Right, fingers crossed this questions come out OK:

Attachment 21250

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

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- Mar 24th 2011, 11:31 AMMickHarfordQuestion using Chebyshev's inequality
Right, fingers crossed this questions come out OK:

Attachment 21250

My inital thought is where do I start? I don't quite understand the question - Mar 24th 2011, 07:11 PMSambit
Since $\displaystyle X_i$'s have mean $\displaystyle \mu$ and variance $\displaystyle \sigma^2$ for each $\displaystyle i$, then $\displaystyle \bar{X}$ has mean $\displaystyle \mu$ and variance $\displaystyle \frac{\sigma^2}{n}$. Consequently, the Chebyshev's inequality can be rewritten for $\displaystyle \bar{X}$ as:

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

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

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