The number of fishes caught by a skilled fisherman each day has a Poisson distribution with mean $2.5$. After $50$ days, what is the mean and variance for the total number of fishes the fisherman would have caught ? I know that the $$p(x=k)=\dfrac{e^{-2.5}2.5^{k}}{k!}$$ Now I can't really figure out what next to do. Will the total number of fishes be $50(2.5)$ ?

Would it be right to say that $$E(X)=\sum_{x=0}^{20}x\dfrac{e^{-2.5}2.5^{x}}{x!}$$ and $$V(X)=\left(\sum_{x=0}^{50}x^{2}\dfrac{e^{-2.5}2.5^{x}}{x!}\right)-(E(X))^{2}$$

  • $\begingroup$ Welcome to Math.SE! Please use MathJax. For some basic information about writing math at this site, see mathjax tutorial and quick reference and equation editing how-to. Please read this post for writing a good question. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 26 at 19:36
  • $\begingroup$ Hint: Let $X_i$ be the number of fish caught on day $i$ (for $i=1$ to $50$). Then $S:= \sum\limits_{i=1}^{50}X_i$ is the number of fish caught over the $50$ days. Can you find the expectation and variance of $S$ using the expected value and variance of the $X_i $, along with your knowledge of how to find mean and variance of a sum of random variables? (You may assume for simplicity that the $X_i$ are independent, this was probably implicitly assumed by the question.) $\endgroup$ – Minus One-Twelfth Feb 28 at 11:55

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