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Disclaimer: This is part of a larger homework question, but I'm getting stuck on one point.

Assume that $n/3$ people voted for a candidate in an election, and in an exit poll $k'/k$ people voted for the candidate. (sample of original population - chosen with replacement).

What is $P[1/6 < k′/k < 1/2]$. I understand that Chebyshev’s inequality needs to be applied, so the answer will be:

$P[1/6 < k′/k < 1/2] = P[|k'/k - \mu| \leq 1/6] \geq 1 - \frac{\sigma^2}{(\frac{1}{6})^2}$

The problem I have relates to the mean and variance of the Bernoulli variables. From the information of the problem, I'm assuming that the mean is $1/3$ since n/3 people voted for the candidate in the election, and then the variance is $1/3*2/3 = 2/9$. When I try to use these values, then the $P[1/6 < k′/k < 1/2] \geq -7$ which makes no sense. What am I misunderstanding?

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Your $\sigma$ is for a single person, not a $k$ person sample, I guess. Also, $1-\ldots$ cannot be $>1$. – Hagen von Eitzen Nov 10 '12 at 15:35
Corrected the issue with the prob being greater than 1, but I'm not sure how to calculate the mean and variance in this case. – Slruh Nov 10 '12 at 15:42
up vote 0 down vote accepted

The expected value for $k'$ is $\frac13 k$ and the variance is $\frac29 k$. Hence the $\mu$ for $\frac{k'}{k}$ is $\frac13$ and standard deviation $\sigma$ for $\frac{k'}k$ is $$\frac{\sqrt{\frac29k}}{k}=\frac13\sqrt{\frac{2}{k}}. $$
Plug this into Chebyshev.

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Thank you! I knew I must have been messing up the mean/variance calculation. – Slruh Nov 10 '12 at 15:55

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