Normal approximation to the poisson

Question

Customers arrive at a travel agency at a mean rate of $11$ per hour. Assuming that the number of arrivals per hour has a poisson distribution, give the probability that more than $10$ customers arrive in a given hour.

I am trying to calculate the probability with the normal approximation to poisson but I am having some error.

So we need to find $P(X>10)=P(X \geq 11)$ and using the approximation rules, we obtain $P(X \geq 10.5)=P(\frac{X-\lambda}{\sqrt{\lambda}} \geq \frac{10.5-11}{\sqrt{11}})=P(Z \geq -.15)$

However, when I use a chart I get $0.5596$ but my issue is when I am calculating on the graphing calculator.

On the graphing calculator: I use normalcdf$(-100,-.15)=0.44$ to get $0.5596$, I have to do $0.5-0.44=0.0596$ and add that to $0.5$, so in result, $0.5+0.0596=0.5596$, but why won't normalcdf$(-100,-.15)=0.5596$

Answer 1

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.