Larson ch 4 Stats

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  • 1. Chapter 4 Discrete Probability DistributionsLarson/Farber 4th ed1

2. Chapter Outline 4.1 Probability Distributions 4.2 Binomial Distributions 4.3 More Discrete Probability Distributions (not in syllabus)Larson/Farber 4th ed2 3. Section 4.1 Probability DistributionsLarson/Farber 4th ed3 4. 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 Larson/Farber 4th ed4 5. 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.Larson/Farber 4th ed5 6. 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. x0Larson/Farber 4th ed123456 7. 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. x0Larson/Farber 4th ed123247 8. Example: Random Variables Decide whether the random variable x is discrete or continuous. 1. x = The number of stocks in the Dow Jones Industrial Average that have share price increases on a given day. Solution: Discrete random variable (The number of stocks whose share price increases can be counted.) x0 Larson/Farber 4th ed12330 8 9. Example: Random Variables Decide whether the random variable x is discrete or continuous. 2. x = The volume of water in a 32-ounce container. Solution: Continuous random variable (The amount of water can be any volume between 0 ounces and 32 ounces) x0 Larson/Farber 4th ed12332 9 10. Textbook Exercises. Page 197 Distinguish between Discrete and Continuous random variable 14. x represents the length of time it takes to go to work. x is a continuous random variable because length of time is a measurement and not a count. Measurements are continuous. 16. x represents number of tornadoes in the month of June in Oklahoma. x is a discrete random variable because number of tornadoes is a count. Counts are discrete.Larson/Farber 4th ed10 11. 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 WordsIn Symbols1. The probability of each value of the discrete random variable is between 0 and 1, inclusive.0 P (x) 12. The sum of all the probabilities is 1.P (x) = 1Larson/Farber 4th ed11 12. What is a PROBABILITY? 0%25%0 Impossible Certain50%75% or .25 Not Very LikelyLarson/Farber 4th ed100% 0r .5 or .75Equally Likely1 Somewhat Likely12 13. 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 and that the sum is 1. Larson/Farber 4th ed13 14. 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 4 30 distribution using a histogram. 5Larson/Farber 4th ed21 14 15. 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. P (1) =24 = 0.16 15030 P (4) = = 0.20 150P (2) =33 = 0.22 15042 P (3) = = 0.28 15021 P (5) = = 0.14 150 Discrete probability distribution: x12345P(x)0.160.220.280.200.14Larson/Farber 4th ed15 16. Solution: Constructing a Discrete Probability Distribution x12345P(x)0.160.220.280.200.14This 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, P(x) = 0.16 + 0.22 + 0.28 + 0.20 + 0.14 = 1.Larson/Farber 4th ed16 17. Solution: Constructing a Discrete Probability Distribution HistogramBecause the width of each bar is one, the area of each bar is equal to the probability of a particular outcome. Larson/Farber 4th ed17 18. Textbook Exercises. Page 198 Problem 22 Blood Donations x being number of donations is countable and hence a discrete random variable. Also, number of donations are mutually exclusive. a. P(X>1) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) = 0.25 + 0.10 + 0.05 + 0.03 + 0.02 = 0.45 b. P(X < 3)Problem 24 x 0 P(x)0.05= P(x = 2) + P(x = 1) + P(x = 0) = 0.25 + 0.25 + 0.30 = 0.80 Missing Probability 1 2 3 ?0.230.214560.170.110.08Since it is a probability distribution sum of all the probabilities is equal to one. missing value = 1 sum of rest of the probabilities Larson/Farber 4th ed = 1 0.85 = 0.1518 19. Mean Mean of a discrete probability distribution = xP(x) Each value of x is multiplied by its corresponding probability and the products are added.Larson/Farber 4th ed19 20. Example: Finding the Mean The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean. xP(x)xP(x)10.161(0.16) = 0.1620.222(0.22) = 0.4430.283(0.28) = 0.8440.204(0.20) = 0.805Solution:0.145(0.14) = 0.70 = xP(x) = 2.94 Larson/Farber 4th ed20 21. Variance and Standard Deviation Variance of a discrete probability distribution 2 = (x )2P(x) Standard deviation of a discrete probability distribution = 2 = ( x ) 2 P ( x)Larson/Farber 4th ed21 22. 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. ( = 2.94) x 10.1620.2230.2840.205Larson/Farber 4th edP(x)0.14 22 23. Solution: Finding the Variance and Standard Deviation Recall = 2.94 xP(x)x(x )2(x )2P(x)10.161 2.94 = 1.94(1.94)2 = 3.7643.764(0.16) = 0.60220.222 2.94 = 0.94(0.94)2 = 0.8840.884(0.22) = 0.19430.283 2.94 = 0.06(0.06)2 = 0.0040.004(0.28) = 0.00140.204 2.94 = 1.06(1.06)2 = 1.1241.124(0.20) = 0.22550.145 2.94 = 2.06(2.06)2 = 4.2444.244(0.14) = 0.594 Variance:2 = (x )2P(x) = 1.616 = 2 = 1.616 1.3 Standard Deviation: Larson/Farber 4th ed23 24. Textbook Exercises. Page 199 Problem 30 Camping Chairs Construct the probability distribution and determine : a. Mean b. Standard Deviation c. Variance Defects, xBatches, fP(x)09595 380 = 0.2501113113 380 = 0.2972870.2293640.1684130.034580.021n = 380 P(x) = 1sample mean x = 1.5 sample standard deviation 1.24 sample variance 2 1.54 Larson/Farber 4th edTo determine the mean, standard deviation and variance we will use TI 83/84. First store the x-values and the P(x) values in two of the lists, say L1 and L2. Make sure that x goes into L1 and P(x) goes into L2. Do not reverse that order. Then using 1-var stats from CALC menu enter L1 comma L2 and hit the ENTER key. You may notice that the sample standard deviation is not shown but you may use population standard deviation to get an estimate. Double check your results with hand calculations.24 25. Expected Value Expected value of a discrete random variable Equal to the mean of the random variable. E(x) = = xP(x)Larson/Farber 4th ed25 26. 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. What is the expected value of your gain?Larson/Farber 4th ed26 27. 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 = $2Larson/Farber 4th ed27 28. Solution: Finding an Expected Value Probability distribution for the possible gains (outcomes) Gain, x$498$248$148$73$2P(x)1 15001 15001 15001 15001496 1500E (x ) = xP (x ) 1 1 1 1 1496 = $498 + $248 + $148 + $73 + ($2) 1500 1500 1500 1500 1500 = $1.35You can expect to lose an average of $1.35 for each ticket you buy. Larson/Farber 4th ed28 29. 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 Larson/Farber 4th ed29 30. Section 4.2 Binomial DistributionsLarson/Farber 4th ed30 31. 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 and a binomial table Graph a binomial distribution Find the mean, variance, and standard deviation of a binomial probability distribution Larson/Farber 4th ed31 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 fo