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Question: A number in the interval $[0, 4]$ is selected randomly. How many picks do you have to make so that the arithmetic mean $X$ satisfies $P[|X-2|\ge 0.1]\lt 0.01$ ?

Answer: I've solved by using Chebychev's theorem, to get $P[2-\frac{1}{10}\lt X\lt2+\frac{1}{10}]\ge 1-\frac{1}{10^2}$. Here we get $\mu=2, \sigma=\frac{1}{100}$. But I don't know how to continue to get the answer of "how many picks do you have to make so that..."

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This looks like homework and so I have added the tag. Could you show some of your work and in particular explain how you came up with $\sigma = \frac{1}{100}$? Also, doesn't Chebychev's inequality use $\sigma^2$ instead of $\sigma$ as you do? –  Dilip Sarwate Oct 7 '11 at 14:51
You are picking numbers with uniform distribution on $[0,4]$. This distribution has mean $2$, variance $16/12$. So $X$, the sample mean, has mean $2$, variance $16/(12n)$, where $n$ is the number of trials. Now try to use Chebyshev. Are you really supposed to use the Chebyshev Inequality? It gives pessimistic estimates. Maybe you are expected to use the fact that if $n$ is largish, $X$ has roughly normal distribution. –  André Nicolas Oct 7 '11 at 14:54
@Dilip: homework is a tag that is supposed to be voluntarily added by an asker, not a tag to be appended by an editor. Please don't do this again. Thanks. :) –  J. M. Oct 10 '11 at 13:31
@J.M. OK, I won't do it again. I am still learning how this site works. –  Dilip Sarwate Oct 10 '11 at 13:47

1 Answer 1

up vote 1 down vote accepted

The point is the following: $X_n = \frac1n\sum\limits_{i=1}^n\xi_i$ where $\xi_i\in U([0,4])$ are iid. The mean of $X_n$ always stays the same: $\mu = 2$, but std. deviation decreases: $$ \operatorname{Var}[X_n] = \frac1{n^2}\sum\limits_{i=1}^n\operatorname{Var}[\xi_i] = \frac4{3n}, $$ so $\sigma_n = \frac2{\sqrt{3n}}$. Using Chebyshev's inequality, we have: $$ \mathsf P\left(|X-2|\geq 0.1\right) \leq \frac{\sigma_n^2}{(0.1)^2} = \frac{4}{0.03 n} $$ and we should make $\frac{4}{0.03n}\leq 0.01$ so $n\geq\frac{40000}3$.

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