Binominal distribution to standard normal distribution

Question

I test my skills in statistics and probabilities and I decided to work with distributions. So, I tried to solve the below problem

Problem

Suppose that a hospital serves in average $80$ citizens daily from a city with $11000$ citizens. In a random day, what is the probability that the hospital serves at most $8$ citizens?

My solution

I defined a random variable $X$ {number of citizens who will be served in one day }.

$X \sim b(x;n=11000,p)$, where \begin{align} p &= \frac{E(X)}{n} = \frac{80}{11000} = 0.07 \end{align}

Provided that $npq = 76.4 > 10$:

$b(x;n=11000,p) \sim N(pq,npq)$

According to the central limit theorem, \begin{align} Z = \frac{X - np}{\sqrt{npq}} = \frac{8-80}{8.74} = -8.23 \end{align} So $P(Z\le -8.23) = 0$.

Where is my fault? I think my reasoning is not correct.

Answer 1

Binomial: If $X \sim Binom(n, p),$ then $E(X) = np = 80,$ and $n = 11000.$ So $p = 80/11000 = 0.00727.$ You seek $P(X \le 8).$ In R statistical software this is about $6.6 \times 10^{-25}.$

n = 11000; p = 80/n
pbinom(8, n, p)
## 6.572574e-25

Poisson: If $Y \sim Pois(\lambda = 80),$ then $P(Y \le 8) = 8.3 \times 10^{-25}.$

ppois(8, 80)
## 8.331982e-25

Normal approximation: The mean is $\mu = 80.$ According to the Poisson distribution, the standard deviation is $\sigma = \sqrt{80} = 8.944.$ According to the binomial distribution, the SD is $\sigma = \sqrt{np(1-p)} = 8.912.$ Then $P(W \le 8) = P(W < 8.5) \approx 0.$ The three answers below are from binomial, Poisson, and the standardized normal in your Question, respectively.

 pnorm(8.5, 80, sqrt(80))
 ## 6.534509e-16
 pnorm(8.5, 80, 8.912)
 ## 5.164271e-16
 pnorm(-8.23)
 ## 9.360672e-17

From a printed table of the standard normal CDF, you can tell only that the integral in @SeanRobertson's Answer (posted while I am typing this), is very nearly 0. Using R (or other statistical software) you can get 'exact' values for any of the normal integrals. But they are all essentially 0 for practical purposes, and they are all approximations.

Note: Unless this problem is intended to explore normal probabilities 'off the table', I'm wondering if the intention was to find $P(X \le 80)$ or $P(X \ge 80).$

Answer 2

Using the appropriate commands in R and Excel, the actual probability is $9.360672 \cdot 10^{-17}$. However, this is extremely small, so an answer of zero would also be acceptable.

When it comes to statistics and the normal distribution, don't expect "exact" answers. Why? Well, the probabilities are generated by this integral:

$$ P(Z < z) = \frac{1}{\sqrt{2\pi}} \int_{-\infty} ^z e^{-\frac{x^2}{2}} \ dx $$

which has no elementary antiderivative. Hence, only estimates can be given.

TL;DR, you're fine.