# Chebyshev's Inequality to solve amount of stock needed

A mail order company offers their first $$1000$$ customers a ladies' or mens' watch. Suppose that both sexes are equally attracted by the offer. How many ladies' and mens' watches are needed in order to ensure that --with a probability of at least $$98$$%, all $$1000$$ customers receive a matching watch.

I was tipped to use Chebyshev's inequality.

My idea:

Define $$X:=$$"Number of male watches stocked"

Note that since "both sexes are equally attracted by the offer", I only need to look at $$500$$ watches.

My problem is that my definition of RV $$X$$ does not allow for me to obtain $$\mathbb E[X]$$ let alone $$\operatorname{Var}(X)$$. I believe that I am using the wrong random variables, but cannot find any other appropriate ones.

• Start with $\{X_i=1\} = \{"\text{matching watch}"\}$. Then $\mathbb P(X_i = 1) = \mathbb P(X_i = 0) = \frac 12$, i.e. Bernoulli distributed with parameter (?), and all $X_i$ are iid. Hence, $S_n = X_1 + ... + X_n$ is the number of (?) and we are looking for $\mathbb P(S_n > 500) \geq 0.98$ and use the en.wikipedia.org/wiki/Central_limit_theorem#Classical_CLT. – wueb Feb 15 '19 at 8:19

Let $$X=500+x$$ male watches stocked, and the same number of female watches stocked. Note that this is not a random variable. Here the number of men or women is random since any customer can be of any sex equiprobable. Denote by $$S_{1000}$$ the number of men among first $$1000$$ customers. As wueb suggests, $$S_{1000}=X_1+\ldots+X_{1000}$$ where $$X_i$$ equals to $$1$$ if $$i$$-th customer is a men and $$X_i=0$$ if it is a women. Then $$1000-S_{1000}$$ is the total number of women.
We need both $$S_{1000}\leq 500+x$$ and $$1000-S_{1000}\leq 500+x$$ with probabilility at least $$0.98$$. These inequalities transform to $$500-x\leq S_{1000}\leq 500+x$$ or $$|S_{1000}-500|\leq x.$$ We need to find $$x$$ such that $$\mathbb P\left(|S_{1000}-500|\leq x \right) \geq 0.98.$$ If it is required to use Chebyshev's inequality, use $$\textrm{Var}(S_{1000})=1000\cdot \textrm{Var}(X_1)=250$$ in r.h.s.
$$\mathbb P\left(|S_{1000}-500|\leq x \right) = 1- \mathbb P\left(|S_{1000}-500|> x \right)=1- \mathbb P\left(|S_{1000}-500|\geq x-1 \right)$$ $$\geq 1-\frac{\textrm{Var}(S_{1000})}{(x-1)^2}\geq 0.98.$$ Find minimal integer $$x$$ satisfying this inequality and $$500+x$$ will be the answer for number of male watches.
It's worth pointing out that if you want to minimise the number of watches you have to stock, then using Chebyshev's inequality is a very poor way of going about it. From NCh's answer, the estimate is $$\ 500 + \sqrt{\frac{250}{0.02}}\approx 611.8\$$, which rounds up to $$\ 612\$$.
However, more precise calculations of the sums $$\ \sum_{i=500-x}^{500+x} \frac{1000\choose i}{2^{1000}}\$$ for $$\ x=36\$$ and $$\ x=37\$$ give: $$\sum_{i=463}^{537} \frac{1000\choose i}{2^{1000}}\approx 0.9823\\ \sum_{i=464}^{536} \frac{1000\choose i}{2^{1000}}\approx 0.9791\ .$$ So you really only need to stock $$\ 537\$$ of each kind of watch to achieve the desired $$\ 98\%\$$ probability.
A much better way to obtain this value is to use the normal approximation to the binomial distribution to obtain an initial estimate (which turns out to be $$\ 536.8\$$), and then check the exact probabilities for a few integer values on either side of this. In this case, the normal approximation is so close that it gives you the right answer immediately.