Question using Chebyshev's inequality

**MickHarford** · March 24th 2011, 11:31 AM

Right, fingers crossed this questions come out OK:

Question using Chebyshev's inequality-q.png

My inital thought is where do I start? I don't quite understand the question

**Sambit** · March 24th 2011, 07:11 PM

Since $X_i$ 's have mean $\mu$ and variance $\sigma^2$ for each $i$ , then $\bar{X}$ has mean $\mu$ and variance $\frac{\sigma^2}{n}$ . Consequently, the Chebyshev's inequality can be rewritten for $\bar{X}$ as:

$P(|\bar{X}-\mu|\geq a)\leq \frac{\sigma^2}{na^2}$

$=>P(\bar{X}\geq \mu+a, \bar{X}\leq \mu-a)\leq\frac{\sigma^2}{na^2}$ (breaking the modulus sign)

So $\frac{\sigma^2}{na^2}$ is the required upper limit.

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