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 's have mean and variance for each , then has mean and variance . Consequently, the Chebyshev's inequality can be rewritten for as:

(breaking the modulus sign)

So is the required upper limit.