 10 3 9 P ( 2) = 3 − 2 ( ) 10 2 1 1 10 81 1 = 3 100 10 81 = 3 1000 = 0.243 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 43 Example: Constructing a Binomial Distribution In a survey, U.S. adults were asked how often they go online. The results are shown in the figure. Six adults who participated in the survey are randomly selected and asked whether they go online several times a day. Construct a binomial probability distribution for the number who respond that they go online several times a day. (Source: Pew Research) Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 44 Solution: Constructing a Binomial Distribution (1 of 2) p = 0.48 and q = 0.52 n = 6, possible values for x are 0, 1, 2, 3, 4, 5 and 6 P ( 0 ) = 6C0 (0.48)0 (0.52)6 = 1(0.48)0 (0.52)6 0.020 P (1) = 6C1 (0.48)1 (0.52)5 = 6(0.48)1 (0.52)5 0.109 P ( 2 ) = 6C2 (0.48) 2 (0.52) 4 = 15(0.48) 2 (0.52) 4 0.253 P ( 3) = 6C3 (0.48)3 (0.52)3 = 20(0.48)3 (0.52)3 0.311 P ( 4 ) = 6C4 (0.48) 4 (0.52) 2 = 15(0.48) 4 (0.52) 2 0.215 P ( 5 ) = 6C5 (0.48)5 (0.52)1 = 6(0.48)5 (0.52)1 0.079 P ( 6 ) = 6C6 (0.48)6 (0.52)0 = 1(0.48)6 (0.52)0 0.012 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 45 Solution: Constructing a Binomial Distribution (2 of 2) Notice in the table that all the probabilities are between 0 and 1 and that the sum of the probabilities is about 1. x P( x) 0 0.020 1 0.109 2 0.253 3 0.311 4 0.215 5 0.079 6 0.012 Blank P of x P( x) = 0.999 sum of P of x = 0.999 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 46 Example: Finding a Binomial Probabilities Using Technology A survey found that 75% of U.S. adults are experiencing digital device fatigue. (This is the state of mental exhaustion caused by concurrent and excessive use of digital devices such as smartphones and tablet computers.) You randomly select 100 adults. What is the probability that exactly 66 are experiencing digital device fatigue? Use technology to find the probability. (Source: The Harris Poll) Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 47 Solution: Finding a Binomial Probabilities Using Technology (1 of 2) Solution Minitab, Excel, StatCrunch, and the T I-84 Plus each have features that allow you to find binomial probabilities. Try using these technologies. You should obtain results similar to these displays. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 48 Solution: Finding a Binomial Probabilities Using Technology (2 of 2) Solution From these displays, you can see that the probability that exactly 66 adults are experiencing digital device fatigue is about 0.011. Because 0.011 is less than 0.05, this can be considered an unusual event. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 49 Example: Finding Binomial Probabilities Using Formulas A survey found that 22% of U.S. adults say that the economy, unemployment, and jobs are the most important problems facing the U.S. today. You randomly select four adults and ask them whether the economy, unemployment, and jobs are the most important problems facing the U.S. today. 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. (Source: Ipsos) Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 50 Solution: Finding Binomial Probabilities Using Formulas (1 of 4) Solution 1. Using n = 4, p = 0.22, q = 0.78, and x = 2, the probability that exactly two adults will respond yes is P ( 2 ) = 4 C2 ( 0.22 ) ( 0.78 ) = 6 ( 0.22 ) ( 0.78 ) 0.177. 2 2 2 2 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 51 Solution: Finding Binomial Probabilities Using Formulas (2 of 4) Solution 2. To find the probability that at least two adults will respond yes, find the sum of P(2), P(3), and P(4). Begin by using the binomial probability formula to write an expression for each probability. P ( 2 ) = 4 C2 ( 0.22 ) ( 0.78 ) = 6 ( 0.22 ) ( 0.78 ) 2 2 2 2 P ( 3) = 4 C3 ( 0.22 ) ( 0.78 ) = ( 0.22 ) ( 0.78 ) 3 1 3 1 P ( 4 ) = 4 C4 ( 0.22 ) ( 0.78 ) = ( 0.22 ) ( 0.78 ) 4 0 4 0 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 52 Solution: Finding Binomial Probabilities Using Formulas (3 of 4) Solution 2. So, the probability that at least two will respond yes is P ( x ) = P ( 2 ) + P ( 3) + P ( 4 ) = 6 ( 0.22 ) ( 0.78 ) 2 +4 ( 0.22 ) ( 0.78 ) + 1( 0.22 ) ( 0.78 ) 3 1 4 2 0 0.212. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 53 Solution: Finding Binomial Probabilities Using Formulas (4 of 4) Solution 3. To find the probability that fewer than two adults will respond yes, find the sum of P(0) and P(1). P ( 0 ) = 4 C0 ( 0.22 ) ( 0.78 ) = ( 0.22 ) ( 0.78 ) 0 4 0 P (1) = 4 C1 ( 0.22 ) ( 0.78 ) = ( 0.22 ) ( 0.78 ) 1 3 1 4 3 So, the probability that fewer than two will respond yes is P ( x 2 ) = P ( 0 ) + P (1) = 1( 0.22 ) ( 0.78 ) + 4 ( 0.22 ) ( 0.78 ) 0.788. 0 4 1 3 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 54 Example: Finding a Binomial Probability Using a Table About 5% of workers (ages 16 years and older) in the United States commute to their jobs by using public transportation (excluding taxicabs). You randomly select eight workers. What is the probability that exactly three of them use public transportation to get to work? Use a table to find the probability. (Source: American Community Survey) Solution: • Binomial with n = 8, p = 0.05, x = 3 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 55 Solution: Finding Binomial Probabilities Using a Table (1 of 2) • A portion of Table 2 is shown According to the table, the probability is 0.005. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 56 Solution: Finding Binomial Probabilities Using a Table (2 of 2) • You can check the result using technology. So, the probability that exactly four of the eight workers carpool to work is 0.005. Because 0.005 is less than 0.05, this can be considered an unusual event. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 57 Example: Graphing a Binomial Distribution Sixty-four percent of people who have survived cancer are ages 65 years or older. You randomly select six people who have survived cancer and ask them whether they are 65 years of age or older. Construct a probability distribution for the random variable x. Then graph the distribution. (Source: American Cancer Society) Solution: • n = 6, p = 0.64, q = 0.36 • Find the probability for each value of x Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 58 Solution: Graphing a Binomial Distribution (1 of 2) x P( x) P of x 0 1 2 3 4 5 6 0.002 0.023 0.103 0.245 0.326 0.232 0.069 Notice in the table that all the probabilities are between 0 and 1 and that the sum of the probabilities is 1. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 59 Solution: Graphing a Binomial Distribution (2 of 2) Histogram: People Who Have Survived Cancer, 65 Years of Age or Older From the histogram, you can see that it would be unusual for none or only one of the survivors to be age 65 years or older because both probabilities are less than 0.05. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 60 Mean, Variance, and Standard Deviation • Mean: = np • Variance: = npq 2 • Standard Deviation: = npq Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 61 Example: 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 Oceanic and Atmospheric Administration) Solution: n = 30, p = 0.56, q = 0.44 Mean: = np = 30 0.56 = 16.8 2 Variance: = npq = 30 0.56 0.44 7.4 Standard Deviation: = npq = 30 0.56 0.44 2.7 Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 62 Solution: Mean, Variance, and Standard Deviation = 16.8 2 7.4 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 less would be unusual. – 16.8 + 2 ( 2.7 ) = 22.2; A June with 23 cloudy days or more would also be unusual. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 63 Copyright This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Copyright © 2023, 2019, 2015 Pearson Education, Inc. All Rights Reserved Slide - 64