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Random Variables and Discrete Probability Distributions

Random Variables and Discrete Probability Distributions. Random Variables…. A random variable is a function or rule that assigns a number to each outcome of an experiment. Basically it is just a symbol that represents the outcome of an experiment.

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Random Variables and Discrete Probability Distributions

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  1. Random Variables and Discrete Probability Distributions

  2. Random Variables… • A random variable is a function or rule that assigns a number to each outcome of an experiment. Basically it is just a symbol that represents the outcome of an experiment. • X = number of heads when the experiment is flipping a coin 20 times. • C = the daily change in a stock price. • R = the number of miles per gallon you get on your auto during a family vacation. • Y = the amount of medication in a blood pressure pill. • V = the speed of an auto registered on a radar detector used on I-20

  3. Two Types of Random Variables… • Discrete Random Variable – usually count data [Number of] • * one that takes on a countable number of values – this means you can sit down and list all possible outcomes without missing any, although it might take you an infinite amount of time. • X = values on the roll of two dice: X has to be either 2, 3, 4, …, or 12. • Y = number of accidents on the UTA campus during a week: Y has to be 0, 1, 2, 3, 4, 5, 6, 7, 8, ……………”real big number” • Continuous Random Variable – usually measurement data [time, weight, distance, etc] • * one that takes on an uncountable number of values – this means you can never list all possible outcomes even if you had an infinite amount of time. • X = time it takes you to drive home from class: X > 0, might be 30.1 minutes measured to the nearest tenth but in reality the actual time is 30.10000001…………………. minutes?) • Exercise: try to list all possible numbers between 0 and 1.

  4. Probability Distributions… • A probability distribution (density function) is a table, formula, or graph that describes the values of a random variable and the probability associated with these values. • – Discrete Probability Distribution • X = outcome of rolling one die • – Continuous Probability Distribution

  5. Discrete Probability Notation… • An upper-case letter will represent the name of the random variable, usually X. • Its lower-case counterpart, x, will represent the value of the random variable. • The probability that the random variable X will equal x is: • P(X = x) or more simply P(x) • X = number of heads in 10 flips of coin • P(X = 5) = P(5) = probability of 5 heads (x) in 10 flips

  6. Discrete Probability Distributions… • Probabilities, P(x),associated with Discrete random variables have the following properties.

  7. Developing Discrete Probability Distributions • Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household (X) from US survey data (Example 7.1)… 1,218 ÷ 101,501 = 0.012 e.g. P(X=4) = P(4) = 0.076 = 7.6%

  8. Questions you might want answered • E.g. what is the probability there is at least one television but no more than three in any given household? “at least one television but no more than three” P(1 ≤ X ≤ 3) = P(1) + P(2) + P(3) = .319 + .374 + .191 = .884

  9. Example • In most college courses, you get as a grade, either an A, B, C, D, or F. For credit purposes, A’s are • given 4 points, B’s are given 3 points, C’s are given 2 points, Ds are given 1 point, and F’s are no points. • Let Xbe the random variable representing the points a student gets. What are the possible values of X? The possible values of X are 4, 3, 2, 1, and 0.

  10. Example A college instructor teaching a large class traditionally gives 10% A’s, 20% B’s, 45% C’s, 15% D’s, and 10% F’s. If a student is chosen at random from the class, the student’s grade on a 4-point scale (A = 4) is a random variable X. Create the distribution of X. What is the probability that a student has a grade point of 3 or better in this class? What is the probability that a student has a grade point of 2 or worse in this class? Draw a probability histogram to picture the probability distribution of the random variable X.

  11. Discrete Probability Distribution • The mean of a discrete random variable is the weighted average of all of its values. The weights are the probabilities. This parameter is also called the expected value of X and is represented by E(X). • The variance is • The standard deviation is

  12. Example A college instructor teaching a large class traditionally gives 10% A’s, 20% B’s, 45% C’s, 15% D’s, and 10% F’s. If a student is chosen at random from the class, the student’s grade on a 4-point scale (A = 4) is a random variable X. Create the distribution of X. Find the mean (expected value) and standard deviation for the probability distribution.

  13. Example • A fair coin is flipped 3 times. Find the mean and standard deviation of the discrete random variable X that counts the number of heads.

  14. Example • The daily lottery costs $1 to play. You pick a 3 digit number. If you win, you win $500. Find the expected value and standard deviation of the lottery.

  15. Example • Choose an American household at random and let the random variable X be the number of persons living in the household. Find the mean and standard deviation of the average American household.

  16. Computing Mean, Variance, and Std. Dev. for Discrete Random Variable • Mean = 1(.25) + 2(.32) + 3(.17) + 4(.15) + 5(.07) + 6(.03) + 7(.01) • = 2.6 • Variance = (1-2.6)2*(.25) + (2-2.6)2*(.32) + . . . + (7-2.6)2*(.01) • = 8.78 • Std. Dev. = SQRT(8.78) = 1.42 • We are as smart as the goddess of statistics now, since we know the true mean, variance, and standard deviation of the population.

  17. Rules for means Example: You play two casino games. Game 1 has an expectation of losing $1 a play and game 2 has anexpectation of losing $2 a play. If you play both games, what is your expectation for both games. Example: You go to a casino and play the same slot machine which averages losing 15 cents a play. You play the machine 100 times and then leave, paying $5 for parking. Find your expectation for your casino visit.

  18. Rules for Variances of Random Variables Example: Suppose two pro bowlers Adam (A) and Bart (B) have the following distribution of scores.

  19. Example • Depending on the attendance of a minor league baseball team, the number of hot dogs and sodas sold at a game is given by the following table. • Find the standard deviation of the number of hot dogs plus the number of sodas sold.

  20. Example: You weight all 30,000 students • Random Variable: X = students weight • Mean(X) = X-Bar = 160 lbs • Variance(X) = s2 = 900 lbs2 • StdDev(X) = s = 30 lbs • ************************************* • You now discover that the scales reported a student’s weight 5 lbs too heavy. The student’s real weights (Y) should have been Y = X – 5. What are the mean and variance of the student’s REAL weights • Mean(Y) = Mean(X) – 5 = 160 – 5 = 155 lbs • Variance(Y) = Variance(X) = 900 • StdDev(Y) = SQRT(900) = 30

  21. Example: You measure the height of all 30,000 students • Random Variable: X = students height in “Feet” • Mean(X) = X-Bar = 5.8 feet • Variance(X) = s2 = 0.09 feet2 • StdDev(X) = s = 0.3 feet • ************************************* • You now discover that the President wanted to measure student’s heights in “Inches” and not “Feet”. The student’s height in “Inches” (Y) should have been Y = 12*X . What are the mean and variance of the student’s heights in Inches? • Mean(Y) = 12*Mean(X) = 12*5.8 = 69.6 inches • Variance(Y) = 122*Variance(X) = 144*(.09) = 12.96 • StdDev(Y) = SQRT(12.96) = 3.6

  22. Laws… • We can derive laws of expected value and variance for the sum of two independent random variables as follows… • E(X + Y) = E(X) + E(Y) • V(X + Y) = V(X) + V(Y) • ************************************************************** • X = weight of right shoes: Mean(X) = .5 lbs and Var(X) = .0004 • Y = weight of left shoes: Mean(Y) = .5 lbs and Var(Y) = .0004 • ************************************************************** • What is the mean and variance of a “Pair” of shoes. P = X +Y • E(P) = E(X + Y) = E(X) + E(Y) = .5 + .5 = 1.0 • V(P) = V(X+Y) = V(X) + V(Y) = .0004 + .0004 = .0008 • NOTE: WEIGHTS OF RIGHT AND LEFT SHOE INDEPENDENT • *************************************************************** • ? How could you determine the mean and variance of the weight of an automobile after you make all the parts but before you assemble the automobile

  23. Binomial Distribution… 2 parameters [n and p] • The binomial distribution is the probability distribution that results from doing a “binomial experiment”. Binomial experiments have the following properties: • Fixed number of trials, represented as n. • Each trial has two possible outcomes, a “success” and a “failure”. • P(success)=p (and thus: P(failure)=1–p), for all trials. • The trials are independent, which means that the outcome of one trial does not affect the outcomes of any other trials.

  24. Success and Failure… • …are just labels for a binomial experiment, there is no value judgment implied. You may define either one of the 2 possible outcomes as “Success” • For example a coin flip will result in either heads or tails. If we define “heads” as success then necessarily “tails” is considered a failure (inasmuch as we attempting to have the coin lands heads up). • Other potential examples of binomial random variables: • A firecracker pops or fails to pop • A patient get an infection during an operation or does not get an infection

  25. Binomial Random Variable… • The random variable of a binomial experiment is defined as the number of successes, X, in the n trials, where the probability of success on a single trial is p. • E.g. flip a fair coin 10 times… • 1) Fixed number of trials n=10 • 2) Each trial has two possible outcomes  {heads (success), tails (failure)} • 3) P(success)= 0.50; P(failure)=1–0.50 = 0.50  • 4) The trials are independent (i.e. the outcome of heads on the first flip will have no impact on subsequent coin flips). • Hence flipping a coin ten times is a binomial experiment since all conditions were met.

  26. Are these examples of binomial distributions? • 1) Tossing 20 coins and counting the number of heads. • 1) Success is a heads, failure is a tails. • 2) n = 20 • 3) Independence is true – coins have no influence on each other • 4) p = .5. • So X is B(20, .5). The possible values of X are the integers from 0 to 20. • 2) Picking 5 cards from a standard deck and counting the number of hearts. We replace the card eachtime and reshuffle. • 1) Success is a heart, failure is anything but a heart. • 2) n = 5. • 3) Independence is true. • 4) p =.25. • So X is B(5, .25). The possible values of X are the integers from 0 to 5.

  27. Are these examples of binomial distributions? • 3) Picking 5 cards from a standard deck and counting the number of hearts without reshuffling. • This is not binomial because of the independence issue. • 4) Choosing a card from a standard deck until you get a heart. • This is not binomial as there are not a fixed number of observations. • 5) It is estimated that 87% of computers users use Explorer as their default web browser. We choose50 computer users and ask their default browser. • 1) Success is Explorer, failure is anything else. • 2) n =50. • 3) Independence seems logical. • 4) p = .87. • So X is B(50, .87). The possible values of X are the integers from 0 to 50.

  28. Binomial Distribution [formula] • The binomial random variable (# of successes in n trials) take on values 0, 1, 2, …, n. Thus, its a discrete random variable. • Once we know a random variable is binomial, we can calculate the probability associated with each value of the random variable from the binomial distribution: • where n = number in sample p = probability of success • k = # successes and n-k = # failures

  29. Problem: Pat Statsdud… • Pat Statsdud failed to study for the next stat exam. Pat’s exam strategy is to rely on luck for the next quiz. The quiz consists of 10 multiple-choice questions (n=10). Each question has five possible answers, only one of which is correct (p=0.2). Pat plans to guess the answer to each question. • What is the probability that Pat gets no answers correct? • P(X=0) = P(0) = • What is the probability that Pat gets two answers correct? • P(X=2) = P(2) =

  30. Pat Statsdud… • n=10, and P(success) = .20 • What is the probability that Pat gets no answers correct? • I.e. # success, x, = 0; hence we want to know P(x=0) Pat has about an 11% chance of getting no answers correct using the guessing strategy.

  31. Pat Statsdud… • n=10, and P(success) = .20 • What is the probability that Pat gets two answers correct? • I.e. # success, x, = 2; hence we want to know P(x=2) Pat has about a 30% chance of getting exactly two answers correct using the guessing strategy.

  32. Example • You toss 5 coins. What is the probability that you get 3 heads?

  33. Cumulative Probability… • “Find the probability that Pat fails the quiz” • If a grade on the quiz is less than 50% (i.e. 5 questions • out of 10), that’s considered a failed quiz. • P(fail quiz) = P(X < 4) = P(0)+P(1)+P(2)+P(3)+P(4) • Called a cumulative probability, that is, P(X ≤ x) • Note: Calculating all these individual probabilities would be tedious and time consuming, however, the Binomial tables at back of book gives you the cumulative probabilities [n=10, p=0.2, x=4]

  34. Pat Statsdud… • Calculate Individual Probabilities and Add Up! • P(X ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4) • We already know P(0) = .1074 and P(2) = .3020. Using the binomial formula to calculate the others: • P(1) = .2684 , P(3) = .2013, and P(4) = .0881 • Hense P(X ≤ 4) = .1074 + .2684 + … + .0881 = .9672 • OR • Use binomial tables at back of book for n=10, p=0.2, and x=4 • OR • Use CALCULATOR!

  35. Calculator • binompdf(number of trials, probability of success, number of success) • Used to compute binomial probabilities for a particular number of successes. • binomcdf(number of trials, probability, number of successes or less) • Used to compute cumulative (from smaller) successes.

  36. Example • Will Guess takes a true-false test of 6 questions and has absolutely no idea of any of the answers. • So, true to his name, he guesses on all of them. If 4 questions correct is passing, what is the probability that hepasses the exam? Suppose the test above is now multiple choice with 4 answers per problem and again, Will Guesses. Find the probability that he passes the test and the expected number of passing students in a school of 1,500 if they all guessed.

  37. Example • In a particular city, 63% of the adults own their home and 37% rent. A sample of 20 adults is taken. • Find the probability that the sample will have at least half home-owners.

  38. Binomial Distribution… • As you might expect, statisticians have determined formulas for the mean, variance, and standard deviation of a binomial random variable. They are: • Previous example: n=10, p=0.2 • μ = n*p = 10*0.2 = 2 • σ2 = n*p*(1-p) = 10*0.2*0.8= 1.6 • σ = SQRT(1.6) = 1.26

  39. Example • A basketball player is traditionally a 72% foul shooter. In a season, he takes 427 foul shots. Find • the mean and standard deviation of the distribution.

  40. Geometric Random Variables • Only two possible outcomes (success or failure) • Probability of success is constant for each trial • Each trial is independent of other trials • Looking for the number of trials to obtain the FIRST success

  41. Geometric Distributions • Geometric distributions: probability distribution of a geometric random variable (all possible outcomes of X before the first success is seen and their possibilities) • P(x = n) = (1 - p)n-1 p You roll two dice and add them. Find the probability that we roll a 7 on the first trial, the second,the third, the 4th, and the 5th. P(7) =

  42. Calculator • geometpdf(probability of success, number of trials) • Use for computing geom. probabilities for a particular number of trials. • geometcdf (probability of success, number of trials) • Use for computing geom. probabilities of a given number of trials or less (cumulative from smaller end)

  43. Example • It is estimated that 45% of people in Fast-Food restaurants order a diet drink with their lunch. • Find the probability that the fourth person orders a diet drink. • Also find the probability that the first diet • drinker of the day occurs before the 5th person.

  44. Mean and Standard deviational Mean of geometric random variable X: Standard deviation of geometric random variable X: • In New York City at rush hour, the chance that a taxicab passes someone and is available is 15%. • How many cabs can you expect to pass you for you to find one that is free? • What is theprobability that more than 10 cabs pass you before you find one that is free?

  45. Poisson Distribution… 1 parameter [μ] • Named for Simeon Poisson, the Poisson distribution is a discrete probability distribution and refers to the number of events (a.k.a. successes) within a specific time period or region of space. For example: • The number of cars arriving at a service station in 1 hour. (The interval of time is 1 hour.) • The number of flaws in a bolt of cloth. (The specific region is a bolt of cloth.) • The number of accidents in 1 day on a particular stretch of highway. (The interval is defined by both time, 1 day, and space, the particular stretch of highway.)

  46. Poisson Probability Distribution… • The probability that a Poisson random variable assumes a value of x is given by: • Note: μ is the only parameter [tell me μ and I can calculate the probabilities] • and e is the natural logarithm base. • FYI:

  47. Example 7.12… • The number of typographical errors in new editions of textbooks varies considerably from book to book. After some analysis he concludes that the number of errors is Poisson distributed with a mean of 1.5 typos per 100 pages. The instructor randomly selects 100 pages of a new book. What is the probability that there are no typos? • That is, what is P(X=0) given that = 1.5? “There is about a 22% chance of finding zero errors”

  48. Poisson Distribution… • As mentioned on the Poisson experiment slide: • The probability of a success is proportional to the size of the interval • Thus, knowing an error rate of 1.5 typos per 100 pages, we can determine a mean value for a 400 page book as: • =1.5(4) = 6 typos / 400 pages.

  49. Example 7.13… • For a 400 page book, what is the probability that there are • no typos? • P(X=0) = “there is a very small chance there are no typos”

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