# Chebyshev's Inequality: sample mean differs from true mean

Let p be the (unknown) true fraction of voters who support a particular candidate A for office. To estimate p, we poll a random sample of n voters. Let Fn be the fraction of voters who support A among n randomly selected voters.

(a) Using Chebyshev’s inequality, calculate an upper bound on the probability that if we poll 100 voters, our estimate Fn differs from p by more than 0.1. (Hint: you may need to use the fact that x(1 − x) ≤ 1/4 for any 0 ≤ x ≤ 1).

What I have tried :

Chebyshev's Inequality : $$P(|X-\mu| \geq k\sigma) \leq 1/k^2$$

So since p is our true fraction of voters : $$\mu = p$$

Now I know I have to use the 100 voters and the fact p differs from 0.1 but I'm not sure how to fit it into my inequality. Also, I have thought of find the standard deviation but since I can't calculate the $$Var(X) = E[X^2] - \mu$$since I can't find E[X^2]

• Here your $X$ is $F_n$. What is the distribution of $F_n$? Binomial. The moments are known for such distribution. The hint uses $x(1-x)$ should have helped you think about the variance of the Bernoulli distribution $p(1-p)$. Mar 27, 2018 at 19:12