The random number $X$ of customers arriving per hour is Poisson distributed with mean $\lambda = 11$. We want $$\Pr[X > 10] = 1 - \Pr[X \le 10] = 1 - \sum_{x=0}^{10} e^{-\lambda} \frac{\lambda^x}{x!}.$$ Since this is tedious to compute by hand, we elect to use a normal approximation to the Poisson. The mean and variance of a Poisson being equal, we then let $Y \sim \operatorname{Normal}(\mu = \lambda, \sigma^2 = \lambda)$, and the above probability is approximately $$\Pr[X > 10] \approx \Pr[Y > 10.5]$$ with a continuity correction. We have $$\Pr[Y > 10.5] = \Pr\left[\frac{Y - \mu}{\sigma} > \frac{10.5 - \lambda}{\sqrt{\lambda}}\right] = \Pr[Z > -0.150756],$$ where $Z$ is a standard normal distribution. But since we are calculating a right-tailed probability, this is (assuming a TI-83 calculator)
normalcdf(-0.150756,1E99,0,1)
which is approximately $0.5599$. The exact probability is obtained using the command
1-poissoncdf(11,10)
which is $0.5401112973$. Without continuity correction, the probability given by the normal approximation would be
normalcdf(-0.301511,1E99,0,1)
which is $0.618488$, too large.
The syntax for normalcdf
is
normalcdf(lower,upper,mean,sd)
which calculates the probability that a normally distributed random variable with mean mean
and standard deviation sd
is between lower
and upper
. So if we want a right-tailed probability of the form $\Pr[Y > 10.5]$, then the value of lower
should be $10.5$, and upper
we choose to be some very large number, say 1E99
(this represents $1 \times 10^{99}$ and is typed into the calculator using 2nd
followed by EE
, the yellow function above the comma button). Note we could have also entered
normalcdf(10.5,1E99,11,11^(1/2))
and gotten the same result, because we can save ourselves the step of standardizing $Y$--the calculator will do it for you if you type in the correct mean and standard deviation. But you still need the continuity correction. But of course, the most precise and easiest command of all is to use poissoncdf
itself, although that may not be the intended purpose of the exercise. It's a good idea to do it anyway, so that you can see how reasonable the normal approximation might be.
Best Answer
Let $X_{\alpha}$ Poisson $\pi(\alpha)$, for $\alpha = 1, 2, \ldots$ The probability mass function of $X_{\alpha}$ is $$f_{X_{\alpha}}(x)=\frac{{\alpha}^x\operatorname{e}^{-x}}{x!} \qquad \alpha = 1, 2, \ldots$$ The moment generating function of $X_{\alpha}$ is $$ M_{X_{\alpha}}=\Bbb{E}\left(\operatorname{e}^{tX_{\alpha}}\right)=\operatorname{e}^{\alpha(\operatorname{e}^{t}-1)}\qquad t\in(-1,1) $$ Now consider a “standardized” Poisson random variable $Z_{\alpha}=\frac{X_{\alpha}-{\alpha}}{\sqrt{\alpha}}$ which has limiting moment generating function $$ \begin{align} \lim_{\alpha\to\infty}M_{Z_{\alpha}}&= \lim_{\alpha\to\infty}\Bbb{E}\left(\exp{(tZ_{\alpha})}\right)\\ &=\lim_{\alpha\to\infty}\Bbb{E}\left(\exp{\left(t\frac{X_{\alpha}-{\alpha}}{\sqrt{\alpha}}\right)}\right)\\ &=\lim_{\alpha\to\infty}\exp(-t\sqrt\alpha)\Bbb{E}\left(\exp{\left(\frac{tX_{\alpha}}{\sqrt{\alpha}}\right)}\right)\\ &=\lim_{\alpha\to\infty}\exp(-t\sqrt\alpha)\exp\left(\alpha(\operatorname{e}^{t/\sqrt\alpha}-1)\right)\\ &=\lim_{\alpha\to\infty}\exp\left(-t\sqrt\alpha +\alpha\left[t\alpha^{-1/2}+\frac{t^2\alpha^{-1}}{2}+\frac{t^3\alpha^{-3/2}}{6}+\cdots\right]\right)\\ &=\lim_{\alpha\to\infty}\exp\left(\frac{t^2}{2}+\frac{t^3\alpha^{-3/2}}{6}+\cdots\right)\\ &=\exp\left(\frac{t^2}{2}\right) \end{align} $$ by using the moment generating function of a Poisson random variable and expanding the exponential function as a Taylor series. This can be recognized as the moment generating function of a standard normal random variable. This implies that the associated unstandardized random variable $X_{\alpha}$ has a limiting distribution that is normal with mean $\alpha$ and variance $\alpha$.