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The random variable X has a Poisson distribution with mean $μ=5.55$. Find the probability that X lies within one standard deviation of the mean.

Using my calculator, I found $P(3.19<X<7.9)$ where $3.19 = 5.55 - \sqrt{5.55}$ and $7.9 = 5.55 + \sqrt{5.55}$ and got $0.683$.

The answer shows:

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Any ideas where I went wrong?

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  • $\begingroup$ How did you get $0.683$? $\endgroup$ – David Oct 31 '16 at 2:04
  • $\begingroup$ normalcdf($5.55 - \sqrt{5.55}$, $5.55 + \sqrt{5.55}$, $5.55$, $\sqrt{5.55}$) $\endgroup$ – StopReadingThisUsername Oct 31 '16 at 2:21
  • $\begingroup$ The question said it's a Poisson distribution and you have calculated a normal distribution. $\endgroup$ – David Oct 31 '16 at 2:26
  • $\begingroup$ My bad...Thanks for spotting it. $\endgroup$ – StopReadingThisUsername Oct 31 '16 at 4:18

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