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I'm having a bit of trouble with Chebyshev's inequality and I was wondering if someone could point me in the right direction as to how to answer this question:

Use chebyshev's theorem to verify that the probability is at least 35/36 that in 900 flips of a balanced coin the proportion of heads will be between 0.40 and 0.60.

By the looks of it, I have to use E(Y)=p, variance(Y)=p(1-p)/n.

I'm not quite sure how to approach this as my lecturer didn't cover this in much depth.

Thank you.

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# Chebyshev's Theorem

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