Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
4 Discrete Probability Distributions x = number of on time arrivals x = number of correct answers Elementary Statistics Larson Farber x = number of employees reaching sales quota Larson/Farber Ch. 4 x = number of points scored in a game Section 4.1 Probability Distributions Larson/Farber Ch. 4 Random Variables A random variable x is the numerical outcome of a probability experiment. x = The number of people in a car x = The gallons of gas bought in a week x = The time it takes to drive from home to school x = The number of trips to school you make per week Larson/Farber Ch. 4 Types of Random Variables A random variable is discrete if the number of possible outcomes is finite or countable. Discrete random variables are determined by a count. A random variable is continuous if it can take on any value within an interval. The possible outcomes cannot be listed. Continuous random variables are determined by a measure. Larson/Farber Ch. 4 Types of Random Variables Identify each random variable as discrete or continuous. 1. x = The number of people in a car 2. x = The gallons of gas bought in a week 3. x = The time it takes to drive from home to school 4. x = The number of trips to school you make per week Larson/Farber Ch. 4 Discrete Probability Distributions A discrete probability distribution lists each possible value of the # of vehicles random variable with x its probability. A survey asks a sample of families how many vehicles each owns. 0 1 2 3 Probability P(x) 0.004 0.435 0.355 0.206 Properties of a probability distribution • Each probability must be between 0 and 1, inclusive. • The sum of all probabilities is 1. Larson/Farber Ch. 4 Probability Histogram Number of Vehicles 0.435 .40 0.355 P(x) .30 0.206 .20 .10 0.004 0 00 11 22 33 x • The height of each bar corresponds to the probability of x. • When the width of the bar is 1, the area of each bar corresponds to the probability the value of x will occur. Larson/Farber Ch. 4 Mean, Variance and Standard Deviation The mean of a discrete probability distribution: The variance of a discrete probability distribution: The standard deviation of a discrete probability distribution: Larson/Farber Ch. 4 Mean (Expected Value) Calculate the mean: Multiply each value by its probability. Add the products x 0 1 2 3 P(x) 0.004 0.435 0.355 0.206 xP(x) The expected value (the mean) is ____________ vehicles. Larson/Farber Ch. 4 Mean (Expected Value) Calculate the mean: Multiply each value by its probability. Add the products x 0 1 2 3 P(x) 0.004 0.435 0.355 0.206 xP(x) 0 0.435 0.71 0.618 1.763 The expected value (the mean) is 1.763 vehicles. Larson/Farber Ch. 4 Variance and Standard Deviation The mean is 1.763 vehicles. x 0 1 2 3 P(x) 0.004 0.435 0.355 0.206 x- μ (x - μ ) P(x)(x P(x) - ) variance The standard deviation is __________ vehicles. Larson/Farber Ch. 4 Variance and Standard Deviation The mean is 1.763 vehicles. x 0 1 2 3 P(x) 0.004 0.435 0.355 0.206 x- μ -1.763 -0.763 0.237 1.237 (x -μ ) 3.108 0.582 0.056 1.530 P(x)(xP(x) - ) 0.012 0.253 0.020 0.315 0.601 variance The standard deviation is 0.775 vehicles. Larson/Farber Ch. 4 Section 4.2 Binomial Distributions Larson/Farber Ch. 4 Binomial Experiments Characteristics of a Binomial Experiment • There are a fixed number of independent trials (n) • Each trial has 2 outcomes, S = Success or F = Failure. • The probability of success is the same for each trial • The random variable x counts the number of successful trials Larson/Farber Ch. 4 Binomial Experiments Notation for Binomial Experiments • n the number of times the trial is repeated • x the random variable represents a count of the number of successes in n trials: x = 0,1,2,3,…..n • p = P(S) the probability of success in a single trial • q = P(F) the probability of failure in a single trial p+q=1 Larson/Farber Ch. 4 Binomial Experiments A multiple choice test has 8 questions each of which has 3 choices, one of which is correct. You want to know the probability that you guess exactly 5 questions correctly. Find n, p, q, and x. A doctor tells you that 80% of the time a certain type of surgery is successful. If this surgery is performed 7 times, find the probability exactly 6 surgeries will be successful. Find n, p, q, and x. Larson/Farber Ch. 4 Binomial Experiments A multiple choice test has 8 questions each of which has 3 choices, one of which is correct. You want to know the probability that you guess exactly 5 questions correctly. Find n, p, q, and x. n=8 p = 1/3 q = 2/3 x=5 A doctor tells you that 80% of the time a certain type of surgery is successful. If this surgery is performed 7 times, find the probability exactly 6 surgeries will be successful. Find n, p, q, and x. n=7 p = 0.80 Larson/Farber Ch. 4 q = 0.20 x=6 Guess the Answers 1. What is the 11th digit after the decimal point for the irrational number e? (a) 2 (b) 7 (c) 4 (d) 5 2. What was the Dow Jones Average on February 27, 1993? (a) 3265 (b) 3174 (c) 3285 (d) 3327 3. How many students from Sri Lanka studied at U.S. universities from 1990-91? (a) 2320 (b) 2350 (c) 2360 (d) 2240 4. How many kidney transplants were performed in 1991? (a) 2946 (b) 8972 (c) 9943 (d) 7341 5. How many words are in the American Heritage Dictionary? (a) 60,000 (b) 80,000 (c) 75,000 (d) 83,000 Larson/Farber Ch. 4 Quiz Results The correct answers to the quiz are: 1. d 2. a 3. b 4. c 5. b Count the number of correct answers. Let the number of correct answers = x. Why is this a binomial experiment? What are the values of n, p and q? What are the possible values for x? Larson/Farber Ch. 4 Binomial Probabilities Find the probability of getting exactly 3 correct on the quiz you took earlier. Write the first 3 correct and the last 2 wrong as SSSFF P(SSSFF) = (.25)(.25)(.25)(.75)(.75) = (.25)3(.75)2 = 0.00879 Since order does not matter, you could get any combination of three correct out of five questions. List these combinations. SSSFF FFSSS SSFSF FSFSS SSFFS FSSFS SFFSS SFSSF SFSFS FSSSF Each of these 10 ways has a probability of 0.00879. P(x = 3) = 10(0.25)3(0.75)2 = 10(0.00879) = 0.0879 Larson/Farber Ch. 4 Combination of n values, choosing x = 10 ways Find the probability of getting exactly 3 correct on the quiz. P(x = 3) = 10(0.25)3(0.75)2= 10(0.00879)= 0.0879 In a binomial experiment, the probability of exactly x successes in n trials is Larson/Farber Ch. 4 Binomial Probabilities In a binomial experiment, the probability of exactly x successes in n trials is Use the formula to calculate the probability of getting none correct, exactly one, two, three, four correct or all 5 correct on the quiz. Larson/Farber Ch. 4 Binomial Probabilities In a binomial experiment, the probability of exactly x successes in n trials is Use the formula to calculate the probability of getting none correct, exactly one, two, three, four correct or all 5 correct on the quiz. P(3) = 0.088 Larson/Farber Ch. 4 P(4) = 0.015 P(5) = 0.001 Binomial Distribution x 0 1 2 3 4 5 Binomial Histogram .396 .40 .30 .294 .237 P(x) 0.237 0.396 0.264 0.088 0.015 0.001 .20 .088 .10 .015 .001 4 5 0 0 Larson/Farber Ch. 4 1 2 3 x Probabilities 1. What is the probability of answering either 2 or 4 questions correctly? x 0 1 2 3 4 5 P(x) 0.237 0.396 0.264 0.088 0.015 0.001 2. What is the probability of getting at least 3 questions correct? 3. What is the probability of getting at least one question correct? Larson/Farber Ch. 4 Probabilities 1. What is the probability of answering either 2 or 4 questions correctly? P( x = 2 or x = 4) = 0.264 + 0.015 = 0. 279 x 0 1 2 3 4 5 P(x) 0.237 0.396 0.264 0.088 0.015 0.001 2. What is the probability of answering at least 3 questions correctly? P(x > 3) = P( x = 3 or x = 4 or x = 5) = 0.088 + 0.015 + 0.001 = 0.104 3. What is the probability of answering at least one question correctly? P(x 1) = 1 - P(x = 0) = 1 - 0.237 = 0.763 Larson/Farber Ch. 4 Parameters for a Binomial Experiment Mean: Variance: Standard deviation: Use the binomial formulas to find the mean, variance and standard deviation for the distribution of correct answers on the quiz. Larson/Farber Ch. 4 Section 4.3 More Discrete Probability Distributions Larson/Farber Ch. 4 The Geometric Distribution A geometric distribution is a discrete probability distribution of the random variable x that satisfies the following conditions: 1. A trial is repeated until a success occurs. 2. The repeated trials are independent of each other. 3. The probability of success p is the same for each trial. The probability that the first success will occur on trial number x is P(x) = (q)x – 1p where q = 1 – p Larson/Farber Ch. 4 The Geometric Distribution A marketing study has found that the probability that a person who enters a particular store will make a purchase is 0.30. •The probability the first purchase will be made by the first person who enters the store 0.30. That is P(1) = 0.30. •The probability the first purchase will be made by the second person who enters the store is (0.70) ( 0.30). So P(2) = (0.70) ( 0.30) = 0.21. •The probability the first purchase will be made by the third person who enters the store is (0.70)(0.70)( 0.30). So P(3) = (0.70) (0.70)(0.30) = 0.147. The probability the first purchase will be made by person number x is P(x) = (.70)x - 1(.30) Larson/Farber Ch. 4 Application A cereal maker places a game piece in its boxes. The probability of winning a prize is one in four. Find the probability you a) Win your first prize on the 4th purchase b) Win your first prize on your 2nd or 3rd purchase c) Do not win your first prize in your first 4 purchases. Larson/Farber Ch. 4 Application a) Win your first prize on the 4th purchase P(4) = (.75)3 (.25) = 0.1055 b) Win your first prize on your 2nd or 3rd purchase P(2) = (.75)1(.25) = 0.1875 P(3) = (.75)2(.25) = 0.1406 So P(2 or 3) = 0.1875 + 0.1406 = 0.3281 c) Do not win your first prize in your first 4 purchases. (.75)4 = 0.3164 or 1 – (P(1) + P(2) + P(3) + P(4)) 1 – ( 0.25 + 0.1875 + 0.1406 + 0.1055) = 0.3164 Larson/Farber Ch. 4 The Poisson Distribution The Poisson distribution is a discrete probability distribution of the random variable x that satisfies the following conditions: 1. Counting the number of times, x, an event occurs in an interval of time, area or volume. 2. The probability an event will occur is the same for each interval. 3. The number of occurrences in one interval is independent of the number of occurrences in other intervals. The probability of exactly x occurrences in an interval is e is approximately 2.718 Larson/Farber Ch. 4 Application It is estimated that sharks kill 10 people each year worldwide. Find the probability: a) Three people are killed by sharks this year b) Two or three people are killed by sharks this year P(3) = 0.0076 P(2 or 3) = 0.0023 + 0.0076 = 0.0099 Larson/Farber Ch. 4