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!
Best Answer
$P(11.1\le X\le28.9)=P(12\le X\le28)$ since $X$ is discrete. So what you calculated is not correct.
Also, they might use normal approximation of a Poisson random variable.
$$\frac{X-\lambda}{\sqrt\lambda}\sim N(0,1)$$