Lect8_L6_L7_handout2

Lect8_L6_L7_handout2 - lecture 8, Discrete Random Variables...

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Unformatted text preview: lecture 8, Discrete Random Variables III Binomial Distribution, ctd Poisson Distribution lecture 8, Discrete Random Variables III Binomial Distribution, ctd Mean and Variance of a Binomial Random Variable If X is a binomial ( n , p ) random variable, then • Mean μ X = np • Variance σ 2 X = npq • Standard deviation σ X = √ npq Proof: X is the sum of n independent Bernoulli(p) random variables X 1 , X 2 , ··· , X n . For i = 1 , ··· , n , P ( X i = 1 ) = p and P ( X i = ) = ( 1- p ) = q ⇒ μ X i = p and σ 2 X i = ( 1- p ) 2 p + (- p ) 2 ( 1- p ) = p ( 1- p ) = pq ⇒ μ X = np and σ 2 X = npq . lecture 8, Discrete Random Variables III Binomial Distribution, ctd Poisson Distribution lecture 8, Discrete Random Variables III Binomial Distribution, ctd Example: True or false questions( ctd) • The number X of questions that an unprepared student answers correctly has a binomial ( 5 , . 5 ) distribution • μ X = np = 5 × . 5 = 2 . 5, σ X = √ npq = √ 5 × . 5 × . 5 = 1 . 12 — sometimes people say that X is around μ X = 2 . 5, give or take σ X = 1 . 12 • The number Y of questions that the prepared student answers correctly has a binomial ( 5 , . 9 ) distribution • μ Y = np = 5 × . 9 = 4 . 5, σ Y = √ npq = √ 5 × . 9 × . 1 = . 67 — so Y is around μ Y = 4 . 5, give or take σ Y = . 67 lecture 8, Discrete Random Variables III Binomial Distribution, ctd Poisson Distribution lecture 8, Discrete Random Variables III Binomial Distribution, ctd Example: True or false questions( ctd) 1 2 3 4 5 0.05 0.10 0.15 0.20 0.25 0.30 Distribution of # correct by the unprepared # correct p(x) 1 2 3 4 5 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Distribution of # correct by the prepared # correct p(x) lecture 8, Discrete Random Variables III Binomial Distribution, ctd Poisson Distribution lecture 8, Discrete Random Variables III Binomial Distribution, ctd Example: Insurance ( ctd) If X is the number of death among policy holders, then the total profit Y is Y = 300 × 10000- 20000 × X . • Recall, X is a binomial(10000, 0.001) r.v. ⇒ μ X = 10000 × . 001 and σ 2 X = 10000 × . 001 × . 999 . • By the rules of expectations/variances, Y has • mean--- μ Y = 300 × 10000- 20000 × μ X = 2 . 8 × 10 6 , and • variance--- σ 2 Y = 20000 2 × σ 2 X = 3 . 996 * 10 9 lecture 8, Discrete Random Variables III Binomial Distribution, ctd Poisson Distribution lecture 8, Discrete Random Variables III Binomial Distribution, ctd Example: Insurance ( ctd) • If we let Z = Y / 10000 be the average profit, then • μ Z =--- μ Y / 10000 = 280, • variance σ 2 Z = σ 2 Y / 10000 2 = 39 . 96, and standard deviation √ 39 . 96 = 6 . 32 • So the average profit is very concentrated around the expected value, and the insurance is faced with very small risk! lecture 8,...
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This note was uploaded on 12/30/2009 for the course ISOM ISOM111 taught by Professor Anthonychan during the Fall '09 term at HKUST.

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Lect8_L6_L7_handout2 - lecture 8, Discrete Random Variables...

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