Answer to Question #155306 in Statistics and Probability for Rohith Kumar

Question

When the first proof of 392 pages of a book of 1200 pages were 

read , the distribution of printing mistakes were found to be as 

follows

No of mistakes in a page (x) 0 1 2 3 4 5

No of pages (f) 275 72 30 7 5 2

Fit a Poisson distribution to the above data and test the goodness of

Answer 1

The PDF for the poisson distribution is

$$f(x)=\frac{e^{-\lambda}\lambda^x}{x!}$$

Where $\lambda$ is the mean. We equate the sample and population moments for the mean to obtain the parameter for the distribution

$$\lambda= E(X)$$

$$E(X)= \frac{1}{\sum frequency}\sum x * frequency$$ $$\sum frequency =275+72+30+7+5+2 =391$$

Therefore

$$E(X)=\frac{1}{391}[0*275+1*72+2*30+3*7+4*5+5*2]$$

$$=\frac{183}{391} =0.4680$$

Since $E(X)=\lambda=0.4680$

The PDF is therefore

$$f(x)=\frac{e^{-0.468}0.468^x}{x!}$$

Given the 391 pages, the expected frequency for the number of errors based on the poisson distribution is

$$X=0: 391* \frac{e^{-0.468}0.468^0}{0!} = 244.865$$

$$X=1: 391* \frac{e^{-0.468}0.468^1}{1!} = 114.597$$

$$X=2: 391* \frac{e^{-0.468}0.468^2}{2!} = 26.816$$

$$X=3: 391* \frac{e^{-0.468}0.468^3}{3!} = 4.183$$

$$X=4: 391* \frac{e^{-0.468}0.468^4}{4!} = 0.489$$

$$X=5: 391* \frac{e^{-0.468}0.468^5}{5!} = 0.046$$

We use the chi square test for the goodness of fit

$$X^2=\sum\frac{(observed - Expected)^2}{Expected}$$

$$=\frac{(275-244.865)^2}{244.865} + \frac{(72-114.597)^2}{114.597} + \frac{(30-26.816)^2}{26.816} + \frac{(7-4.183)^2}{4.183} + \frac{(5-0.489)^2}{0.489} + \frac{(2-0.046)^2}{0.046}$$

$$X^2=146.434$$

Since the sample size is greater than 200,(i.e. 391 pages were evaluated) the chi square critical value given a level of significance of 0.05 is

$X^2 _{critical\ value}=233.994$

We conclude that the model is a good fit since the calculated $X^2: 146.434 < X^2 _{critical}$