Poisson random variable and standard deviation
I am reading Probability and Statistics for Engineering and the Sciences.
Exercise 81, Chapter 3 says:
Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter $\mu=20$ (suggested in the article “Dynamic Ride Sharing: Theory and Practice,” J. of Transp. Engr., 1997: 308–312). What is the probability that the number of drivers will
- Be within 2 standard deviations of the mean value?
I know that the standard deviation for a Poisson random variable has the same value of the mean (which in this case is $\mu=20$).
Therefore I computed the cumulative density function of the Poisson random variable from $0$ to $60.5$ (2 Standard deviations should be $40$).
However I get as result $0.99999999999986222$ while the correct result seems to be $0.945$. What am I missing?
$\endgroup$2 Answers
$\begingroup$The standard deviation is the square root of variance. The variance is $\mu$, and the standard deviation should be $\sqrt{\mu}$.
$\endgroup$ $\begingroup$The mean value and variance of a Poisson are the same. The standard deviation is $\sqrt\mu$. Therefore,
$$P(\mid X-\mu\mid\leq 2\sqrt \mu)=P(\mu-2\sqrt\mu\leq X\leq \mu+2\sqrt \mu)=P(11.05\leq X\leq 28.94)=0.9447$$
approximately.
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