The Poisson Distribution

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From before: \(P(X = 0)\) given than the mean number of occurrences is 2.

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Compute the cumulative distribution functions for the values \(k=\{0,1,2\}\), given that the mean number of occurrences is 2

• The Poisson distribution can sometimes be used to approximate the binomial distribution
• When the number of observations n is large, and the success probability p is small, the \(Bin(n,p)\) distribution approaches the Poisson distribution with the parameter given by \(m = np\).
• This is useful since the computations involved in calculating binomial probabilities are greatly reduced in complexity.
• As a rule of thumb, n should be greater than 50 with p very small, such that np should be less than 5.
• We set = np (other notation m = np) and use the Poisson tables.
• If the value of p is very high, the definition of what constitutes a success" orfailure” can be switched. \end{itemize}

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• Suppose we sample 1000 items from a production line that is producing, on average, \(0.1\%\) defective components.

• Using the binomial distribution, the probability of exactly 3 defective items in our sample is \[P(X = 3) = ^{1000}C_{3} \times 0.001^{3} \times 0.999^{997}\]

• Lets compute each of the component terms individually.

• [\(\ast\)] \(^{1000}C_{3}\) \[^{1000}C_{3} = \frac{1000 \times 999 \times 998}{3 \times 2 \times 1} = 166,167,000\]

• [\(\ast\)] \(0.001^3\) \[0.001^3 = 0.000000001\]

• [\(\ast\)] \(0.999^{997}\) \[0.999^{997} = 0.36880\] \end{itemize}

• Multiply these three values to compute the binomial probability \(P(X = 3) = 0.06128\)

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• Lets use the Poisson distribution to approximate a solution.
• First check that \(n \geq 50\) and \(np < 5\) (Yes to both).
• We choose as our parameter value \(m = np = 1000 \times 0.001 = 1\) \end{itemize} \[P(X = 3) = \frac{e^{-1} \times 1^3}{3!} = \frac{e^{-1}}{6} = \frac{0.36787}{6} = 0.06131 \] Compare this answer with the Binomial probability P(X = 3) = 0.06128. Very good approximation, with much less computation effort.

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The Poisson Approximation of the binomial ditribution

Example

\[P(X \geq 2) = 1 - (0.134+0.27) = 0.596\]

\[P(X=1) = 200 0.01 0.99 199\]

\[P(X=1) = 0.270\]

Poisson Approximaitions

\[X \sim \mbox{Binomial}(200, 0.01) \]

P(X= k)=e-kk!

\[P(X \geq 2) = 1- (0.135+0.27) = 0.595\]

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