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Suppose that our editor makes an average of two and one-half typographical errors per page. You select a 30-page chapter on this text at random and let C denote the total number of typos in the chapter.

(a) Find $\mu_C$

Alright this looks like a really simple question, but I'm so confused. It says in my text:

Mean and Variance of a Poisson Random Variable

Suppose that Z is a Poisson random variable with parameter $\lambda$. Then

$E[Z]=Var[Z] =\lambda$

So $\mu_C$ is just the expected value and at first I thought the solution is just 2.5, but its really $2.5*30=75$. Can someone please explain what I am misunderstanding with lambda and the theorem in my book? Thank you in advance.

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up vote 3 down vote accepted

The mean per page is $2.5$, so the mean per $30$ pages is $(30)(2.5)$.

This really has nothing much to do with the Poisson. Let $X_1$ be the number of typos on the first page, $X_2$ the number of typos on the second, and so on up to $30$. The total number $T$ of typos is given by $$T=X_1+X_2+\cdots+X_{30}.$$ The mean of a sum is always the sum of the means ("linearity of expectation"). It follows that $E(T)=30E(X_1)=75$.

Where the Poisson comes in is not for the computation of the mean. But it is important for the distribution of $T$. Since the sum of independent Poissons is Poisson, $T$ has Poisson distribution.

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