Probability Theory The Binomial and Poisson Distributions. Sections 5.2 and 5.3

Transcription

1 Probability Theory The Binomial and Poisson Distributions Sections 5.2 and 5.3

2 Models for count data The binomial distributions provide a theoretical model for count data having a fixed maximum Examples: # girls among three children # defects in a lot of one hundred items # recoveries among ten sick patients The Poisson distributions provide a theoretical model for open-ended counts Examples: # accidents on I81 the day before Thanksgiving # help-line calls on a typical weekday afternoon # flaws in a section of carpet

3 The binomial setting* Observe a fixed number of trials, n The trials are independent Each trial has the same two possible outcomes: S = success or F = failure Constant success probability, p = P(S), across trials * Also called a binomial experiment

4 Properties of the binomial setting The sample space consists of 2 n possible outcomes Example: n = 5 trials 2 n = 32 possible outcomes S = { SSSSS, SSSSF, SSSFS, SSFSS, SFSSS, FSSSS, SSSFF, SSFSF, SSFFS, SFSSF, SFSFS, SFFSS, FSSSF, FSSFS, FSFSS, FFSSS, SSFFF, SFSFF, SFFSF, SFFFS, FSSFF, FSFSF, FSFFS, FFSSF, FFSFS, FFFSS, SFFFF, FSFFF, FFSFF, FFFSF, FFFFS, FFFFF }

5 Properties of the binomial setting (continued) P( outcome ) = p #S (1 p) #F Example: n = 5 trials, p = P(S) = 0.3 P(FFSSF) = p 2 (1 p) 3 = (0.3) 2 (0.7) 3 = P(SFSSF) = p 3 (1 p) 2 = (0.3) 3 (0.7) 2 = P(SFFFS) = p 2 (1 p) 3 = (0.3) 2 (0.7) 3 = P(FFFFF) = p 0 (1 p) 5 = (0.3) 0 (0.7) 5 =

6 Properties of the binomial setting (continued) The number of outcomes with k successes is n choose k n factorial where n! = n (n 1) (and 0! = 1) Example: n = 5 trials, k = 2 successes A = { SSFFF, SFSFF, SFFSF, SFFFS, FSSFF, FSFSF, FSFFS, FFSSF, FFSFS, FFFSS }

7 Binomial random variables The random variable X = # successes is a binomial random variable Probabilities are: Example: n = 5 trials, p = P(S) = 0.3 Binomial distribution k p k (1 p) n k P(X = k)

8 Binomial mean and standard deviation If X is a binomial random variable of n trials with p = P(S) then µ X = n p and σ X = {n p (1 p)} n = 10, p = 0.10 n = 10, p = 0.50 n = 10, p = 0.75 µ X = 1.0, σ X = 0.95 µ X = 5.0, σ X = 1.58 µ X = 7.5, σ X = 1.37 Observe: Skewed distribution unless p = 0.5

9 Example: Color blindness 8% of white male population is colorblind Sample n = 140 white males, p = P( color blind ) = 0.08 What are µ X and σ X? µ X = n p = (140)(0.08) = 11.2 σ X = {n p (1 p)} = {(140)(0.08)(0.92)} = 3.21

10 Options for finding binomial probabilities Use the formula: Use Excel: =binomdist(k, n, p, 0) for P(X = k) =binomdist(k, n, p, 1) for P(X k) Use a Normal approximation Rule of thumb: valid if np 10 and n(1 p) 10

11 Example: Color blindness 8% of white male population is colorblind Sample n = 140 white males, p = P( color blind ) = 0.08 What is P(X 5)? =binomdist(5, 140, 0.08, 1) = Note: np = (140)(0.08) = n(1 p) = (140)(0.92) =

12 The Poisson setting Count success points* within a fixed region or time -interval, etc., Success-point counts are independent between nonoverlapping subunits The mean count of success points in any subunit is proportional to the size of the subunit The probability of two or more success points in the same subunit goes to zero as the subunit shrinks * Also called Poisson events

13 Examples Flaws in a patch of carpet: ATM visits per hour:

14 Poisson random variables The random variable X = # success points is a Poisson random variable Probabilities are: µ = mean count of success points e (the exponential constant) Poisson distribution Example: µ = 2.5 k P(X = k)

15 Poisson mean and standard deviation If X is a Poisson random variable with µ mean success points µ X = µ and σ X = µ µ = 2.5 µ = 15 µ = 30 µ X = 2.5, σ X = 1.58 µ X = 15, σ X = 3.87 µ X = 30, σ X = 5.48 Observe: Skewed distribution when µ is small

16 Options for finding Poisson probabilities Use the formula: Use Excel: =poisson(k, µ, 0) for P(X = k) =poisson(k, µ, 1) for P(X k)

17 Example: Industrial accidents Manufacturing center averages 7 accidents per month X = # accidents this month Poisson with µ = 7 What is P(X = 6)? =poisson(6, 7, 0) = What is P(X 6)? =poisson(6, 7, 1) =

18 Scaling the mean count of success points If µ is the mean success-point count per unit, then aµ is the mean success-point count per a units That is: Suppose X counts success points in a one unit region Suppose Y counts the same phenomenon s success points in a region of a units. If X is Poisson then Y is Poisson with µ Y = aµ X

19 Example: Industrial accidents (continued) Manufacturing center averages 7 accidents per month X = # accidents this year Poisson with µ = (12)(7) = 84 What is P(X 75)? =poisson(75, 84, 1) = What is P(75 X 90)? P(X 90) P(X 74) = = { =poisson(90, 84, 1) } { =poisson(74, 84, 1) } = =

20 Independent sums of Poisson r.v.s If X and Y are independent Poisson random variables, then Z = X + Y is a Poisson random variable with mean µ Z = µ X + µ Y. Example: Counts of a car s paint flaws X = # flaws on sq. yard of car s roof, µ X = 0.7 Y = # flaws on sq. yard of car s side panel, µ Y = 1.4 Z = X + Y = # flaws in both areas combined, µ Z = µ X + µ Y = = 2.1

21 Probability Theory Conditional Probability Section 5.4

22 General multiplication rule A conditional probability, P(B A), is the probability of some event, B, under the condition that some other event, A, has definitely occurred General multiplication rule: P(A and B) = P(A) P(B A) Extended version for multiple events: P(A and B and C) = P(A) P(B A) P(C A and B) etc.

23 Implications of the general multiplication rule A definition of conditional probability is P(B A) = P(A and B) / P(A) If A and B are independent, P(A and B) = P(A) P(B), hence P(B A) = P(B) To derive: rearrange the general multiplication rule

24 Example: Unemployment rates Draw a single individual from a certain population (cnt./1000) Employment status Row Gender Emp. Unemp. NA totals Female Male Column totals P( female ) = 9116 / = P( employed and female ) = 4313 / = P( employed female ) = 4313 / 9116 = P( employed ) = 8240 / =

25 Example: Prosperity and education Draw a single household from a certain population A = household is educated B = household is prosperous Known profile of the population: P(A) = 0.254, P(B) = 0.134, P(A and B) = Are A and B independent? P(B A) = P(A and B) / P(A) Not independent. = / = = P(B)

26 Tree diagrams Tree diagrams can help to organize thinking about conditional probabilities Example: High-school to professional athlete P(A) College A Professional B High school S B c B P(B A c ) A c B c

27 General multiplication rule from a tree diagram Multiply along a complete path for the general multiplication rule, P(A and B) = P(A) P(B A) Example: High-school to professional athlete P(A and B) = P(A) P(B A) = (0.05)(0.017) = S 0.05 A B B c P(A c and B) = P(A c ) P(B A c ) B = (0.95)(0.0001) = A c B c

28 A total probability rule from a tree diagram Probability of a terminal event is the sum of the product of probabilities along all complete paths leading to it Example: High-school to professional athlete P(B) = P(B and A) + P(B and A c ) = P(A) P(B A) + P(A c ) P(B A c ) = (0.05)(0.017) + (0.95)(0.0001) S A B B c B = A c B c

29 Reverse conditional probabilities Bayes s rule: Example: High-school to professional athlete P(A) P(B A) = (0.05)(0.017) = P(A c ) P(B A c ) = (0.95)(0.0001) =

30 Example: Medical screening Tree diagram for breast cancer screening of women in twenties Women in 20s S Cancer Incidence Specificity Sensitivity A A c Cancer Diagnosis B B c B B c False negative False positive

31 Example: Medical screening (continued) Given a cancer diagnosis, what is the probability of a cancer incidence? S A B B c B A c 0.9 B c

32 Example: Defective parts Tree diagram for parts from two different suppliers Defective Supplier A 0.02 B Part S B c B A c 0.95 B c

33 Example: Defective parts (continued) Given a defective part, what is the probability it came from Supplier A? S A B B c B A c 0.95 B c