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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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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
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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