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UW-Madison STAT 301 - 4.1b - Discrete (Binomial, Poisson)

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Slide 1Slide 2~ The Binomial Distribution ~Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Example: Blood Type probabilities, revisitedExample: Blood Type probabilities, revisitedExample: Blood Type probabilities, revisitedExample: Blood Type probabilities, revisitedExample: Blood Type probabilities, revisitedExample: Blood Type probabilities, revisitedSlide 18Example: Deaths in WisconsinExample: Deaths in WisconsinClassical Discrete Probability DistributionsCHAPTER 4• 4.1 - Discrete Models General distributions Classical: Binomial, Poisson, etc.• 4.2 - Continuous Models General distributions Classical: Normal, etc.POPULATIONDiscrete random variable X Examples: shoe size, dosage (mg), # cells,… Pop values xProbabilitiesf (x)Cumul ProbsF (x)x1f (x1) f(x1) x2f (x2) f(x1) + f(x2) x3f (x3) f(x1) + f(x2) + f(x3)⋮ ⋮⋮1Total 1all 2 2all ( )( ) ( )xxx f xx f xms m== -��MeanVarianceXTotal Area = 1Recall…Recall…~ The Binomial Distribution ~~ The Binomial Distribution ~Used only when dealing with binary outcomes (two categories: “Success” vs. “Failure”), with a fixed probability of Success () in the population.Calculates the probability of obtaining any given number of Successes in a random sample of n independent “Bernoulli trials.”Has many applications and generalizations, e.g., multiple categories, variable probability of Success, etc.4For any randomly selected individual, define a binary random variable:pp=�=�- =�1 if Male, with prob 0.40 if Female, with prob 1 0.6YPOPULATION40% Male, 60% FemaleRANDOMSAMPLE n = 100Discrete random variableX = # Males in sample(0, 1, 2, 3, …, 99, 100)How can we calculate the probability ofP(X = 0), P(X = 1), P(X = 2), …, P(X = 99), P(X = 100)?P(X = x), for x = 0, 1, 2, 3, …,100?f(x) =F(x) = P(X ≤ x), for x = 0, 1, 2, 3, …,100?How can we calculate the probability ofP(X = x), for x = 0, 1, 2, 3, …,100?f(x) =x f (x)x1f (x1) x2f (x2) x3f (x3) ⋮ ⋮1P(X = x), for x = 0, 1, 2, 3, …,100?f(x) =f(25) = P(X = 25)?How can we calculate the probability ofF(x) = P(X ≤ x), for x = 0, 1, 2, 3, …,100?5For any randomly selected individual, define a binary random variable:pp=�=�- =�1 if Male, with prob 0.40 if Female, with prob 1 0.6YPOPULATION40% Male, 60% FemaleRANDOMSAMPLE n = 100Discrete random variableX = # Males in sample(0, 1, 2, 3, …, 99, 100)Example:Solution: Model the sample as a sequence of independent coin tosses, with 1 = Heads (Male), 0 = Tails (Female), where P(H) = 0.4, P(T) = 0.6Solution:.… etc….permutations of 25 among 100 …etc…etc…etc…There are 100 possible open slots for H1 to occupy.X = 25 Heads: { H1, H2, H3,…, H25 }For each one of them, there are 76 possible open slots left for H25 to occupy.How many possible outcomes of n = 100 tosses exist with X = 25 Heads?1 2 3 4 5. . . . . . 97 98 99 100. . . . . .For each one of them, there are 99 possible open slots left for H2 to occupy.For each one of them, there are 98 possible open slots left for H3 to occupy.For each one of them, there are 77 possible open slots left for H24 to occupy.Hence, there are ?????????????????????? possible outcomes. 100  99  98  …  77  76 How many possible outcomes of n = 100 tosses exist?1002… This value is the number of permutations of the coins, denoted 100P25. HOWEVER…permutations of 25 among 100 How many possible outcomes of n = 100 tosses exist with X = 25 Heads?1 2 3 4 5. . . . . . 97 98 99 100. . . . . .How many possible outcomes of n = 100 tosses exist?1002HOWEVER…100  99  98  …  77  76 X = 25 Heads: { H1, H2, H3,…, H25 }This number unnecessarily includes the distinct permutations of the 25 among themselves, all of which have Heads in the same positions.For example:. . . . . .. . . . . .. . . . . .We would not want to count these as distinct outcomes.permutations of 25 among 100 How many possible outcomes of n = 100 tosses exist with X = 25 Heads?1 2 3 4 5. . . . . . 97 98 99 100. . . . . .How many possible outcomes of n = 100 tosses exist?1002HOWEVER…100  99  98  …  77  76 X = 25 Heads: { H1, H2, H3,…, H25 }This number unnecessarily includes the distinct permutations of the 25 among themselves, all of which have Heads in the same positions.How many is that? By the same logic…...25  24  23  …  3  2  1 “25 factorial” - denoted 25!“100-choose-25” - denoted or 100C25 This value counts the number of combinations of 25 Heads among 100 coins.� �� �� �10025100  99  98  …  77  7625  24  23  …  3  2  1 100!_25! 75!=100 nCr 25 on your calculator.1 2 3 4 5. . . . . . 97 98 99 100. . . . . .How many possible outcomes of n = 100 tosses exist with X = 25 Heads?What is the probability of each such outcome?0.4 0.6 0.6 0.4 0.6 . . . . . . 0.6 0.4 0.4 0.6Answer: Via independence in binary outcomes between any two coins,0.4  0.6  0.6  0.4  0.6  …  0.6  0.4  0.4  0.6 = .25 75(0.4) (0.6)Therefore, the probability P(X = 25) is equal to…….� �� �� �1002525 75(0.4) (0.6)How many possible outcomes of n = 100 tosses exist?1002Question: What if the coin were “fair” (unbiased), i.e.,  = 1 –  = 0.5 ? Answer:� �� �� �10025Recall that, per toss, P(Heads) =  = 0.4 P(Tails) = 1 –  = 0.6Recall that, per toss, P(Heads) =  = 0.4 P(Tails) = 1 –  = 0.6 How many possible outcomes of n = 100 tosses exist with X = 25 Heads?1 2 3 4 5. . . . . . 97 98 99 100. . . . . .What is the probability of each such outcome?0.4 0.6 0.6 0.4 0.6 . . . . . . 0.6 0.4 0.4 0.6Answer: Via independence in binary outcomes between any two coins,0.4  0.6  0.6  0.4  0.6  …  0.6  0.4  0.4  0.6 = .25 75(0.4) (0.6)Therefore, the probability P(X = 25) is equal to…….� �� �� �1002525 75(0.4) (0.6)How many possible outcomes of n = 100 tosses exist?1002Question: What if the coin were “fair” (unbiased), i.e.,  = 1 –  = 0.5 ? Answer:� �� �� �100250.5 0.5 0.5 0.5 0.5 . . . . . . 0.5 0.5 0.5 0.50.5  0.5  0.5  0.5  0.5  …  0.5  0.5  0.5  0.5 = 100(0.5) = 0.5 1 –  = 0.5100(0.5)100(1/ 2)100 2This is the “equally likely”


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UW-Madison STAT 301 - 4.1b - Discrete (Binomial, Poisson)

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