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Section 5.1 Discrete Random Variables With helpful added content by D.R.S., University of Cordele HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Discrete Random Variables Probability Distribution A probability distribution is a table or formula that gives the probabilities for every value of the random variable X, where 0 P X x 1 and P X xi 1. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Discrete is our focus for now Discrete A countable number of values (outcomes) • “Red”, “Yellow”, “Green” • “Improved”, “Worsened” • 2 of diamonds, 2 of hearts, … etc. • What poker hand you draw. • 1, 2, 3, 4, 5, 6 rolled on a die • Total dots in rolling two dice Continuous • Will talk about continuous probability distributions in future chapters. Examples of Probability Distributions Rolling a single die Total of rolling two dice Value Probability Value Prob. Value Prob. 1 1/6 2 1/36 8 5/36 2 1/6 3 2/36 9 4/36 3 1/6 4 3/36 10 3/36 4 1/6 5 4/36 11 2/36 5 1/6 6 5/36 12 1/36 6 1/6 7 6/36 Total Total 1 (Note that it’s a two-column chart but we had to typeset it this way to fit it onto the slide.) 1 Example of a Probability Distribution http://en.wikipedia.org/wiki/Poker_probability Draw this 5-card poker hand Probability Royal Flush 0.000154% Straight Flush (not including Royal Flush) 0.00139% Four of a Kind 0.0240% Full House 0.144% Flush (not including Royal Flush or Straight Flush) 0.197% Straight (not including Royal Flush or Straight Flush) 0.392% Three of a Kind 2.11% Two Pair 4.75% One Pair 42.3% Something that’s not special at all 50.1% Total (inexact, due to rounding) 100% Exact fractions avoid rounding errors (but is it useful to readers?) Draw this 5-card poker hand Royal Flush Probability 4 / 2,598,960 Straight Flush (not including Royal Flush) Four of a Kind 36 / 2,598,960 624 / 2,598,960 Full House 3,744 / 2,598,960 Flush (not including Royal Flush or Straight Flush) 5,108 / 2,598,960 Straight (not including Royal Flush or Straight Flush) 10,200 / 2,598,960 Three of a Kind 54,912 / 2,598,960 Two Pair 123,552 / 2,598,960 One Pair 1,098,240 / 2,598,960 Something that’s not special at all 1,302,540 / 2,598,960 Total (exact, precise, beautiful fractions) 2,598,600 / 2,598,600 Example of a probability distribution “How effective is Treatment X?” Outcome Probability The patient is cured. The patient’s condition improves. There is no apparent effect. The patient’s condition deteriorates. 85% 10% 4% 1% Total Confirm – is it = 1 ? Discrete Random Variables Random variable A random variable is a variable whose numeric value is determined by the outcome of a probability experiment. The value is determined by chance, Or it “could be” determined by “chance”. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.1: Creating a Discrete Probability Distribution Create a discrete probability distribution for X, the sum of two rolled dice. Solution To begin, let’s list all of the possible values for X. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.1: Creating a Discrete Probability Distribution (cont.) HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.1: Creating a Discrete Probability Distribution (cont.) When rolling two dice, there are 36 possible rolls, each giving a sum between 2 and 12, inclusive. To find the probability distribution, we need to calculate the probability for each value. 1 P X 2 because there is only one way to get a 36 sum of 2: 2 because you may get the sum of 3 in P X 3 36 two ways: or HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.1: Creating a Discrete Probability Distribution (cont.) Continuing this process will give us the following probability distribution. Sum of Two Rolled Dice x 2 3 4 5 6 7 8 9 10 11 12 P(X = x) 1 36 2 3 36 36 4 36 5 36 6 36 5 36 4 36 3 36 2 36 1 36 Check for yourself that the probabilities listed are the true values for the probability distribution of X, the sum of two rolled dice. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.1: Creating a Discrete Probability Distribution (cont.) Note that all of the probabilities are numbers between 0 and 1, inclusive, and that the sum of the probabilities is equal to 1. Check this for yourself. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Another Example Start with a frequency distribution General layout • Outcome Count of occurrences A specific made-up example How many children live here? Number of households 0 50 1 100 2 150 3 80 4 40 5 20 6 or more 10 Total responses 450 Include a Relative Frequency column General layout • Outcome Count of occurrences A specific simple example Relative Frequency =count ÷ total # of child ren Number of households Relative Frequency 0 50 0.108 1 110 0.239 2 150 0.326 3 80 0.174 4 40 0.087 5 20 0.043 6+ 10 0.022 Total 460 1.000 You can drop the count column General layout • Outcome Relative Frequency =count ÷ total A specific simple example # of children Relative Frequency 0 0.108 1 0.239 2 0.326 3 0.174 4 0.087 5 0.043 6+ 0.022 Total 1.000 Answer Probability Questions What is the probability … • …that a randomly selected household has exactly 3 children? • …that a randomly selected household has children? • … that a randomly selected household has fewer than 3 children? • … no more than 3 children? A specific simple example # of children Relative Frequency 0 0.108 1 0.239 2 0.326 3 0.174 4 0.087 5 0.043 6+ 0.022 Total 1.000 Answer Probability Questions Referring to the Poker probabilities table “What is the probability of drawing a Four of a Kind hand?” “What is the probability of drawing a Three of a Kind or better?” “What is the probability of drawing something worse than Three of a Kind?” “What is the probability of a One Pair hand twice in a row? (after replace & reshuffle?)” HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Tossing coin and counting Heads One Coin Four Coins How many heads Probability How many heads Probability 0 1/2 0 1/16 1 1/2 1 4/16 Total 1 2 6/16 3 4/16 4 1/16 Total 1 Graphical Representation Histogram, for example Four Coins How many heads Probability 0 1/16 1 4/16 2 6/16 6/16 3 4/16 4/16 4 1/16 Total 1 Probability 1/16 0 1 2 3 4 heads Shape of the distribution Histogram, for example Probability Distribution shapes matter! • This one is a bell-shaped distribution • Rolling a single die: its graph is a uniform distribution 6/16 • Other distribution shapes can happen, too 3/16 1/16 0 1 2 3 4 heads Remember the Structure Required features • The left column lists the sample space outcomes. • The right column has the probability of each of the outcomes. • The probabilities in the right column must sum to exactly 1.0000000000000000000. Example of a Discrete Probability Distribution # of children Relative Frequency 0 0.108 1 0.239 2 0.326 3 0.174 4 0.087 5 0.043 6+ 0.022 Total 1.000 The Formulas MEAN: 𝜇 = 𝑋 ∙ 𝑃(𝑋) VARIANCE: 𝜎 2 = 𝑋 2 ∙ 𝑃(𝑋) − 𝜇2 STANDARD DEVIATION: 𝜎 = 𝜎 2 Practice Calculations Rolling one die Statistics Value Probability 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 Total 1 • The mean is 𝜇 = 3.5 • The variance is 𝜎 2 = 2.92 • The standard deviation is 𝜎 = 1.71 Practice Calculations Statistics Total of rolling two dice • The mean is 𝜇 = 7 2 • The variance is 𝜎 = 5.83 • The standard deviation is 𝜎 = 2.42 Value Prob. Value Prob. 2 1/36 8 5/36 3 2/36 9 4/36 4 3/36 10 3/36 5 4/36 11 2/36 6 5/36 12 1/36 7 6/36 Total 1 Practice Calculations One Coin Statistics How many heads Probability 0 1/2 1 1/2 Total 1 • The mean is 𝜇 = 0.5 • The variance is 𝜎 2 = 0.25 • The standard deviation is 𝜎 = 0.5 Practice Calculations Statistics Four Coins • The mean is 𝜇 = 2 2 • The variance is 𝜎 = 1.00 • The standard deviation is 𝜎 = 1.00 How many heads Probability 0 1/16 1 4/16 2 6/16 3 4/16 4 1/16 Total 1 TI-84 Calculations • Put the outcomes into a TI-84 List (we’ll use L1) • Put the corresponding probabilities into another TI-84 List (we’ll use L2) • 1-Var Stats L1, L2 • You can type fractions into the lists, too! • Excel calculations The Mean of a Probability Distribution is easy: =SUMPRODUCT, just like for a frequency distribution. But computing the variance and standard deviation in Excel probably requires using the primitive formulas. Expected Value Expected Value The expected value for a discrete random variable X is equal to the mean of the probability distribution of X and is given by E X xi P X xi Where xi is the ith value of the random variable X. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.2: Calculating Expected Values Suppose that Randall and Blake decide to make a friendly wager on the football game they are watching one afternoon. For every kick the kicker makes, Blake has to pay Randall $30.00. For every kick the kicker misses, Randall has to pay Blake $40.00. Prior to this game, the kicker has made 18 of his past 23 kicks this season. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.2: Calculating Expected Values (cont.) a. What is the expected value of Randall’s bet for one kick? b. Suppose that the kicker attempts four kicks during the game. How much should Randall expect to win in total? HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.2: Calculating Expected Values (cont.) Solution a. There are two possible outcomes for this bet: Randall wins $30.00 (x = 30.00) or Randall loses $40.00 (x = -40.00). If the kicker has made 18 of his past 23 kicks, then we assume that the probability that he will make a kick—and that Randall will win 18 the bet—is . 23 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.2: Calculating Expected Values (cont.) By the Complement Rule, the probability that the kicker 18 5 will miss—and Randall will lose the bet—is 1 - . 23 23 Randall’s Bet for One Kick x $30.00 -$40.00 P(X = x) 18 23 5 23 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.2: Calculating Expected Values (cont.) Then we calculate the expected value as follows. E X xi P X xi But use TI-84 1-Var Stats (giving it TWO lists, one with values and the other with frequencies. 18 5 30.00 -40.00 23 23 540 200 23 23 340 23 $14.78 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.2: Calculating Expected Values (cont.) We see that the expected value of the wager is $14.78. Randall should expect that if the same bet were made many times, he would win an average of $14.78 per bet. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.2: Calculating Expected Values (cont.) b. We know that over the long term Randall would win an average of $14.78 per bet. So for four attempted kicks, we multiply the expected value for one bet by four: 14.78 4 59.12. If he and Blake place four bets, then Randall can expect to win approximately $59.12. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.3: Calculating Expected Values Peyton is trying to decide between two different investment opportunities. The two plans are summarized in the table below. The left column for each plan gives the potential earnings, and the right columns give their respective probabilities. Which plan should he choose? HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.3: Calculating Expected Values (cont.) Investment Plans Plan A Plan B Earnings Probability Earnings Probability $1200 0.1 $1500 0.3 $950 0.2 $800 0.1 $130 0.4 -$100 0.2 -$575 0.1 -$250 0.2 -$1400 0.2 -$690 0.2 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.3: Calculating Expected Values (cont.) Solution It is difficult to determine which plan will yield the higher return simply by looking at the probability distributions. Let’s use the expected values to compare the plans. Let the random variable X A be the earnings for Plan A, and let the random variable X B be the earnings for Plan B. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.3: Calculating Expected Values (cont.) For Investment Plan A: E X A xi P X A xi Again, use TI-84 1-Var Stats with two lists is recommended, instead of this by-hand calculation. 1200 0.1 950 0.2 130 0.4 -575 0.1 -1400 0.2 120 190 52 - 57.5 - 280 $24.50 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.3: Calculating Expected Values (cont.) For Investment Plan B: E XB xi P XB xi Again, use TI-84 1-Var Stats with two lists is recommended, instead of this by-hand calculation. 1500 0.3 800 0.1 -100 0.2 -250 0.2 -690 0.2 450 80 - 20 - 50 - 138 $322.00 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.3: Calculating Expected Values (cont.) From these calculations, we see that the expected value of Plan A is $24.50, and the expected value of Plan B is $322.00. Therefore, Plan B appears to be the wiser investment option for Peyton. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Variance and Standard Deviation for a Discrete Probability Distribution Variance and Standard Deviation for a Discrete Probability Distribution The variance for a discrete probability distribution of a random variable X is given by s 2 xi 2 P X xi - m2 2 xi - m P X xi Where x i is the i th value of the random variable X and μ is the mean of the probability distribution. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Formula Variance and Standard Deviation for a Discrete Probability Distribution (cont.) The standard deviation for a discrete probability distribution of a random variable X is the square root of the variance, given by the following formulas. s s2 2 2 x P X x m i i 2 xi - m P X xi HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.4: Calculating the Variances and Standard Deviations for Discrete Probability Distributions Which of the investment plans in the previous example carries more risk, Plan A or Plan B? Solution To decide which plan carries more risk, we need to look at their standard deviations, which requires that we first calculate their variances. Let’s calculate the variance separately for each investment plan. To do this, we will use a table to organize our calculations as we compute the variance. We will use the expected values that we calculated in the previous example as the means. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.4: Calculating the Variances and Standard Deviations for Discrete Probability Distributions (cont.) Again, use TI-84 1-Var Stats with two lists is recommended, instead of this by-hand calculation. For Investment Plan A, m E X A $24.50. Investment Plan A x - 2 x - P X x 2 x P(X = x) x - $1200 0.1 1175.50 1,381,800.25 138,180.025 $950 0.2 925.50 856,550.25 171,310.05 $130 0.4 105.60 11,130.25 4452.1 -$575 0.1 -599.50 359,400.25 35,940.025 -$1400 0.2 -1424.50 2,029,200.25 405,840.05 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.4: Calculating the Variances and Standard Deviations for Discrete Probability Distributions (cont.) 2 xi - m P X A xi 138,180.025 171,310.05 4452.1 35, 940.025 405,840.05 755,722.25 2 2 755,722.25 $869.32 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Again, use TI-84 1-Var Stats with two lists is recommended, instead of this by-hand calculation. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.4: Calculating the Variances and Standard Deviations for Discrete Probability Distributions (cont.) Again, use TI-84 1-Var Stats with two lists is recommended, instead of this by-hand calculation. For Investment Plan B, E XB $322.00. Investment Plan B x - 2 P X x x P(X = x) x - x - 2 $1500 0.3 1178 1,387,684 416,305.2 $800 0.1 478 228,484 22,848.4 -$100 0.2 422 178,084 35,616.8 -$250 0.2 572 327,184 65,436.8 -$690 0.2 1012 1,024,144 204,828.8 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.4: Calculating the Variances and Standard Deviations for Discrete Probability Distributions (cont.) 2 xi - m P XB xi 416,305.2 22,848.4 35,616.8 65,436.8 204,828.8 745,036 Again, use TI-84 1-Var Stats 2 2 with two lists is recommended, instead of this by-hand calculation. 745,036 $863.15 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5.4: Calculating the Variances and Standard Deviations for Discrete Probability Distributions (cont.) What do these results tell us? Comparing the standard deviations, we see that not only does Plan B have a higher expected value, but its profits vary slightly less than those of Plan A. We may conclude that Plan B carries a slightly lower amount of risk than Plan A. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Expected Value Problems The Situation • 1000 raffle tickets are sold • You pay $5 to buy a ticket • First prize is $2,000 • Second prize is $1,000 • Two third prizes, each $500 • Three more get $100 each • The other ____ are losers. What is the “expected value” of your ticket? The Discrete Probability Distr. Outcome Win first prize Net Value Probability $1,995 1/1000 Win second prize $995 1/1000 Win third prize $495 2/1000 Win fourth prize $95 3/1000 Loser $ -5 993/1000 Total 1000/1000 Expected Value Problems Statistics • The mean of this probability is $ - 0.70, a negative value. • This is also called “Expected Value”. • Interpretation: “On the average, I’m going to end up losing 70 cents by investing in this raffle ticket.” The Discrete Probability Distr. Outcome Win first prize Net Value Probability $1,995 1/1000 Win second prize $995 1/1000 Win third prize $495 2/1000 Win fourth prize $95 3/1000 Loser $ -5 993/1000 Total 1000/1000 Expected Value Problems Another way to do it • Use only the prize values. • The expected value is the mean of the probability distribution which is $4.30 • Then at the end, subtract the $5 cost of a ticket, once. • Result is the same, an expected value = $ -0.70 The Discrete Probability Distr. Outcome Net Value Probability Win first prize $2,000 1/1000 Win second prize $1,000 1/1000 Win third prize $500 2/1000 Win fourth prize $100 3/1000 Loser Total $0 993/1000 1000/1000 Expected Value Problems The Situation • We’re the insurance company. • We sell an auto policy for $500 for 6 months coverage on a $20,000 car. • The deductible is $200 What is the “expected value” – that is, profit – to us, the insurance company? The Discrete Probability Distr. Outcome Net Value Probability No claims filed _______ An $800 fender bender 0.004 An $8,000 accident 0.002 A wreck, it’s totaled 0.002 An Observation The mean of a probability distribution is really the same as the weighted mean we have seen. Recall that GPA is a classic instance of weighted mean – Grades are the values – Course credits are the weights Think about the raffle example – Prizes are the values – Probabilities of the prizes are the weights