1. 1 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Chapter 14 Random Variables and Probability Models

2. 2 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 2 Objectives: The student will be able to: 47. Recognize random variables. 48. Find the probability model for a discrete random variable. 49. Find and interpret in context the mean (expected value) and the standard deviation of a random variable. 50. Tell if a situation involves Bernoulli trials. 51. Know the appropriate conditions for using a Binomial model. 52. Calculate binomial probabilities. 53. Find and interpret in context the mean and standard deviation of a Binomial model.

3. 3 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 14.1 Expected Value: Center

4. 4 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Probability Model An insurance company pays $10,000 if you die or $5,000 for a disability. The amount the company pays is a random variable: a numeric value based on the outcome of a random event. It is a discrete random variable, since we can list all the possible outcomes A continuous random variable is a random variable that is not discrete. The collection of all possible values and their probabilities is called a probability model.

5. 5 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Expected Value The expected value is the average amount that is likely to occur if there are many trials. Expected Value Formula: The company expects to pay an average of about $20 per policy per year.

6. 6 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Valentine’s Day Discount Deal At the end of the meal, lovers can play for a discount. From 4 aces: Black Ace: No discount Ace of Hearts: $20 discount Ace of Diamonds: Pick another ace – Black Ace: No Discount – Ace of Hearts: $10 discount Find a probability model and interpret the expected value.

7. 7 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Valentine’s Day Discount Deal P(X = 20) = P(A♥) = 1/4 P(X = 10) = P(A♦, then A♥) = P(A♦) × P(A ♥ | A ♦) = 1/4 × 1/3 = 1/12 P(X = 0) = 1 – (1/4 + 1/12) = 2/3

8. 8 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Valentine’s Day Discount Deal Expected Value: The restaurant should expect the average discount to be about $5.83 per couple.

9. 9 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 14.2 Standard Deviation

10. 10 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Standard Deviation of a Probability Model Standard Deviation Consider data from many outcomes of a random variable. The variance and standard deviation of these outcomes will measure the spread of the data. Theoretically: The standard deviation is the square root of the variance:

11. 11 Copyright © 2014, 2012, 2009 Pearson Education, Inc. About the Standard Deviation Not just the standard deviation of the X values A weighted average Measures how outcomes will likely be spread out if many are selected Will be large if there is a high probability of both small values and large values

12. 12 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 12 Using the TI-83 to find E(X) and SD(X) for a probability model Much of what you have to do for discrete random data is very much like what we did to get the descriptive statistics for data involving frequencies only here we will use decimals, probabilities. Example: A recent study involving households and the number of children in the household showed that 18% of the households had no children, 39% had one, 24% had two, 14% had three, 4% has four and 1% had five. Find the expected value, mean number of children per household, and the standard deviation of this distribution. Place the values of the variable in L1: 0,1,2,3,4,5 and the associated probabilities in L2:.18,.39,.24,.14,.04 and.01.

13. 13 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 13 Using the TI-83 Then do Stat -> Calc -> 1-var stat(L1, L2) The mean is 1.5 which is also the sum…. the calculator is doing what we would do if we used the formula… multiply each value by its corresponding probability and then add the separate products. The standard deviation is 1.1180. Notice the n=1. That just means that the individual probabilities added up to one which they should if this is a proper distribution.

14. 14 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 14 Practice 4) A day trader buys an option on a stock that will return $100 profit if it goes up today and loses $200 if it goes down. If the trader thinks there is a 75% chance that the stock will go up, what is his expected value of the option? 6,14) You roll a die. If it comes up a 6, you win $100. If not, you get to roll again. If you get a 6 the second time you win $50, if not, you lose. Create a probability model for this game Find the expected amount you’ll win How much would you be willing to pay to play this game? Find the standard deviation of the amount you might win

15. 15 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 24) Your company bids for two contracts. You believe the probability that you get contract #1 is 0.8. If you get contract #1, the probability that you get contract #2 is 0.2, and if you do not get contract #1, the probability that you get #2 will be 0.3. a) Are the two contracts independent? Explain b) Find the probability that you get both contracts c) Find the probability that you get no contract d) Let X be the number of contracts you get. Find the probability model for X e) Find the expected value and standard deviation of X Slide 1- 15

16. 16 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Refurbished Computers, Expected Value, and Standard Deviation The problem You just shipped two computers. Out of the 15 in stock, 4 were refurbished. If one of the two is refurbished, you lose $100. If both are, you lose $1000 due to angry customer. Find and interpret the expected value and standard deviation.

17. 17 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Refurbished Computers Accidentally Shipped Think→ Plan: Find the expected loss and standard deviation due to shipping refurbished computers. Variable: Let X = amount of loss. Plot: Make a picture. Model:

18. 18 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Refurbished Computers Accidentally Shipped Show→ Mechanics: Use your calculator to find the expected value and standard deviation. Tell → Conclusion: I expect this mistake to cost the firm $98.90 with a standard deviation of $226.74. The large standard deviation comes from the large range of possible values.

19. 19 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 14.3 Combining Random Variables THIS IS FYI – not on the exam!

20. 20 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Adding or Subtracting a Constant Adding or subtracting a constant to all the data values shifts the expected value by that constant. Adding or subtracting a constant to all the data values has no effect on the standard deviation.

21. 21 Copyright © 2014, 2012, 2009 Pearson Education, Inc. A Coupon on Top of the Valentine’s Discount The Valentine’s Discount E(X) = $5.83, SD(X) = $8.62 If everybody brings a coupon for $5 off, what are the new expected value and standard deviation? E(X + 5) = $5.83 + $5 = $10.83 SD(X + 5) = $8.62

22. 22 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Multiplying by a Constant There is a double the rewards special Valentines Day discount. The rewards are now $40 and $20 instead of $20 and $10. What are the new expected value and standard deviation? With the double rewards special the restaurant expects an average discount of $11.66 and a standard deviation of $17.24.

23. 23 Copyright © 2014, 2012, 2009 Pearson Education, Inc. The Addition Rule Two couples try the Valentines Day discount. For each: E(X) = $5.83 and SD(X) = $8.62. What is the combined expected value and SD? Let the total discount be. Notice that since the variables are independent, the standard deviation, $12.19, is less than the standard deviation, $17.24, of the double discount.

24. 24 Copyright © 2014, 2012, 2009 Pearson Education, Inc. The Subtraction Rule The rule for subtraction only makes sense after a moment of thought. Roll two dice. Each die outcome will have the same SD. Subtracting the standard deviations gives 0, but the standard deviation of the differences is not 0. The range of the differences is −5 to 5, larger than the range for a single die: 1 to 6.

25. 25 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Subtracting Discounts A competing restaurant also has a game style discount: E(X) = $10, SD = $15. How much more can you expect to save compared with the Valentines Day Discount: E(X) = $5.83, SD = $8.62? E(W – X) = 10 – 5.83 = $4.17 What is the standard deviation of the differences? The competing restaurant’s discount averages $4.17 more than the Valentines Day Discount. The standard deviation for the difference is $17.30.

26. 26 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Adding Random Variables The probability model for X + X + X is not the same as the model for 3X. If X can be 0 or 5, then 3X can be 0 or 15, while X + X + X can be 0, 5, 10, or 15. It is better to write X 1 + X 2 + X 3.

27. 27 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Valentine’s Day Discount for 40 Couples. Find the expected value and standard deviation for the total. The restaurant can expect a total discount of $233.20 and a standard deviation of $54.52.

28. 28 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 14.4 The Binomial Model

29. 29 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 29 Bernoulli Trials The basis for the Binomial probability model we will examine in this chapter is the Bernoulli trial. We have Bernoulli trials if: there are two possible outcomes (success and failure). the probability of success, p, is constant. the trials are independent. Examples: – Flipping a coin (where heads is success) – rolling a die (where getting a “6” is success) – throwing free throws in a basketball game – drawing a card from a deck of cards with replacement (where drawing an Ace is success)

30. 30 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Do we have Bernoulli Trials? You are rolling 5 dice and need to get at least two 6’s to win the game We record the eye colors found in a group of 500 people A city council of 11 Republicans and 8 Democrats picks a committee of 4 at random. What is the probability that they choose all Democrats? A 2002 Rutgers University study found that 74% of high school students have cheated on a test at least once. Your local high school principle conducts a survey and gets responses that admit to cheating from 322 of 481 students. How likely is it that in a group of 120 the majority may have type A blood, given that Type A is found in 43% of the population? Slide 1- 30

31. 31 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 31 The Geometric Model – (just an application of the multiplication rule) A single Bernoulli trial is usually not all that interesting. A Geometric probability model tells us the probability for a random variable that counts the number of Bernoulli trials until the first success. Example: lets draw cards from a standard deck with replacement and consider drawing a heart “success.” – Do we have Bernoulli trials? Would we have Bernoulli trials if we were drawing without replacement? – What is the probability p of success? What is the probability q of failure? – What is the probability that the first heart is the 3 rd card drawn? ○ i.e. first success occurs on trial 3. ○ This is the P(fail)*P(fail)*P(succeed) = 3/4 * 3/4 * 1/4

32. 32 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 32 The Binomial Model A Binomial model tells us the probability of having x successes in n Bernoulli trials. Example: If success is drawing a heart (drawing with replacement), what is the probability that if we draw and replace 3 cards that we drew exactly one heart? Two parameters define the Binomial model: n, the number of trials; and, p, the probability of success. We denote this Binom(n, p). Example: If we flip a coin 6 times what is the probability of getting heads exactly 3 times?

33. 33 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 33 The Binomial Model (cont.) In n trials, there are ways to have k successes. Read n C k as “n choose k,” and is called a combination. Example: How many ways are there to roll a die five times and roll a 6 three of those times? Note: n! = n x (n – 1) x … x 2 x 1, and n! is read as “n factorial.”

34. 34 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 34 The Binomial Model (cont.)

35. 35 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Searching for Hope Solo’s Card 20% of the cereal boxes have Hope’s card. What is the expected number of boxes to open to get a Hope card? Bernoulli Trial Two outcomes: success or failure The probability of success, p, is the same for each trial. The trials are independent.

36. 36 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Probability of Getting 2 Hopes in 5 Trials (by hand) Bernoulli trials: Millions of boxes, sample size 5. P(X = 2) from Binom(n, p), n = 5, p = 0.2, q = 0.8. 2 successes, 3 failures. No quite 0.2 2 × 0.8 3. Must consider all orders of 2 successes and 3 failures. Number of ways of picking k items from n: P(X = 2) = 10 × 0.2 2 × 0.8 3 = 0.2048

37. 37 Copyright © 2014, 2012, 2009 Pearson Education, Inc. The Binomial Model Summary n = Number of trials p = Probability of success q = 1 – p = Probability of failure X = Number of successes Mean = np Standard Deviation =

38. 38 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 38 Using the TI-83/84 Compute the P(X=x) using: Distr  binompdf(n, p, x) Compute the P(X<=x) using: Distr  binomcdf(n, p, x) This is a cumulative probability and is the sum of P(X=0) + P(X=1) + … + P(X=x)

39. 39 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 39 Using the TI-83/84 -- Practice Suppose a light bulb company has a 20% defective rate. Consider taking a sample of 6 bulbs. 1) What is the probability of getting exactly 1 defective bulb in that group of 6? (Even though a defect isn't pleasant at times, it is considered a success in this experiment since that is where our focus is!) If we computed this probability long hand we would do 6 C 1 (.20) 1 (.80) 5 On the calculator: 2nd Distr...(#0) for binompdf(6,.2, 1)... enter to get.393216 binompdf gives you the probability at a particular x. The pdf must be followed by n (total number of trials), p (probability of a success), x (number of successes you are interested in) binomcdf, which we use next, will compute cumulative probabilities.

40. 40 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 40 Using the TI 2) What is the probability of getting at most 2 defective light bulbs? This means P(0) + P(1) + P(2) OR use the cumulative binomial button: 2nd Distr...#A for binomcdf(6,.2, 2)...enter to get.90112 3) What is the probability of getting at least two defective light bulbs? At least two means two or more which is the same as adding the probabilities of 2 to 3 to 4 etc... OR 1 minus the complement of "at least two" which is 1 minus the cdf to 1 1 - binomcdf (6,.2,1) =.34464 4) What is the probability of getting from two to four defective light bulbs? You could do the pdf for 2 + pdf for 3 + pdf for 4 or be a little creative and do binomcdf(6,.2,4) - binomcdf(6,.2,1) =.34304

41. 41 Copyright © 2014, 2012, 2009 Pearson Education, Inc. The Binomial Model StatCrunch: Stat → Calculators → Binomial Fill in n, p, x. Decide on the inequality. Compute.

42. 42 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Using StatCrunch To use StatCrunch to calculate Binomial Probabilities (or to view the binomial probability histogram) go to Stat -> Calculators -> Binomial Enter the appropriate n, p, and Prob statement. Then click "Calculate“ – For example, if you want to compute the probability of observing at least one "6" in 5 rolls of the die, – n = 5 – p = 0.1667 – Prob (X=>1) = 0.598 Slide 1- 42

43. 43 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Practice 20) An Olympic archer is able to hit the bull’s-eye 80% of the time. Assume each shot is independent of the others. If she shoots 6 arrows, what’s the probability of the following Her first bull’s-eye comes on the 3 rd arrow (note: this uses the Geometric not the Binomial Distribution) She misses the bull’s-eye at least once Her first bull’s-eye comes on the fourth or fifth arrow (Geometric) She gets exactly 4 bull’s-eyes She gets at least 4 bull’s-eyes She gets at most 4 bull’s-eyes Slide 1- 43

44. 44 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Notice that when the Binomial Model applies we are given E(X) and SD(X)… n = Number of trials p = Probability of success q = 1 – p = Probability of failure Mean = np Standard Deviation =

45. 45 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 17) If you flip a fair coin 100 times Intuitively how many heads do you expect? Use the formula for expected value to verify your intuition 18) An American roulette wheel has 38 slots, of which 18 are red, 18 are black, and 2 are green. If you spin the wheel 38 times Intuitively how many times do you expect the ball to land in a green slot? Use the formula for expected value to verify your intuition Slide 1- 45

46. 46 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Consider the same archer… How many Bull’s-eyes do you expect her to get? With what standard deviation? Suppose our archer shoots 10 arrows Find the mean and standard deviation of the number of bull’s- eyes you may get What’s the probability that she never misses? What the probability that there are no more than 8 bull’s-eyes What’s the probability that there are exactly 8 bull’s-eyes What’s the probability that she hits the bull’e-eye more often than she misses Slide 1- 46

47. 47 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Practice 6) Suppose 75% of all drivers always wear their seatbelts. Lets investigate how many of the drivers might be belted among six cars waiting at a traffic light. Describe how you’ll simulate the number of seatbelt wearing drivers among the six cars Run 30+ trials Based on the simulation estimate the probabilities that there are exactly no belted drivers, one, two, three, etc. Calculate the actual probability model Compare the distribution of outcomes in the simulation to the actual model Slide 1- 47

48. 48 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Spam and the Binomial Model 91% of all email is spam. Your inbox has 25 emails. Find the mean, standard deviation, and the probability that 1 or 2 of the emails are not spam. n = 25, p = 1 – 0.91 = 0.09, q = 0.91 Mean: np = (25)(0.09) = 2.25 Standard Deviation = P(X = 1 or X = 2) = P(X =1) + P(X = 2) = 0.2340 + 0.2777 = 0.5117 There is about a 51% chance of 1 or 2 emails that are not spam.

49. 49 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Universal Blood Donors 6% of all people are O −. If 20 donate at the blood drive, find the mean and standard deviation and find the probability that 2 or 3 are O −. Think →Plan: Either success (O − ) or Failure (not O − ) p = 0.06, independent Therefore binomial Variable: X = no. of O −, n = 20 people Model: X is Binom(20, 0.06)

50. 50 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Universal Blood Donors 6% of all people are O −. If 20 donate at the blood drive, find the mean and standard deviation and find the probability that 2 or 3 are O −. Show →Mechanics: Mean = np = (20)(0.06) = 1.2 SD = Use StatCrunch to find P(X = 2 or 3) = P(X = 2) + P(X = 3) = 0.2246 + 0.0860 = 0.3106

51. 51 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Universal Blood Donors 6% of all people are O −. If 20 donate at the blood drive, find the mean and standard deviation and find the probability that 2 or 3 are O −. Tell → Conclusion: In Groups of 20 randomly selected blood donors, I expect to find an average of 1.2 universal blood donors with a standard deviation of 1.06. There is about a 31% chance that 2 or 3 of the 20 donors are O −.

52. 52 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 14.5 Modeling the Binomial with the Normal Model

53. 53 Copyright © 2014, 2012, 2009 Pearson Education, Inc. The Trouble with Large Sample Sizes 6% of all people are O −. The Tennessee Red Cross has 32,000 donors and needs at least 1850 that are O −. Will they run out? The computations involve ridiculously large numbers. “At least” requires P(X = 1850), P(X = 1851), all the way up to P(X = 32,000). Mean = np = 1920

54. 54 Copyright © 2014, 2012, 2009 Pearson Education, Inc. The Solution for Large Sample Sizes The Tennessee Red Cross has 32,000 donors and needs at least 1850 that are O −. Will they run out (less than)? Mean = np = 1920 The normal model with the same mean and standard deviation is a very good approximation. There is about a 5% chance that they will run out.

55. 55 Copyright © 2014, 2012, 2009 Pearson Education, Inc. How Large is “Large Enough” The Success/Failure Condition A Binomial is approximately Normal if we expect at least 10 successes and 10 failures. np  10, nq  10 This comes from the binomial being skewed for a small number of successes or failures expected.

56. 56 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Example: Spam and the Normal Approximation to the Binomial Only 151 of 1422 emails got through your spam filter. Might the filter be too aggressive? What is the probability that no more than 151 of the emails are real messages? These emails represent less than 10% of all emails. np = (1422)(0.09) = 127.98  10 nq = (1422)(0.91) = 1294.02  10 Yes, the Normal model is a good approximation.

57. 57 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Example: Spam and the Normal Approximation to the Binomial What is the probability that no more than 151 of the emails are real messages?  = np = 127.98 There is over a 98% chance that no more than 151 of them were real messages. The filter may be working.

58. 58 Copyright © 2014, 2012, 2009 Pearson Education, Inc. If your candidate is favored by approximately 53% of the population, but only 100 people vote, what is the probability that your candidate wins? In other words, your candidate needs at least 51 votes of 100 votes. Assume each voter is independent. Do we have Bernoulli trials? What is the probability of “success” for a trial? What the probability that at least 51 of 100 voters vote for your candidate? Slide 1- 58

59. 59 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 59 Approximating with the Normal Model When we use the Normal model to approximate the Binomial model, we are using a continuous random variable to approximate a discrete random variable. So, when we use the Normal model, we no longer calculate the probability that the random variable equals a particular value, but only that it lies between two values. Ex. For our election example: μ = np = 100*.53 =53 σ = √(npq) = √(100*.53*.47) =4.99 P(at least 51 votes) ~ Normalcdf(51,100, 53, 4.99)

60. 60 Copyright © 2014, 2012, 2009 Pearson Education, Inc. What Can Go Wrong? Probability models are still just models. They just approximate reality. Always question whether the model is accurate. If the model is wrong, so is everything else. Check the mean and standard deviation for reasonableness. Be sure that the probabilities sum to 1. Don’t assume you have Bernoulli trials without checking. Check for only 2 possible outcomes, independence, consistency of probabilities, and the 10% rule.

61. 61 Copyright © 2014, 2012, 2009 Pearson Education, Inc. What Can Go Wrong? Don’t assume everything is Normal. Check there the distribution is not skewed and that there is a bell shape. Don’t use the normal approximation for small n. np> 10, nq > 10 Watch out for variables that are not independent. The arithmetic rules do not work for variables that are not independent. Always check for independence.