2022-09-15

McDonald's

There is a McDonald's downstairs from the office.

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The business is quite good from 6am to 10am every day, but I am worried about one thing: how many hamburger should be prepared so that they are not wasted and can be fully supplied?

The boss counted the number of hamburger sold every day for a week (the data was narrowed down for easy calculation and explanation)：

$\begin{array}{c|c} \qquad\qquad&\qquad sales\qquad\\\hline\color{SkyBlue}{monday}& 3 \\ \hline \color{blue}{Tuesday}& 7 \\ \hline \color{orange}{Wednesday}&4\\\hline \color{Goldenrod}{Thursday}&6\\ \hline \color{green}{Friday}&5\\\end{array}\\$

Average sales：

$\overline{X}=\frac{3+7+4+6+5}{5}=5\\$

By definition the mean is a good choice, but if you prepare 5 hamburgers per day, from the statistics table, there are not enough to sell at least two days, $40\%$ of the time is not enough to sell.

$\begin{array}{c|c}\qquad\qquad&\qquad sales\qquad&\quad 5 hamburger\\\hline\color{SkyBlue}{monday}& 3 \\\hline \color{blue}{Tuesday}& 7&\color{red}{not\ enough} \\ \hline \color{orange}{Wednesday}&4\\ \hline \color{Goldenrod}{Thursday}&6&\color{red}{not\ enough}\\\hline \color{green}{Friday}&5\\\end{array}\\$

You "McDonald's" is not XIAOMI, what [hunger marketing](https://www.zhihu.com/search?q=%E9%A5%A5%E9%A5%BF%E8%90%A5%E9%94%80& search_source=Entity&hybrid_search_source=Entity&hybrid_search_extra=%7B%22sourceType%22%3A%22answer%22%2C%22sourceId%22%22 3A429569625%7D) huh? The boss, of course, knew this and grabbed a pen and paper to start thinking.

2 The owner's thinking

The owner tries to abstract the business hours as a line segment by expressing this time in terms of $T$ T.

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Then place the three buns of $\color{SkyBlue}{Monday}$ on the line segment according to the time of sale:

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Divide T equally into four time periods：

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At this point, on each time period, either (a) hambeger is sold, or not sold：

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In each time period, it's a bit like flipping a coin, either heads (sold) or tails (not sold)：

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The probability of selling 3 hamburgers in T is the same as flipping a coin 4 times (4 time periods), of which 3 times are heads (3 sold).

Such a probability is calculated by means of a binomial distribution as：
$\binom{4}{3}p^3(1-p)^1$

However, if the seven hamburgers of $\color{blue}{Tuesday}$ are placed on the line segment, dividing it into four segments is not enough：

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From the graph, each time period, there are 3 sold, 2 sold, 1 sold, it is no longer simply "sold, not sold". Can not be applied to the binomial distribution.

The solution to this problem is also very simple: divide T into 20 time periods, then each time period becomes a coin flip again.

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Thus, the probability of selling 7 Hamburgers in T is (equivalent to flipping a coin 20 times and getting 7 heads)：

$\binom{20}{7}p^7(1-p)^{13}$

To ensure that only "sells and no sells" occur within a time period, simply cut the time into n parts：

$\binom{n}{7}p^7(1-p)^{n-7}$

The finer the better, expressed in terms of limits.

$\lim_{n\ \to\ \infty} \ \binom{n}{7}p^7(1-p)^{n-7}$

More abstractly, the probability of selling k hamburgers in time T is:

$\lim_{n\ \to\ \infty} \ \binom{n}{k}p^k(1-p)^{n-k}$

3 p calculation:

"So", the boss tapped the table with his pen, "there is only one problem left, how to find the probability p?"

Under the above assumptions, the problem has been transformed into a binomial distribution. The expectation of the binomial distribution is：

$E(X)=np=\mu$

so：

$p= \frac{\mu}{n}$

4 Poisson distribution

if $p=\frac{\mu}{n}$ then：

$\lim_{n\ \to\ \infty} \binom{n}{k}p^k(1-p)^{n-k}=\lim_{n\ \to\ \infty}\binom{n}{k}\left(\frac{\mu}{n}\right)^k(1-\frac{\mu}{n})^{n-k}$

Let's calculate this limit：

$\begin{align}\lim_{n\ \to\ \infty} \binom{n}{k} \left(\frac{\mu}{n} \right)^k(1-\frac{\mu}{n})^{n-k} &= \lim_{n\ \to\ \infty} \frac{n(n-1)(n-2)\ \cdots(n-k+1)}{k!} \frac{\mu^k}{n^k} \left(1- \frac{\mu}{n}\right)^{n-k} \\ &=\lim_{n\ \to\ \infty} \frac{\mu^k}{k!}\frac{n}{n}\ \cdot \frac{n-1}{n} \cdots \frac{n-k+1}{n} \left(1-\frac{\mu}{n}\right)^{-k} \left(1-\frac{\mu}{n} \right)^n\ \end{align}$
note：

$\lim_{n\ \to\ \infty}\ \frac{n}{n}\cdot\ \frac{n-1}{n}\ \cdots\ \frac{n-k+1}{n}\ \left(1-\frac{\mu}{n}\right)^{-k}=1$

$\lim_{n \ \to \ \infty}\ \left(1-\ \frac{\mu}{n}\right)^n = e^{-\mu}$

then：

$lim_{n\ \to\ \infty}\ \binom{n}{k} \left(\frac{\mu}{n}\right)^k(1-\frac{\mu}{n})^{n-k}=\frac{\mu^k}{k!}e^{-\mu}$

The above is the probability density function of the Poisson distribution, i.e., the probability of selling k hamburgers in time T is：

$P(X=k)=\frac{\mu^k}{k!}e^{-\mu}$

Normally, we would change the symbol， $\mu=\lambda$ ，then：

$P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda}$

This is the textbook probability density function of the Poisson distribution。

5 hamburger store's problem solving

The boss still frowned, do not know $\mu$ ?

It does not matter, just now not calculated the sample mean: the

$\overline{X}=5$

It can be used to approximate：

$\overline{X}\approx\mu$

so：

$P(X=k)=\frac{5^k}{k!}e^{-5}$

Drawing the curve of the probability density function is：

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As you can see, if you prepare 8 hamburgers per day, the probability of having enough to sell is to add up the probability of the first 8：

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In this way 93% of the cases are enough, occasionally sell out of stock also help the brand image.

The boss counted out a head of sweat, "Then it's settled!"

6 Binomial and Poisson distributions

Given the relationship between the binomial distribution and the Poisson distribution, it is a natural corollary that when the binomial distribution has a small p, the two are closer to.
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7 Summary

This story tells us to study hard ah

There are many Poisson distributions in life. For example, the half-life in physics, we only know how much time to expect the decay of half of the substance, but because of uncertainty principle, we have no way to know exactly which atom will decay at what time? So the Poisson distribution can be used to calculate it.

There are also problems such as traffic planning and so on.