Chapter 3 Discrete Random Variables and Probability Distributions 3.2 - Probability Distributions for Discrete Random Variables 3.3 - Expected Values 3.4 - The Binomial Probability Distribution 3.5 - Hypergeometric and Negative Binomial Distributions 3.6 - The Poisson Probability Distribution

Discrete random variable X Examples: shoe size, dosage (mg), # cells,… Recall… POPULATION Discrete random variable X Examples: shoe size, dosage (mg), # cells,… Pop values x Probabilities p(x) Cumul Probs F (x) x1 p(x1) x2 p(x2) p(x1) + p(x2) x3 p(x3) px1) + p(x2) + p(x3) ⋮ 1 Total X Total Area = 1 Mean Variance

Classical Discrete Probability Distributions Binomial ~ X = # Successes in n trials, P(Success) =  Poisson ~ As above, but n large,  small, i.e., Success RARE Negative Binomial ~ X = # trials for k Successes, P(Success) =  Geometric ~ As above, but specialized to k = 1 Hypergeometric ~ As Binomial, but  changes between trials Multinomial ~ As Binomial, but for multiple categories, with 1 + 2 + … + last = 1 and x1 + x2 + … + xlast = n

~ 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.

For x = 0, 1, 2, 3, …, n X = # Successes in sample x p(x) p(0) 1 p(1) For any randomly selected individual, define a binary random variable: POPULATION Discrete random variable X = # Successes in sample (x = 0, 1, 2, 3, …, n) RANDOMSAMPLE size n x p(x) p(0) 1 p(1) … n p(n) F(x) F(0) F(1) … 1 For x = 0, 1, 2, 3, …, n

Reformulate this as n independent coin tosses. For any randomly selected individual, define a binary random variable: POPULATION Discrete random variable X = # Heads in sample (x = 0, 1, 2, 3, …, n) Reformulate this as n independent coin tosses. RANDOMSAMPLE size n .… etc…. .… etc…. Each such sequence has probability There are such sequences of n tosses, with x Heads.

X = # Successes in sample For any randomly selected individual, define a binary random variable: POPULATION “Success” vs. “Failure” RANDOM SAMPLE of n “Bernoulli trials.” Discrete random variable X = # Successes in sample (x = 0, 1, 2, 3, …, n) independent, with constant probability () per trial Then X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability mass function” p(x) = , x = 0, 1, 2, …, n.

Example: Blood Type probabilities, revisited Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Check: 1. Independent outcomes? Reasonably assume that outcomes “Type O” vs. “Not Type O” between two individuals are independent of each other.  2. Constant probability  ? Suppose n = 10 individuals are to be selected at random from the population. Probability table for X = #(Type O) From table,  = P(Type O) = .461 throughout population.  Binomial model applies?

Example: Blood Type probabilities, revisited p(x) = (.461)x (.539)10 – x Example: Blood Type probabilities, revisited R: dbinom(0:10, 10, .461) Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 x p(x) F (x) (.461)0 (.539)10 = 0.00207 0.00207 1 (.461)1 (.539)9 = 0.01770 0.01977 2 (.461)2 (.539)8 = 0.06813 0.08790 3 (.461)3 (.539)7 = 0.15538 0.24328 4 (.461)4 (.539)6 = 0.23257 0.47585 5 (.461)5 (.539)5 = 0.23870 0.71455 6 (.461)6 (.539)4 = 0.17013 0.88468 7 (.461)7 (.539)3 = 0.08315 0.96783 8 (.461)8 (.539)2 = 0.02667 0.99450 9 (.461)9 (.539)1 = 0.00507 0.99957 10 (.461)10 (.539)0 = 0.00043 1.00000 Suppose n = 10 individuals are to be selected at random from the population. Probability table for X = #(Type O) Binomial model applies. X ~ Bin(10, .461) Also, can show mean  =  x p(x) = and variance  2 =  (x – ) 2 p(x) = n = 4.61 = (10)(.461) n (1 – ) = 2.48

Example: Blood Type probabilities, revisited p(x) = (.461)x (.539)10 – x Example: Blood Type probabilities, revisited R: dbinom(0:10, 10, .461) Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 x p(x) F (x) (.461)0 (.539)10 = 0.00207 0.00207 1 (.461)1 (.539)9 = 0.01770 0.01977 2 (.461)2 (.539)8 = 0.06813 0.08790 3 (.461)3 (.539)7 = 0.15538 0.24328 4 (.461)4 (.539)6 = 0.23257 0.47585 5 (.461)5 (.539)5 = 0.23870 0.71455 6 (.461)6 (.539)4 = 0.17013 0.88468 7 (.461)7 (.539)3 = 0.08315 0.96783 8 (.461)8 (.539)2 = 0.02667 0.99450 9 (.461)9 (.539)1 = 0.00507 0.99957 10 (.461)10 (.539)0 = 0.00043 1.00000 Suppose n = 10 individuals are to be selected at random from the population. Probability table for X = #(Type O) Binomial model applies. X ~ Bin(10, .461) Also, can show mean  =  x p(x) = and variance  2 =  (x – ) 2 p(x) = n = 4.61 n (1 – ) = 2.48

n = 10 p = .461 pmf = function(x)(dbinom(x, n, p)) N = 100000 x = 0:10 bin.dat = rep(x, N*pmf(x)) hist(bin.dat, freq = F, breaks = c(-.5, x+.5), col = "green") axis(1, at = x) axis(2)

Example: Blood Type probabilities, revisited Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Therefore, p(x) = x = 0, 1, 2, …, 1500. RARE EVENT! Suppose n = 10 individuals are to be selected at random from the population. Probability table for X = #(Type AB–) n = 1500 individuals are to Binomial model applies. X ~ Bin(10, .461) X ~ Bin(1500, .007) Also, can show mean  =  x p (x) = and variance  2 =  (x – ) 2 p(x) = n = 10.5 n (1 – ) 2.48 = 10.43

Example: Blood Type probabilities, revisited Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Therefore, p(x) = x = 0, 1, 2, …, 1500. Is there a better alternative? RARE EVENT! RARE EVENT! Suppose n = 10 individuals are to be selected at random from the population. Probability table for X = #(Type AB–) n = 1500 individuals are to Long positive skew as x  1500 …but contribution  0 Binomial model applies. X ~ Bin(10, .461) X ~ Bin(1500, .007) Also, can show mean  =  x p (x) = and variance  2 =  (x – ) 2 p(x) = n = 10.5 n (1 – ) 2.48 = 10.43

Example: Blood Type probabilities, revisited Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Therefore, p(x) = x = 0, 1, 2, …, 1500. Is there a better alternative? Poisson distribution x = 0, 1, 2, …, where mean and variance are  = n and  2 = n RARE EVENT!  Suppose n = 10 individuals are to be selected at random from the population. Probability table for X = #(Type AB–) n = 1500 individuals are to = 10.5 Binomial model applies. X ~ Bin(1500, .007) X ~ Poisson(10.5) Also, can show mean  =  x p(x) = and variance  2 =  (x – ) 2p(x) = n = 10.5 Notation: Sometimes the symbol  (“lambda”) is used instead of  (“mu”). n (1 – ) = 10.43

Example: Blood Type probabilities, revisited Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Rh Factor Blood Type + – O .384 .077 .461 A .323 .065 .388 B .094 .017 .111 AB .032 .007 .039 .833 .166 .999 Therefore, p(x) = x = 0, 1, 2, …, 1500. Is there a better alternative? Poisson distribution x = 0, 1, 2, …, where mean and variance are  = n and  2 = n RARE EVENT! Suppose n = 1500 individuals are to be selected at random from the population. Probability table for X = #(Type AB–) = 10.5 X ~ Poisson(10.5) Ex: Probability of exactly X = 15 Type(AB–) individuals = ? Poisson: Binomial: (both ≈ .0437)

Example: Deaths in Wisconsin

Example: Deaths in Wisconsin Assuming deaths among young adults are relatively rare, we know the following: Average 584 deaths per year λ = Mortality rate (α) seems constant. Therefore, the Poisson distribution can be used as a good model to make future predictions about the random variable X = “# deaths” per year, for this population (15-24 yrs)… assuming current values will still apply. Probability of exactly X = 600 deaths next year P(X = 600) = 0.0131 R: dpois(600, 584) Probability of exactly X = 1200 deaths in the next two years Mean of 584 deaths per yr  Mean of 1168 deaths per two yrs, so let λ = 1168: P(X = 1200) = 0.00746 Probability of at least one death per day: λ = = 1.6 deaths/day P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + … True, but not practical. P(X ≥ 1) = 1 – P(X = 0) = 1 – = 1 – e–1.6 = 0.798