# Applications of Chebyshev's theorem

I'm told that the diameter of particlesare uniformly distributed between 5 and 21 micrometers and asked to use Chebyshev's inequality to compute how many particles are needed so that the average diameter is within 3.2 micrometers of the true population mean with at least 99% probability.

I took my interval to be $$(5, 21)$$ and wrote two equations for this. $$5=\mu -k\sigma$$ and $$21=\mu+k\sigma$$, where $$\sigma=3.2$$. I was able to solve my two equations to yield $$\mu=13$$ and $$k=2.5$$. Chebyshev's inequality is given to me as

$$P(X_n-\mu \mid k) = \frac{\sigma^2}{k^2n}$$

So,

$$.99=1-\frac {3.2^2}{2.5^2n} \Rightarrow n=163 \ \mathrm{particles}$$

This answer is incorrect, it should be at least $$209$$ particles. It's very possible I misunderstood the inequality. It's also likely I did not understand what I'm given in the question. I'm not quite sure where to go from here.

Consider Chebyschev's inequality as following: $$P(|\overline{X}_n-\mu|\geq \varepsilon)\leq \frac{E[|\overline{X}_n-\mu|^2]}{\varepsilon^2}$$ We can then state that (recall $$P(A^c)=1-P(A)$$) $$P(|\overline{X}_n-\mu|< \varepsilon)\geq 1-\frac{E[|\overline{X}_n-\mu|^2]}{\varepsilon^2}$$ Our constraint is $$1-\frac{E[|\overline{X}_n-\mu|^2]}{\varepsilon^2}>\alpha$$ Now recall that the variance of the random sample mean is variance divided by $$n$$. So $$\frac{\sigma^2}{n}<(1-\alpha)\varepsilon^2\implies n>\frac{\sigma^2}{\varepsilon^2(1-\alpha)}=\frac{(b-a)^2}{12\varepsilon^2(1-\alpha)}\approx 208.33$$