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A polling agency wishes to take a sample of voters in a given state large enough that the probability is only $0.01$ that they will find the proportion favoring a certain candidate to be less than $50$ percent when in fact it is $52$ percent. How large a sample should be taken ?

I tried as follows :

Let $N$ ~ The number of people in favor of a certain candidate.

So , $N$ ~ $Bin(n,0.52)$ ,

Now according to me the question says :

$P(N < 0.5n) = 0.01$ and $n$ is required.

=> $P((N-\mu)^2 < (0.5n-\mu)^2) = 0.01$

(Applying Chebyshev's inequality)

$\geq 1 - \dfrac{Var(N)}{(0.5n-0.52n)^2} = 0.01$

=> $1 - \dfrac{(0.52)(0.48)n}{(0.5n-0.52n)^2} = 0.01$

which gives $n \approx 630$

Is this correct ?

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    Of course, Chebyshev is possibly more pessimistic than required. And isn't $\operatorname{Var}(N)=npq=0.52\cdot 0.48\cdot n$? – 2017-01-26
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    Edited. I tried normal approximation but thought how would I know if $n$ is that lage enough or not. @HagenvonEitzen – 2017-01-26

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