1. Discrete Probability Distribution Calculations Mean, Variance, Standard Deviation Expected Value

2. 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 childrenRelative Frequency 00.108 10.239 20.326 30.174 40.087 50.043 6+0.022 Total1.000

3. The Formulas

4. TI-84 Calculations Put the outcomes into a TI-84 List (we’ll use L 1 ) Put the corresponding probabilities into another TI-84 List (we’ll use L 2 ) 1-Var Stats L 1, L 2 You can type fractions into the lists, too!

5. Practice Calculations Rolling one die ValueProbability 11/6 2 3 4 5 6 Total1 Statistics

6. Practice Calculations StatisticsTotal of rolling two dice ValueProb.ValueProb. 21/3685/36 32/3694/36 43/36103/36 54/36112/36 65/36121/36 76/36Total1

7. Practice Calculations One Coin How many headsProbability 01 / 2 1 Total1 Statistics

8. Practice Calculations StatisticsFour Coins How many headsProbability 01/16 14/16 26/16 34/16 41/16 Total1

9. 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. OutcomeNet ValueProbability Win first prize $1,9951/1000 Win second prize $9951/1000 Win third prize $4952/1000 Win fourth prize $953/1000 Loser$ -5993/1000 Total1000/1000

10. 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. OutcomeNet ValueProbability Win first prize $1,9951/1000 Win second prize $9951/1000 Win third prize $4952/1000 Win fourth prize $953/1000 Loser$ -5993/1000 Total1000/1000

11. 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. OutcomeNet ValueProbability Win first prize $2,0001/1000 Win second prize $1,0001/1000 Win third prize $5002/1000 Win fourth prize $1003/1000 Loser$ 0993/1000 Total1000/1000

12. 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