Binominal distribution to standard normal distribution

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.

• You read this the wrong way. To say "at least" is to say "greater than or equal to." – Sean Roberson Jan 12 '17 at 21:20
• I mean, at most. – Dimitris Dimitriadis Jan 12 '17 at 21:22
• Then you're okay. Just because a probability is near zero doesn't mean you did anything wrong. – Sean Roberson Jan 12 '17 at 21:24
• I didn't find the -8.23 to statistical tables and that's why I supposed that P(Z<= -8.23) = 0 – Dimitris Dimitriadis Jan 12 '17 at 21:25
• Many tables do not have entries beyond $|z| > 4$ because the probability is so small. If you want something in scientific notation, try using the appropriate function in MATLAB, R, or Excel. – Sean Roberson Jan 12 '17 at 21:27

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 $P(X \ge 80).$

• I like your answer. However, I don't understand why the probability is calculated with Poisson distribution. Did you use the assumption that n->inf and p->0 ? – Dimitris Dimitriadis Jan 12 '17 at 21:54
• When population sizes are large, and pretty obviously estimates, as for your 11,000, it is common to use the Poisson approximation where the Poisson mean $\lambda$ is taken to be the population mean. If I had not been trying to show you various approaches (since you are doing self-study), I might have shown only the Poisson answer. Now that excellent software is available for getting exact binomial and Poisson probabilities, the normal distribution is less frequently used in practice. It remains popular in textbooks because of the theoretical importance of the normal distribution. – BruceET Jan 12 '17 at 22:07
• Perfect, perfect !! I understood it, thank you ;) – Dimitris Dimitriadis Jan 12 '17 at 22:11

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.

• Nice direct answer to the exact question. (+1) However, modern computational methods can give results of normal integrals to more decimal places than one ever needs in practice. – BruceET Jan 12 '17 at 22:10