Poisson Distribution Probability

Question

If the average number is 16, what is the probability that a given trial will trap between 12 and 20 atoms, inclusive? What would this probability be if the distribution is instead Gaussian, with both a mean and a variance of 16?

I got the formula which is $Pk = \cfrac{u^k}{k!}\times e^{-u}$ where $u = 16$ = average but not sure what k is?

Thank you for the help

Answer 1

The probability mass function for a Poisson random variable $X$ with mean $\lambda$ is $$\Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots.$$ Applied to your question, if the mean number of atoms trapped in a trial is $\lambda = 16$, then the probability that the observed number of atoms trapped is at least $12$ and no more than $20$ is $$\Pr[12 \le X \le 20] = \sum_{x=12}^{20} \Pr[X = x] = \sum_{x=12}^{20} e^{-16} \frac{16^x}{x!}.$$ This is not too many terms to compute with a hand calculator.

For the normal approximation, we observe that a Poisson distribution has its variance equal to its mean; i.e., $$\operatorname{Var}[X] = \operatorname{E}[X] = \lambda,$$ so we would choose a normal distribution whose mean and variance are $\mu = \sigma^2 = 16$. Let $$Y \sim \operatorname{Normal}(\mu = 16, \sigma = 4),$$ and with a continuity correction for this approximation, we would calculate $$\Pr[12 \le X \le 20] \approx \Pr[11.5 \le Y \le 20.5] = \Pr\left[\frac{11.5 - 16}{4} \le \frac{Y - \mu}{\sigma} \le \frac{20.5 - 16}{4}\right] = \Pr[-1.125 \le Z \le 1.125],$$ where $Z$ is a standard normal random variable. We would then use a table to compute this probability, which is approximately $0.739441$. By comparison, the exact probability using the Poisson calculation is $0.741175$. So our approximation is reasonably good!

Exercise: What would the normal approximation have been if we did not employ the continuity correction; i.e., what is $\Pr[12 \le Y \le 20]$? Note that this is in fact the question that is being asked in the second part!

Answer 2

u is NOT 16, u is each 12, 13, ..., 20 and k is 16. Sum those. For the Gaussian distribution, use the standard normal distribution with $z= \cfrac{x- 16}{4}$.

Answer 3

$X=k$ is equal to the number of atoms., $u=16$

$ P(X=k) = \frac{u^k}{k!} \cdot e^{-u} $

https://en.wikipedia.org/wiki/Poisson_distribution#Definition

You are searching the probability:

$ P(12) + P(13) + \ldots P(19) + P(20)$

I don't see any shortcut, on how to compute it fast, so I guess, you have to sum those $9$ values up.

Answer 4

For the second part of your question regarding the Gaussian $P(12\lt x \lt 20) = P\left(\cfrac{12-16}{4} \lt z \lt \cfrac{20-16}{4}\right)\approx 0.683$. This is neglecting continuity correction.