Answer to Question #98396 in Statistics and Probability for chennareddy

Question

Fit a poisson distribution with the following data
X 0 1 2 3 4 5
F 142 156 69 27 5 1

Answer 1

Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is $\lambda$ . Then, the Poisson probability is:

$$P(x;\lambda)={e^{-\lambda}\lambda^x \over x!}$$

$$\begin{array}{c:c:c:c:c:c:c:c} X & 0 & 1 & 2 & 3 & 4 & 5 & Total\\ \hline F & 142 & 156 & 69 & 27 & 5 & 1& 400 \end{array}$$

$$\text{mean of given distribution}={\sum FX \over \sum F}=$$

$$={142(0)+156(1)+69(2)+27(3)+5(4)+1(5) \over 400}=1$$

This is the parameter $\lambda$ of the Poisson distribution $:\lambda=1$

Required Poisson distribution is

$$N\cdot{e^{-\lambda}\lambda^x \over x!}, \text{where } N=\sum F=400$$

$$\begin{array}{c:c:c} x & P(x) & Theoretical\ Frequency \\ \hline 0 & 147.15 & 147 \\ \hline 1 & 147.15 & 147 \\ \hline 2 & 73.58 & 74\\ \hline 3 & 24.53 & 25 \\ \hline 4 & 6.13 & 6 \\ \hline 5 & 1.23 & 1 \\ \hline & & Total=400 \\ \end{array}$$