slides10 - Lecture Stat 302 Introduction to Probability -...

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Unformatted text preview: Lecture Stat 302 Introduction to Probability - Slides 10 AD Feb. 2010 AD () Feb. 2010 1 / 23 Discrete Random Variables A discrete r.v. X takes at most a countable number of possible values f x 1 , x 2 , ... g with p.m.f. p ( x i ) = P ( X = x i ) where p ( x i ) & 0 and ∞ ∑ i = 1 p ( x i ) = 1 . Expected value/mean μ = E ( X ) = ∞ ∑ i = 1 x i p ( x i ) . Variance Var ( X ) = E & ( X ¡ μ ) 2 ¡ = E ¢ X 2 £ ¡ μ 2 . AD () Feb. 2010 2 / 23 Bernoulli and Binomial Distributions Assume you have an experiment w.p. success p and w.p. failure ( 1 & p ) . Bernoulli r.v.: You set X = 1 if success and X = 0 if failure so P ( X = 1 ) = p ( 1 ) = p , P ( X = ) = p ( ) = 1 & p . Assume now you have n independent experiments, each w.p. success p and w.p. failure ( 1 & p ) . Binomial r.v.: Set X = number of successes among the n experiments, then X 2 f , 1 , 2 , ..., n g X = X 1 + X 2 + ¡¡¡ + X n where X k is the Bernoulli r.v. associated to experiment i and P ( X = k ) = & n k ¡ p k ( 1 & p ) n & k . AD () Feb. 2010 3 / 23 Example Example : A survey from Teenage Research Unlimited (Northbrook, Ill.) found that 30% of teenage consumers receive their spending money from part-time jobs. If &ve teenagers are selected at random, &nd the probability that at least three of them will have part-time jobs. Answer : Let X be the number of teenagers having a part-time job among 5 teenagers, then X is Binomial of parameters n = 5, p = . 3 so P ( X & 3 ) = P ( X = 3 ) + P ( X = 4 ) + P ( X = 5 ) = . 132 + . 028 + . 002 = . 162 Example : What is the probability of obtaining 45 or fewer heads in 100 tosses of a fair coin? Solution : Let X be the number of heads, then X is binomial of parameters n = 100, p = . 5 and P ( X ¡ 45 ) = ∑ 45 k = & n k ¡ p k ( 1 ¢ p ) k = . 184. AD () Feb. 2010 4 / 23 Mean of the Binomial Random Variable The mean/expected value of X is given by E ( X ) = np Proof 1 . We have X = X 1 + X 2 + &&& + X n where X i is a Bernoulli r.v. w.p. p so E ( X ) = E ( X 1 + &&& + X n ) = E ( X 1 ) + &&& + E ( X n ) = np . Proof 2 . We have E ( X ) = ∑ n k = k P ( X = k ) = ∑ n k = k & n k ¡ p k ( 1 ¡ p ) n ¡ k = ∑ n k = 1 k n ( n ¡ 1 ) ! k ( k ¡ 1 ) ! ( n ¡ k ) ! p p k ¡ 1 ( 1 ¡ p ) n ¡ k = np ∑ n k = 1 ( n ¡ 1 ) ! ( k ¡ 1 ) ! ( n ¡ k ) ! p k ¡ 1 ( 1 ¡ p ) n ¡ k = np ∑ m l = m ! l ! ( m ¡ l ) ! p l ( 1 ¡ p ) m ¡ l ( m n ¡ 1 , l k ¡ 1 ) = np AD () Feb. 2010 5 / 23 Variance of the Binomial Random Variable The variance of the binomial random variable is Var ( X ) = np ( 1 & p ) . Proof 1 . We have Var ( X ) = E & X 2 ¡ & E ( X ) 2 = E & X 2 ¡ & ( np ) 2 where X 2 = ( X 1 + ¡¡¡ + X n ) 2 = n ∑ k = 1 X 2 k + n ∑ k , l = 1 ; k 6 = l X k X l so E & X 2 ¡ = n ∑ k = 1 E & X 2 k ¡ + n ∑ k , l = 1 ; k 6 = l E ( X k X l ) ....
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This note was uploaded on 10/21/2010 for the course STAT Stat302 taught by Professor 222 during the Spring '10 term at UBC.

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slides10 - Lecture Stat 302 Introduction to Probability -...

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