4. Probability distributions - 4. Probability Distributions...

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4. Probability Distributions Probability : With random sampling or a randomized experiment, the probability an observation takes a particular value is the proportion of times that outcome would occur in a long sequence of observations. Usually corresponds to a population proportion (and thus falls between 0 and 1) for some real or conceptual population.
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Basic probability rules Let A, B denotes possible outcomes P(not A) = 1 – P(A) For distinct (separate) possible outcomes A and B, P(A or B) = P(A) + P(B) If A and B not distinct, P(A and B) = P(A)P(B given A) For “independent” outcomes, P(B given A) = P(B), so P(A and B) = P(A)P(B).
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Happiness (2008 GSS data) Income Very Pretty Not too Total ------------------------------- Above Aver. 164 233 26 423 Average 293 473 117 883 Below Aver. 132 383 172 687 ------------------------------ Total 589 1089 315 1993 Let A = above average income, B = very happy P(A) estimated by 423/1993 = 0.212 (a “marginal probability”), P(not A) = 1 – P(A) = 0.788 P(B) estimated by 589/1993 = 0.296 P(B given A) estimated by 164/423 = 0.388 (a “conditional prob.”) P(A and B) = P(A)P(B given A) estimated by 0.212(0.388) = 0.082 (which equals 164/1993, a “joint probability”)
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P(B) = 0.296, P(B given A) = 0.388, so A and B not independent. If A and B were independent, then P(A and B) = P(A)P(B) = 0.212(0.296) = 0.063 But, P(A and B) = 0.082. Inference questions: In this sample, A and B are not independent, but can we infer anything about whether events involving happiness and events involving family income are independent (or dependent) in the population? If dependent, what is the nature of the dependence? (does higher income tend to go with higher happiness?)
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Success of New England Patriots Before season, suppose we let A1 = win game 1 A2 = win game 2, … , A16 = win game 16 If Las Vegas odds makers predict P(A1) = 0.55, P(A2) = 0.45, then P(A1 and A2) = P(win both games) = (0.55)(0.45) = 0.2475, if outcomes independent If P(A1) = P(A2) = … = P(A16) = 0.50 and indep. (like flipping a coin), then P(A1 and A2 and A3 … and A16) = P(win all 16 games) = (0.50)(0.50)…(0.50) = (0.50) 16 = 0.000015.
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Probability distribution of a variable Lists the possible outcomes for the “random variable” and their probabilities Discrete variable : Assign probabilities P(y) to individual values y, with 0 ( ) 1, ( ) 1 P y P y Σ =
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Example: Randomly sample 3 people and ask whether they favor (F) or oppose (O) legalization of same-sex marriage y = number who “favor” (0, 1, 2, or 3) For possible samples of size n = 3, Sample y Sample y (O, O, O) 0 (O, F, F) 2 (O, O, F) 1 (F, O, F) 2 (O, F, O) 1 (F, F, O) 2 (F, O, O) 1 (F, F, F) 3
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If population equally split between F and O, these eight samples are equally likely. Probability distribution of
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4. Probability distributions - 4. Probability Distributions...

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