# Using Chebyshev's inequality to obtain lower bounds

I need help with a question I found in Master Stats. I'm unaware of Chebyshev's inequality hence I can't do this question, can anyone help.

Q) A company produces planks whose length is a random variable of mean 2.5m and standard deviation 0.1m. Use Chebyshev's inequality to obtain a lower bound on the probability that the length of planks does not differ more than 0.5m from the mean length.

Thanks in advance

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Have you looked up Chebyshev's inequality? en.wikipedia.org/wiki/Chebyshev's_inequality – Daan Michiels May 10 '12 at 10:34

## 1 Answer

Let $X$ be a random variable with mean $\mu$ and standard deviation $\sigma>0$. Then the Chebyshev Inequality says that if $k>0$, then $$P(|X-\mu| \ge k\sigma)\le \frac{1}{k^2}.$$ In our case, $X$ is the length of a plank chosen at random from the company's production. Then $\mu=2.5$ and $\sigma=0.1$. We want to find $k$ such that $k\sigma=0.5$. Thus $k=\frac{0.5}{\sigma}=\frac{0.5}{0.1}=5$. We conclude that $$P(|X-2.5| \ge 0.5) \le \frac{1}{5^2}.$$

It follows that $$P(|X-2.5| < 0.5) \ge 1-\frac{1}{5^2}.\tag{\ast}$$ This is not quite what we want, since we want to find a number $p$ such that $P(|X-2.5| \le 0.5)\ge p$. One can argue that the probability that the difference is absolutely exactly $0.5$ is $0$, so that $(\ast)$ gives us the inequality we want. That gives a lower bound of $1-\frac{1}{5^2}=0.96$. There is a probability of at least $0.96$ that the plank does not differ by more than $0.5$ from the mean $2.5$.

Typically, the Chebyshev Inequality gives very conservative estimates. In our case, though Chebyshev says that $P(|X-2.5|\ge 0.5) \le \frac{1}{5^2}$, the actual probability is likely to be substantially smaller than $\frac{1}{5^2}$. Thus the lower bound of $0.96$ is likely conservative. Informally, more than $96\%$ of the production will be "within spec."

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but isn't the formula for Chebyshev Inequality P(|X−μ|≥k)≤σ$^2$/k$^2$ or is the formula you used just a rearrangement? – methuselah May 15 '12 at 13:06
I used the equivalent version $P(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}$. So my $k$ is the number of "standard deviation units." You can see that (in this case, as always) we get the same thing. Your $k$ is $0.5$, and with your $k$ we have $\sigma^2/k^2=(0.1)^2/(0.5)^2=1/5^2$. – André Nicolas May 15 '12 at 13:14