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Lets say a production machine makes one fault unity per every $500$ units. Assume that it is making $50$ units per package. Let $S$ be a random variable corresponding to the number of faulty units that the machine makes in a single package.

I calculated the expected number of faulty units per page is $E(S) = \frac{50}{500}$

However, I am not able to wrap my mind around calculating the probability of $S = someNumber$

Also, how exactly can this normal distribution be modeled as a poisson distribution?

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2 Answers 2

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The Poisson distribution takes a an average rate, which you've found, and tells you the probability that a number of occurrences will happen. For a Poisson distribution, the density function is given by $$\mathbb P(S=s)=\frac{\lambda^se^{-\lambda}}{s!}$$ where $s$ is a non-negative integer. In this case, we are concerned with the number of faults per $50$ units so $\lambda=\frac{1}{500}\frac{\text{faults}}{1\text{unit}}= 0.1 \frac{\text{faults}}{50\text{units}} \implies \lambda= 0.1$. You can then calculate the probability of $s$ faults happening with the formula above! Note that the maximum number of faults that can occur is $50$, so theoretically $s$ cannot be any positive integer and to be precise you should be using the binomial distribution with $n=50$ and the probability of a fault $p=1/500$, however, the Poisson distribution is derived by taking $n \to \infty$ so for large samples the Poisson distribution is a good approximation.

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    $\begingroup$ I see! Thanks, for the informative answer! $\endgroup$
    – Jonathan
    Nov 18, 2016 at 6:16
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    $\begingroup$ Also, quick correction. Shouldn't $\lambda = 1/10$? $\endgroup$
    – Jonathan
    Nov 18, 2016 at 21:42
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    $\begingroup$ @Christian I agree $\lambda=0.1$ $\endgroup$ Nov 18, 2016 at 22:41
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    $\begingroup$ @Alex I understand what you're saying, but I think you need to re-read the question. 1 in 500 fails, so indeed E(S) = 1/10. Here's what I think the correction should be: $\lambda = \frac{1 fault}{500 units} = \frac{0.1 faults}{50 units}$ $\endgroup$
    – knrumsey
    Nov 21, 2016 at 19:41
  • $\begingroup$ You're correct! $\endgroup$
    – user275377
    Nov 22, 2016 at 4:06
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Alex's answer is acceptable, because of the relationship between the Poisson and the Binomal distribution. But formally speaking, the Binomial distribution is more appropriate here.

A binomial distribution with $n$ trials and probability $p$ of failure has probability mass function:

$P(S=x) = \binom{n}{x}p^x(1-p)^{n-x}$ The expected value of a binomial rv is $np$, so we still have $\mathbb{E}(S) = 50\times\frac{1}{500}$ as desired.

You will find however, that since $n$ is somewhat large, and $p$ is pretty small, the Poisson approximation will provide very similar probabilities. Formally speaking however, only the Binomial distribution provides "exact" probabilities.

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