standard deviation probability of a poission distribution Just looking for a bit of help on the topic.
The question is basically this:
Number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter µ = 20.
What is the probability that the number of drivers will be within two standard deviations of the mean value?
Because it's a Poisson distribution, the expected value and the variance are the same, so mean value = µ = 20.
Because the variance is 20, the standard deviation is the squareroot of 20 = 4.47. Two standard deviations is then 4.47*2 = 8.9
Two standard deviations from the mean is (20-8.9 <= x <= 20+8.9)
So we're looking for P(11 <= x <= 29).
I put it in matlab like so poisscdf(29,20) - poisscdf(11,20) and I get 0.9568.... the answer however is .945. I'm just wondering where I went wrong!
 A: $P(11.1\le X\le28.9)=P(12\le X\le28)$ since $X$ is discrete. So what you calculated is not correct.
Also, they might use normal approximation of a Poisson random variable.
$$\frac{X-\lambda}{\sqrt\lambda}\sim N(0,1)$$
A: If $X \sim Pois(\lambda=20),$ then 
$$P(12 \le X \le 28) = P(X \le 28) - P(X \le 11) = 
\sum_{i=12}^{28} e^{-20}20^i/i!.$$
In R statistical software this can be evaluated exactly as follows:
diff(ppois(c(11,28), 20))
## 0.9442797
sum(dpois(12:28, 20))
## 0.9442797

The normal approximation (with continuity correction) gives:
diff(pnorm(c(11.5,28.5), 20, sqrt(20)))
## 0.9426531

Perhaps they used the normal approximation directly, without regard
to the fact that the Poisson distribution takes only integer values:
diff(pnorm(c(11.1,28.9), 20, sqrt(20)))
## 0.9534201

I think the purpose of this exercise may be to illustrate that the
Empirical Rule works pretty well for the Poisson distribution,
putting approximately 95% of its probability within two standard
deviations of the mean.
I guess they intended for you to use a normal approximation, unless
some kind of software was explicitly mentioned. Below
is a plot that compares $Pois(\lambda=20)$ with $Norm(\mu=20, \sigma=\sqrt{20}).$
For the Poisson probability, you want the sum of the heights of the bars
between the vertical red dotted lines. For the normal approximation, you
want the area under the normal curve between these lines.

A: Your general reasoning is correct, but the details in the calculations are not. I am not sure what inputs Matlab's poisscdf takes. If it accepts real numbers then pass it the exact result of your addition/subtraction. Do not truncate/round.
If you want to use the cdf formula yourself (and I recommend you do, if you are learning) then you need to truncate, but be careful: the formula for the cdf uses the floor function, not the rounding function.
$$P(X \le k)  = \frac{\Gamma(\lfloor(k+1)!\rfloor,\lambda)}{\lfloor k! \rfloor}$$
For your example this means:
$$P(11.1\le X\le28.9) = \frac{\Gamma(\lfloor(28.9+1)!\rfloor,20)}{\lfloor 28.9! \rfloor} - \frac{\Gamma(\lfloor(11.1+1)!\rfloor,20)}{\lfloor 11.1! \rfloor} = \frac{\Gamma(29!,20)}{28!} - \frac{\Gamma(12!,20)}{11!}$$
Wolfram Alpha gives me a result of $0.94428$. I am not sure if there is a rouding discrepancy between Wolfram and Matlab, or you have rounded Matlab's result incorrectly in the description of the question.
