Probability flow chart Are there only two outcomes possible (success/fail)? Bernoulli trials: Are we finding quantity of successes or location of first success? location Geometric Probability 1 = = 2 quantity Binomial Probability = = Yes No Continued on next slide Are there only two outcomes possible (success/fail)? No Is it a question involving an “and/or”? No No Simple Probability Is it a question involving a “given”? Yes Are the events disjoint? No Yes Draw a tree diagram if possible ( ∩ ) = () Draw a Venn diagram Yes ∪ = + − ( ∩ ) ∩ = () ∙ (|) ∪ = + () ∩ = () ∙ (|) Practice FR #1 You are up for your annual job performance review. You estimate that there’s a 30% chance you’ll get a promotion, a 40% chance of a raise, and a 25% chance of getting both a promotion and a raise. a)Find the probability that you get a raise or a promotion. b)Find the probability that you got neither. c)Are the raise and the promotion independent events? d)Are the raise and the promotion disjoint events? a) 0.45 or 45% Promotion Raise b) 0.55 or 55% c) The events are not 0.05 0.25 independent because 0.15 P(Pr|R)=0.25/0.4 0.55 which isn’t equal to P(Pr) of 0.3. d) The events are not disjoint since it is possible to get both a promotion and a raise. Practice FR #2 Assume that 75% of the AP Stat students studied for this test. If 40% of those who study get an A, but only 10% of those who don’t study get an A, what is the probability that someone who gets an A actually studied for the test? A ∩ = 0.75 0.4 = 0.3 0.6 Not A ∩ = 0.75 0.6 = 0.45 0.1 A ∩ = 0.25 0.1 = 0.025 0.9 Not A ∩ = 0.25 0.9 = 0.225 0.4 Yes 0.75 0.25 No ( ∩ ) 0.3 = = = .923 () 0.3 + 0.025 Practice FR #3 Assume the heights of high school basketball players are normally distributed. For girls the mean is 70 inches with a standard deviation of 0 inches, while boy players have a mean height of 74 inches with a standard deviation of 4.5 inches. a) On average, how much taller is a boy player than a girl player? b) What will be the standard deviation of the difference in heights between a boy player and a girl player? c) On what percentage of the team would you expect the girl to be taller than the boy? a) E(B-G) = E(B)-E(G) = 74-70 = 4 inches b) SD B − G = (4.5)2 (3)2 = 5.4 inches c) Let D = B-G and we want the probability that a girl is taller than a boy, so B-G<0. P(D<0) = P( < 0−4 ) 5.4 = P(z < -0.7407) = 0.2294 We would expect about 23% of the time, a girl will be taller than a boy. Practice FR #4 The National Association of Retailers reports that 62% of all purchases are now made by credit card. You think this is true at your store as well. On a typical day you make 20 sales. a) Explain why your sales can be considered Bernoulli trials. b) What is the probability that your 4th customer is the first one who uses a credit card? c) What is the probability model for the number of customers who use a credit card on a typical day? d) What is the probability that on a typical day at least half of your customers use a credit card? a)Bernoulli trials have only two possible outcomes (success = used credit card; failure = doesn’t use credit card). Trials are independent (one transaction does not influence the next transaction). The probability of success stays the same on every trial (62% of all purchases). b)P(X=4) =0.034 c)Model: Binom(20, 0.62) with mean = 12.4 and standard deviation = 2.17 d)P X ≥ 10 = 1 − P X = 0 + P X = 1 + Practice FR #5 New York public health officials report that currently 22% of adults smoke. They hope that newly increased state cigarette taxes will reduce this rate. They plan to check in December by selecting a random sample of 1200 New Yorkers to estimate again the percentage of adults who smoke. a)Verify that a Normal model is a useful approximation for the binomial in this situation. b)In that December sample, how many smokers would it take to convince you that the percentage of NY adults who smoke had decreased significantly? Explain. a)np = 1200 0.22 = 264 ≥ 10 nq = (1200)(0.78) = 936 ≥ 10 Also, 1200 < 10% of all New Yorkers. (The population size isn’t given, but it’s BIG! Much bigger than 12,000.) b)It would be unusual to see the number of adult smokers be lower than 2 standard deviations below the mean. The standard deviation is 14.35 and the mean is 264, so 264 – 2(14.35) = 235.3. I would conclude that the estimate of 22% of adults smoke had decreased significantly if there were fewer than 235 smokers in the December sample of 1200 NY adults.