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I was given this question as HW:

A survey reveals that on average 1 computer out of 800 computers crashes during a severe thunderstorm. A certain company had 4,000 working computers when the area was hit by a severe thunderstorm. Compute the expected value and variance of the number of crashed computers in the storm.

As far as the expected value goes, I get $\frac{4000}{800}=5$.

But how can I get variance? Are there unstated assumptions that I am missing here? How can I know anything about the distribution just based on the mean?

MPW
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soandos
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    The question may be asking you to model using a binomial distribution with $p = 0.00125$ and $n=4000$ – user130512 Mar 11 '14 at 14:27
  • @user130512 how did you get that? – soandos Mar 11 '14 at 14:33
  • You may be expected to use the Poisson approximation, variance is $\lambda$. But binomial is better. We are assuming independence, which seems physically unreasonable. – André Nicolas Mar 11 '14 at 14:34
  • @AndréNicolas Is the question underspecified, or can I just pick a distribution, defend it, and thats it? – soandos Mar 11 '14 at 14:35
  • The question is definitely underspecified – user130512 Mar 11 '14 at 14:36
  • Call a computer crashing a success. Probability $p$ of success is $\frac{1}{800}$, number $n$ of independent (?) trials (computers) is $4000$. – André Nicolas Mar 11 '14 at 14:37
  • If one makes the assumption of independence, the problem is fairly well specified. One can poke fun at the independence assumption. We then have a binomial $n$ large $p$ small, $np=5$ medium. Standard setup for using the Poisson approximation to the binomial. – André Nicolas Mar 11 '14 at 14:40
  • @soandos; Could they just be expecting npq? $4000 \cdot \frac{ 1}{800} \cdot \frac{799}{800}=\frac{799}{160} $ – bobbym Mar 11 '14 at 14:49

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If you check the property of binomial distribution, you can find that the variance for a binomial distribution with mean np (in this case $n = 4000$, $p=1/800$, $q=799/800$) is $npq=4000*(1/800)*(799/800)$

TYZ
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