Chapter 3 Discrete Random Variables and Probability Distributions  3.1 - Random Variables.2 - Probability Distributions for Discrete Random Variables.3.

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Presentation transcript:

Chapter 3 Discrete Random Variables and Probability Distributions  Random Variables.2 - Probability Distributions for Discrete Random Variables.3 - Expected Values.4 - The Binomial Probability Distribution.5 - Hypergeometric and Negative Binomial Distributions.6 - The Poisson Probability Distribution

Infinite Population of decks of 52 cards (Assume each is fair) Random Sample n = 5 (w/ or w/o replacement) Random Variable X = # Spades in sample (x = 0, 1, 2, 3, 4, 5) Outcomes of “X = 2” (0, 0, 0, 1, 1) (0, 0, 1, 0, 1) (0, 0, 1, 1, 0) (0, 1, 0, 0, 1) (0, 1, 0, 1, 0) (0, 1, 1, 0, 0) (1, 0, 0, 0, 1) (1, 0, 0, 1, 0) (1, 0, 1, 0, 0) (1, 1, 0, 0, 0) Binary r.v. 10 combinations  Calculate P(X = 2) Binomial distribution   = P(Spades) = 13/52

Binary r.v. Outcomes of “X = 2” (0, 0, 0, 1, 1) (0, 0, 1, 0, 1) (0, 0, 1, 1, 0) (0, 1, 0, 0, 1) (0, 1, 0, 1, 0) (0, 1, 1, 0, 0) (1, 0, 0, 0, 1) (1, 0, 0, 1, 0) (1, 0, 1, 0, 0) (1, 1, 0, 0, 0) Random Sample n = 5 (w/ or w/o replacement) Random Sample n = 5 (w/ replacement) Infinite Population of decks of 52 cards (Assume each is fair)  Calculate P(X = 2) Finite Population N = 52 (Assume fair) 10 combinations Random Variable X = # Spades in sample (x = 0, 1, 2, 3, 4, 5) Binomial distribution   = P(Spades) = 13/52

Outcomes of “X = 2” (0, 0, 0, 1, 1) (0, 0, 1, 0, 1) (0, 0, 1, 1, 0) (0, 1, 0, 0, 1) (0, 1, 0, 1, 0) (0, 1, 1, 0, 0) (1, 0, 0, 0, 1) (1, 0, 0, 1, 0) (1, 0, 1, 0, 0) (1, 1, 0, 0, 0) Random Sample n = 5 (w/ replacement) Finite Population N = 52 (Assume fair)  Calculate P(X = 2) 10 combinations Random Variable X = # Spades in sample (x = 0, 1, 2, 3, 4, 5) Binomial distribution   = P(Spades) = 13/52 Random Sample n = 5 (w/o replacement) Binary r.v.

Random Sample n = 5 (w/o replacement) ?????? Outcomes of “X = 2” (0, 0, 0, 1, 1) (0, 0, 1, 0, 1) (0, 0, 1, 1, 0) (0, 1, 0, 0, 1) (0, 1, 0, 1, 0) (0, 1, 1, 0, 0) (1, 0, 0, 0, 1) (1, 0, 0, 1, 0) (1, 0, 1, 0, 0) (1, 1, 0, 0, 0) Finite Population N = 52 (Assume fair)  Calculate P(X = 2) = Random Variable X = # Spades in sample (x = 0, 1, 2, 3, 4, 5) Binomial distribution   = P(Spades) = 13/52 Binary r.v.

Sample Space All combinations of 5 cards from 52 # Finite Population N = 52 (Assume fair)  Calculate P(X = 2) Random Sample n = 5 (w/o replacement) # combinations of 2 Spades from 13 Spades # combinations of 3 Non-Spades from 39 Non-Spades =  R command: dhyper(2, 13, 39, 5) Random Variable X = # Spades in sample (x = 0, 1, 2, 3, 4, 5) x, s, N – s, n

 N – s = # Failures Discrete random variable X = # Successes in sample (x = 0, 1, 2, 3, …,, n) Then for any x = 0, 1, 2,…, the pmf f(x) is given by the following… See textbook for  and  2.

POPULATION N = 100 POPULATION N = 100 s = 4 defectives Random Sample (w/o replacement), n = 12 Random Sample (w/o replacement), n = 12 Discrete random variable X = # defectives in sample (x = 0, 1, 2, 3, 4)

See textbook for  and  2. Random Sample (w/ or w/o replacement) Infinite Population of “Successes” and “Failures” Then for any x = s, s+1, s+2,…, the pmf f(x) is… Sample size NOT specified!

Random Sample (w/ replacement)   = P(Spades) = 13/52 Random Variable X = # trials for s = 6 Spades (x = 6, 7, 8, 9, 10,…)

Random Sample (w/ replacement)   = P(Spades) = 13/52 Random Variable X = # trials for s = 6 Spades (x = 6, 7, 8, 9, 10,…) Random Variable X = # trials for s = 1 Spades (x = 1, 2, 3, 4, 5,…)

Random Variable X = # trials for s = 1 Spades (x = 1, 2, 3, 4, 5,…) Random Sample (w/ replacement)   = P(Spades) = 13/52 (x – 1) Failures 1 Success!

Random Variable X = # trials for s = 1 Spades (x = 1, 2, 3, 4, 5,…) Random Sample (w/ replacement)   = P(Spades) = 13/52 x f(x)f(x) Exercise Graph cdf F(x)F(x)