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An event occurs with a mean of 15 times per hour and a standard deviation of 10 times/h. From the population a sample of 100 individuals is drawn. What is the probability that the sample mean will be between 14 and 16?

I want to solve this problem assuming that the event follows a Poisson distribution (I can get the right answer when assuming a normal distribution).

My R input is:

> 1-ppois(16, lambda=15, lower.tail=FALSE)-ppois(14, lambda=15)
[1] 0.1984695

The first ppois call gives the right tail area and the second call the lower tail area, which I both subtract from 1.

What is wrong with my calculation?

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Your calculation is the probability of a Poisson with $\lambda=15$ resulting in either a 15 or 16 (this can also be computed as: sum(dpois(15:16, 15))). If you want 14 included in there then you need to change the 14 to 13 (otherwise you are subtracting out the probability of 14 as well).

But that is the probability of a single observation and the question asks about the mean of a sample of size 100, which is very different. The probability that the mean is between 14 and 16 is the same as the probability that the sum is between 1,400 and 1,600 and the distribution of the sum (assuming iid) will be Poisson (if the individuals are Poisson) with mean 1,500, which gives:

> sum( dpois(1400:1600, 1500) )
[1] 0.9905373

or using the normal approximation to the Poisson:

> pnorm(16, 15, sqrt(15)/10) - pnorm(14, 15, sqrt(15)/10)
[1] 0.9901767

This is quite a bit different from the about 0.68 assuming the normal, because this uses a standard deviation of $\sqrt{15} \approx. 3.8$ instead of the 10 stated in the problem (which means that the Poisson really does not apply for the given question.

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The Poisson distribution is a poor choice in this case because it has the strict assumption that $\mu = \sigma^{2} = \lambda$,[1] but in your case $\mu = 15 \ne \sigma^{2} = 100$ (so what is $\lambda$ supposed to be?). Perhaps a different count model, such as the negative binomial (pnbinom in R) would be appropriate?


[1] Emphasizing the numerical values of $\mu$ and $\sigma^{2}$, and ignoring Stéphane Laurent's point about the interpretability of the different units for these two measures.

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    $\begingroup$ As I said, I'm serious. Here is the explanation: the mean and the variance do not have the same unit of measure. Hence there's no sense to claim they are equal. $\endgroup$ – Stéphane Laurent Jul 23 '14 at 21:26
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    $\begingroup$ @StéphaneLaurent - I was thinking about this the other day too. In many sources, it is stated that mean ($\lambda$) is equal to variance ($\sigma^2$), but as you say, the units are completely different. Did you find an answer to this conundrum anywhere? $\endgroup$ – Ben Feb 17 '18 at 12:21
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    $\begingroup$ @Ben That depends if you work in mathematical statistics or in practical statistics :) Pure mathematicians don't care about units. $\endgroup$ – Stéphane Laurent Feb 17 '18 at 14:19
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    $\begingroup$ Yes, but it is nevertheless always valid to compare $3$ to $3^2$, whether that number counts apples, oranges, or anything else. Moreover, the very definition of cardinal numbers (counts) is in terms of a direct, one-to-one comparison of two sets. The implicit units of a count measurement therefore have nothing whatsoever to do with the nature of the elements of those sets. $\endgroup$ – whuber Feb 18 '18 at 14:23
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    $\begingroup$ @whuber TY (more or less as always :), there's a tension: the need to interpret those units of count, and, as you point out the need for the logic of cardinal numbers. $\endgroup$ – Alexis Feb 18 '18 at 17:24

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