Chapt 4c - 4.7 Poisson Distribution 131 E bracketleftbig K...

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Unformatted text preview: 4.7 Poisson Distribution 131 E bracketleftbig K 2 bracketrightbig = e- summationdisplay k = 1 k k- 1 ( k- 1 ) ! = e- e ( 1 + ) = 2 + The variance of K is given by 2 K = E bracketleftbig K 2 bracketrightbig- ( E [ K ] ) 2 = 2 + - 2 = 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 one-hour 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 = np , which means that p = / n . Substituting for p in the PMF we obtain p X ( x ) = parenleftbigg n x parenrightbiggparenleftbigg n parenrightbigg x parenleftbigg 1- n parenrightbigg n- x = n ( n- 1 )( n- 2 )( n- 3 )...( n- x + 1 ) x ! n x x parenleftbigg 1- 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 x parenleftbigg 1- n parenrightbigg n parenleftbigg 1- n parenrightbigg- x = parenleftbigg 1- 1 n parenrightbiggparenleftbigg 1- 2 n parenrightbiggparenleftbigg 1- 3 n parenrightbigg ......
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Chapt 4c - 4.7 Poisson Distribution 131 E bracketleftbig K...

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