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I am trying to solve the following exercise

A fabric that is dedicated to the production of dairy products produces a certain type of cheese in units which weight (in kg) is a random variable of expected value $\mu=2$ and variance $\sigma=0.04$

How many units would be necessary so as the probability of satisfying a demand of $5000$ kg is at least $0.95$?

This is what I did up to now:

If $X$ is the random variable that models the weight of a unit of cheese, then we want to know the minimum $n$, where $n$ is the units of cheese so that $$P\left(\sum_{i=1}^n X_i \geq 5000\right) \geq 0.95.$$

I don't know how to apply Chebyshev's inequality here, I would appreciate any suggestions. Thanks in advance.

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  • $\begingroup$ is 1 unit equal to 1kg? $\endgroup$ – Brandon Nov 27 '16 at 2:31
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Intuition: as you get more and more units, the mean tends to get higher. The values also tend to not fall too far away from the mean, since the variance is finite. So once the mean is a lot higher than 5000, it will be very likely to see results higher than 5000.

Details: if $X_i$ are independent and identically distributed with mean $\mu$ and variance $\sigma^2$, then $\sum_{i=1}^n X_i$ has mean $n\mu$ and variance $n\sigma^2$. Now Chebyshev's inequality essentially says that events of the form $\left | \sum_{i=1}^n X_i - n \mu \right | \leq \epsilon$ are likely if $\epsilon$ is large relative to $\sqrt{n} \sigma$.

So to use this to bound the probability that $\sum_{i=1}^n X_i \geq c$, you're going to want $n \mu>c$, so that there is an $\epsilon>0$ such that the interval $[n\mu-\epsilon,n\mu+\epsilon]$ is contained in the interval $[c,\infty)$. Once that happens,

$$P \left (\sum_{i=1}^n X_i \geq c \right ) \geq P \left ( \left | \sum_{i=1}^n X_i-n \mu \right | \leq \epsilon \right )$$

whenever $\epsilon \leq n\mu-c$. Chebyshev's inequality can now be applied to the right side.

So can you proceed with your particular problem?

A bit of a reality check: the result that you will get out of this problem setup will be far worse than would ever show up in an application. With the $\mu$ and $c$ in your problem, you'll be with a large enough sum of random variables that the central limit theorem gives a good estimate (unless the distribution in question is quite unusual). Consequently it will be enough to have $n \mu - 2 \sqrt{n} \sigma \geq c$ by virtue of the 68-95-99.7 rule for the normal distribution.

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