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Unformatted text preview: 4.7 Poisson Distribution 131 E bracketleftbig K 2 bracketrightbig = λ e λ ∞ summationdisplay k = 1 k λ k 1 ( k 1 ) ! = λ e λ e λ ( 1 + λ) = λ 2 + λ The variance of K is given by σ 2 K = E bracketleftbig K 2 bracketrightbig ( E [ K ] ) 2 = λ 2 + λ λ 2 = λ The Poisson distribution has many applications in science and engineering. For example, the number of telephone calls arriving at a switchboard during various intervals of time and the number of customers arriving at a bank during various intervals of time are usually modeled by Poisson random variables. Example 4.16 Messages arrive at a switchboard in a Poisson manner at an aver age rate of six per hour. Find the probability for each of the following events: (a) Exactly two messages arrive within one hour. (b) No message arrives within one hour. (c) At least three messages arrive within one hour. Solution Let K be the random variable that denotes the number of messages arriving at the switchboard within a onehour interval. The PMF of K is given by p K ( k ) = parenleftbigg 6 k k ! parenrightbigg e 6 k = , 1 , 2 ,... a. The probability that exactly two messages arrive within one hour is p K ( 2 ) = parenleftbigg 6 2 2 ! parenrightbigg e 6 = 18 e 6 = . 0446 b. The probability that no message arrives within one hour is p K ( ) = parenleftbigg 6 ! parenrightbigg e 6 = e 6 = . 0024 132 Chapter 4 Special Probability Distributions c. The probability that at least three messages arrive within one hour is P [ K ≥ 3 ] = 1 P [ K < 3 ] = 1 { p K ( ) + p K ( 1 ) + p K ( 2 ) } = 1 e 6 braceleftbigg 6 ! + 6 1 1 ! + 6 2 2 ! bracerightbigg = 1 e 6 { 1 + 6 + 18 } = 1 25 e 6 = . 9380 trianglesolid 4.7.1 Poisson Approximation to the Binomial Distribution Let X be a binomial random variable with parameter ( n , p ) and PMF p X ( x ) = parenleftbigg n x parenrightbigg p x ( 1 p ) n x x = , 1 , 2 ,..., n Since the PMF involves evaluating n ! , which can become very large even for mod erate values of n , we would like to develop an approximate method for the case of large values of n . We know that E [ X ] = np . Therefore, let λ = np , which means that p = λ/ n . Substituting for p in the PMF we obtain p X ( x ) = parenleftbigg n x parenrightbiggparenleftbigg λ n parenrightbigg x parenleftbigg 1 λ n parenrightbigg n x = n ( n 1 )( n 2 )( n 3 )...( n x + 1 ) x ! n x λ x parenleftbigg 1 λ n parenrightbigg n x = n x parenleftbigg 1 1 n parenrightbiggparenleftbigg 1 2 n parenrightbiggparenleftbigg 1 3 n parenrightbigg ... parenleftbigg 1 x 1 n parenrightbigg x ! n x λ x parenleftbigg 1 λ n parenrightbigg n parenleftbigg 1 λ n parenrightbigg x = parenleftbigg 1 1 n parenrightbiggparenleftbigg 1 2 n parenrightbiggparenleftbigg 1 3 n parenrightbigg ......
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 Fall '07
 Carlton
 Normal Distribution, Poisson Distribution, Variance, Probability theory, Exponential distribution

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