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CHAPTER 4

CHAPTER 4. 4.1 - Discrete Models General distributions Classical: Binomial, Poisson, etc. 4.2 - Continuous Models General distributions Classical: Normal, etc. Recall…. POPULATION. Discrete random variable X Examples : shoe size, dosage (mg), # cells,…. X.

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CHAPTER 4

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  1. CHAPTER 4 4.1 - Discrete Models General distributions Classical: Binomial, Poisson, etc. 4.2 - Continuous Models General distributions Classical: Normal, etc.

  2. Recall… POPULATION Discrete random variable X Examples: shoe size, dosage (mg), # cells,… X Mean Variance Total Area = 1

  3. ~ 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.

  4. For any randomly selected individual, define a binary random variable: POPULATION 40% Male, 60% Female RANDOMSAMPLE n = 100 Discrete random variable X = # Males in sample (0, 1, 2, 3, …, 99, 100) How can we calculate the probability of How can we calculate the probability of P(X = x), for x = 0, 1, 2, 3, …,100? f(x) = P(X = x), for x = 0, 1, 2, 3, …,100? P(X = 0), P(X = 1), P(X = 2), …, P(X = 99), P(X = 100)? f(x) = F(x) = P(X≤x), for x = 0, 1, 2, 3, …,100?

  5. For any randomly selected individual, define a binary random variable: POPULATION 40% Male, 60% Female RANDOMSAMPLE n = 100 Discrete random variable X = # Males in sample (0, 1, 2, 3, …, 99, 100) Example: How can we calculate the probability of f(25) = P(X = 25)? P(X = x), for x = 0, 1, 2, 3, …,100? f(x) = Solution: F(x) = P(X≤x), for x = 0, 1, 2, 3, …,100? 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) = 0.6 .… etc….

  6. How many possible outcomes of n = 100tosses exist? How many possible outcomes of n = 100tosses exist withX = 25Heads? … X = 25Heads: { H1, H2, H3,…, H25 } HOWEVER… permutations of 25 among 100 There are 100 possible open slots for H1 to occupy. 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. …etc…etc…etc… For each one of them, there are 77 possible open slots left for H24 to occupy. For each one of them, there are 76 possible open slots left for H25 to occupy. Hence, there are ?????????????????????? possible outcomes. 100  99  98  …  77  76 This value is the number of permutations of the coins, denoted 100P25.

  7. How many possible outcomes of n = 100tosses exist? How many possible outcomes of n = 100tosses exist withX = 25Heads? X = 25Heads: { H1, H2, H3,…, H25 } 100  99  98  …  77  76 HOWEVER… permutations of 25 among 100 This number unnecessarily includes the distinct permutations of the 25 among themselves, all of which have Heads in the samepositions. For example: We would not want to count these as distinct outcomes.

  8. How many possible outcomes of n = 100tosses exist? How many possible outcomes of n = 100tosses exist withX = 25Heads? X = 25Heads: { H1, H2, H3,…, H25 } 100  99  98  …  77  76 HOWEVER… permutations of 25 among 100 This number unnecessarily includes the distinct permutations of the 25 among themselves, all of which have Heads in the samepositions. 25  24  23  …  3  2  1 How many is that? By the same logic…... “25 factorial” - denoted 25! 100  99  98  …  77  76 25  24  23  …  3  2  1 = 100!_ 25! 75! 100 nCr 25 on your calculator. “100-choose-25” - denoted or 100C25 This value counts the number of combinations of 25 Heads among 100 coins.

  9. How many possible outcomes of n = 100tosses exist? How many possible outcomes of n = 100tosses exist withX = 25Heads? Answer: What is the probability of each such outcome? Recall that, per toss, P(Heads) =  = 0.4 P(Tails) = 1 –  = 0.6 Answer: Via independencein binary outcomes between any two coins, 0.4  0.6  0.6  0.4  0.6  …  0.6  0.4  0.4  0.6 = . Therefore, the probability P(X = 25) is equal to……. Question: What if the coin were “fair” (unbiased), i.e.,  = 1 –  = 0.5 ?

  10. How many possible outcomes of n = 100tosses exist? How many possible outcomes of n = 100tosses exist withX = 25Heads? This is the “equally likely” scenario! Answer: What is the probability of each such outcome? Recall that, per toss, P(Heads) =  = 0.4 P(Tails) = 1 –  = 0.6  = 0.5 1 –  = 0.5 Answer: 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 = . 0.5  0.5  0.5  0.5  0.5  …  0.5  0.5  0.5  0.5 = Therefore, the probability P(X = 25) is equal to……. Question: What if the coin were “fair” (unbiased), i.e.,  = 1 –  = 0.5 ?

  11. For any randomly selected individual, define a binary random variable: POPULATION 40% Male, 60% Female “Success” vs. “Failure”  “Success” 1 –  “Failure” RANDOMSAMPLE n = 100 Discrete random variable X = # “Successes” in sample (0, 1, 2, 3, …, n) Discrete random variable X = # Males in sample (0, 1, 2, 3, …, n) Discrete random variable X = # Males in sample (0, 1, 2, 3, …, 99, 100) size n Example: What is the probability P(X = 25)? x x = 0, 1, 2, 3, …,100 n Solution: F(x) = P(X≤x), for x = 0, 1, 2, 3, …,100? Solution:Model the sample as a sequence of n=100 independent coin tosses, with 1=Heads(Male), 0=Tails(Female). n Bernoulli trials with P(“Success”) = , P(“Failure”) = 1 – . independent, with constant probability () per trial Then X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability function” f(x) = , x = 0, 1, 2, …, n. .… etc….

  12. Example: Blood Type probabilities, revisited 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?

  13. f(x) = (.461)x (.539)10 – x Example: Blood Type probabilities, revisited R: dbinom(0:10, 10, .461) 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  =  xf (x) = and variance  2 =  (x – ) 2f (x) = n = 4.61 = (10)(.461) n (1 – ) = 2.48

  14. f(x) = (.461)x (.539)10 – x Example: Blood Type probabilities, revisited R: dbinom(0:10, 10, .461) 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  =  xf (x) = and variance  2 =  (x – ) 2f (x) = n = 4.61 n (1 – ) = 2.48

  15. Example: Blood Type probabilities, revisited Therefore, f(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  =  xf (x) = and variance  2 =  (x – ) 2f (x) = = 10.5 n n (1 – ) = 10.43 2.48

  16. Example: Blood Type probabilities, revisited Therefore, f(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 = 10.5  Binomial model applies. X ~ Bin(1500, .007) X ~ Poisson(10.5) Also, can show mean  =  xf (x) = and variance  2 =  (x – ) 2f (x) = = 10.5 n Notation: Sometimes the symbol  (“lambda”) is used instead of  (“mu”). n (1 – ) = 10.43

  17. Example: Blood Type probabilities, revisited Therefore, f(x) = x = 0, 1, 2, …, 1500. 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 = 10.5 X ~ Poisson(10.5) Ex: Probability of exactlyX = 15 Type(AB–) individuals = ? Binomial: Poisson: (both ≈ .0437)

  18. More notation facts: In addition to counting the Successes in a random sample of n trials coming from a population with rareP(Success) = , the Poisson distribution may also be used to count the Successes in a time interval of duration T, with rareP(Success) = α, known as the “Poisson rate.” 0 T X = # “clicks” on a Geiger counter in normal background radiation. BinomialPoisson ↔α n↔T μ↔λ

  19. Example: Deaths in Wisconsin

  20. Example: Deaths in Wisconsin Assuming deaths among young adults are relatively rare, we know the following: λ = • Average 584 deaths per year • Mortality rate (α) seems constant. Therefore, the Poissondistribution 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 0.0131 P(X = 600) = R: dpois(600, 584) • Probability of exactly X = 1200 deaths in the next twoyears so let λ= 1168: Mean of 584 deaths per yr  Mean of 1168 deaths per twoyrs, 0.00746 P(X = 1200) = • Probability of at least one death per day: λ = = 1.6 deaths/day P(X = 1) + P(X = 2) + P(X = 3) + … P(X ≥ 1) = True, but not practical. 0.798 = 1 – e–1.6 = = 1 – P(X ≥ 1) = 1 – P(X = 0)

  21. 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 See Lecture Notes!

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