Continuous distributions. Lecture 6: Probability. Probabilities from continuous distributions. From histograms to continuous distributions

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1 Lecture 6: Probability Below is a histogram of the distribution of heights of US adults. The proportion of data that falls in the shaded bins gives the probability that a randomly sampled US adult is between 180 cm and 185 cm (about 5 11 to 6 1 ). Statistics 101 Mine Çetinkaya-Rundel September 15, height (cm) Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 From histograms to continuous distributions From histograms to continuous distributions Since height is a continuous numerical variable its probability density function is a smooth curve. Probabilities from continuous distributions Probabilities from continuous distributions Therefore, the probability that a randomly sampled US adult is between 180 cm and 185 cm can also be estimated as the shaded area under the curve height (cm) height (cm) Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24

2 By definition... Probabilities from continuous distributions Since continuous probabilities are estimated as the area under the curve, the probability of a person being exactly 180 cm (or any exact value) is defined as 0. Are s more likely to use marijuana when their used drugs? The table below gives the distribution of marijuana use in 445 college s based whether or not their used drugs. Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Marginal probability Joint probability What is the probability that the uses marijuana? What about the probability that a used drugs? P( uses) = = 0.49 P( used) = = 0.47 marginal probability Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 What is the probability that both the child and the used drugs? P( uses and used) = = 0.28 joint probability Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24

3 (cont.) If used drugs, what is the probability that their child uses marijuana? We know that 47% of the sample is comprised of who used drugs. We also know that 28% of the sample is comprised of s and who used drugs. We re interested in finding out what percent of 47% is 28%? P( uses given used) = P( uses used) = (Bayes Theorem) P(A B) = P(A and B) P(B) conditional probability Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 - a simpler approach If used drugs, what is the probability that their child uses marijuana? P( uses used) = = 0.60 At a large apartment complex, 58% of the units have a washer and dryer, 32% have double parking, and 20% have both washer and dryer and double parking. A unit with double parking just became available at this apartment complex, what is the probability that it also has washer and dryer? (a) 0.20 (b) (c) (d) Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24

4 General multiplication rule (graded) At a large apartment complex, 58% of the units have a washer and dryer, 32% have double parking, and 20% have both washer and dryer and double parking. What percent of apartments have neither double parking nor washer and dryer? (a) 0.10 (b) (c) 0.20 (d) 0.30 Earlier we ve seen the multiplication rule for independent events: If A and B are independent, P(A and B) = P(A) P(B) Just now we saw Bayes theorem for calculating conditional probabilities: P(A and B) P(A B) = P(B) Bayes theorem doesn t require that the events be independent. By rearranging the above formula we obtain the following: General multiplication rule P(A and B) = P(A B) P(B) Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Let s reorganize those smokers... If events A and B are independent, which of the below is correct? (a) P(A B) = P(A and B) (b) P(A B) = P(A) (c) P(A B) = P(B) (d) P(A B) = 0 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24

5 Another look at those smokers... Tree diagrams P( uses) = = 0.49 P( doesn t use) = = 0.51 P( used uses) = = 0.57 P( used doesn t use) = = 0.38 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Tree diagrams (cont.) Are you next? Sampling from a small population If used drugs, what is the probability that their child uses marijuana? Have you taken AP statistics in high school? (a) Yes (b) No P( uses and used) P( uses used) = P( used) 0.28 = = Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24

6 Sampling from a small population Sampling from a small population Are you next? 1 What is the probability that a randomly called on in class has taken AP statistics in high school? 2 If sampling without replacement, what s the probability that the next randomly called on in class has taken AP statistics in high school? 3 If sampling without replacement, what s the probability that the next randomly called on in class has not taken AP statistics in high school? What s the probability of being dealt the following hand from a well shuffled full deck: first two hearts in a row, and then a black card? (a) (b) (c) (d) Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Let s bet... Random variables Random variables Expected value and standard deviation According to wunderground.com there s 20% chance of rain tonight in Durham. I ll bet you $5 that it will rain tonight. That means, if it rains I win, and you give me $5. If it doesn t rain, you win, and I give you $5. What s your expected winnings from this game? Event Winnings Probability Winnings Probability Rain $ $ = 1 Doesn t rain $ $ = 4 Total = $3 probability model expected value E(X ) Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24 Let X = winnings. X P(X ) X P(X ) (X E(X )) 2 (X E(X )) 2 P(X ) = 1 ( 5 3) 2 = = = 4 (5 3) 2 = = 3.2 E(X ) = 3 V (X ) = 16 SD(X ) = 16 = 4 Your expected winnings is $3, give or take $4, i.e. if we played this game many times, on average you would win $3, give or take $4. Should you play this game? Should I? Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24

7 Random variables Expected value and standard deviation (cont.) Expected value of a random variable n E(X ) = X i P(X i ) i Standard deviation of a random variable SD(X ) = n (X i E(X )) 2 P(X i ) i Statistics 101 (Mine Çetinkaya-Rundel) Lecture 6: Probability September 15, / 24