# Using Chebyshev's Theorem - statistics

I can state and prove Chebyshev's inequality however I'm having a lot of trouble actually using it. For example, in the following question, I don't understand how the values for $k \sigma$ and $\dfrac{1}{k^2}$ are found?

When a fair tetrahedronal (four-sided) die is thrown, the probability of a one is obviously $\dfrac{1}{4}$. Use Chebyshev's inequality to estimate $n$, the number of times the die must be thrown in order that the probability will be at least $0.8$ that the proportion of ones is between $0.15$ and $0.35$.

The Chebyshev Inequality I use is of the form:

$P(|y-\mu| \geq k \sigma) \leq \dfrac{1}{k^2}$

I thought $k \sigma$ might be the average between the difference of $0.15$ and $0.35$?

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This is not Chebyshev inequality, please have a new look at your notes. And when you will have a correct definition, you may want to add your own thoughts. – Did May 8 '14 at 9:55

What you are looking for is the proportion of ones is between $0.15$ and $0.35$.
Having $0.15 < y < 0.35$ is the same as $-0.1 < y - 0.25 < 0.1$ , which is exactly $|y - 0.25| < 0.1$ .

You found $\mu = 0.25$, that is $|y - \mu| < 0.1$ hence you want : $$P(|y - 0.25| < 0.1) > 0.8$$

Know look at your inequality and notice that's exactly the same as : $$P(|y - 0.25| \ge 0.1) \le 0.2$$ where $\frac{1}{k^2} = 0.2$ and $k\sigma = 0.1$.

Throwing the dice is distributed as $Bin(n,p)$ where $p = 1/4$ and $\sigma^2 = np(1-p)$.

Can you find $n$ now?

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Sorry, I really don't see where you obtained the $k \sigma =0.1$ from? And is the $\dfrac{1}{k^2}$ just equal to the difference between your values? I've only been given an example to work from, not actually explained how to do it? (Thank you by the way) – sarahusher May 8 '14 at 10:05
@sarahusher: I've edited it, hope it explains more clearly. – user88595 May 8 '14 at 10:09
The notation for binomials is Bin(n,p), not what you wrote. – Did May 8 '14 at 10:10
okay, I understand the 0.1 now! Thank you, but why is $\dfrac{1}{k^2} = 0.2$ ? – sarahusher May 8 '14 at 10:18
Oh I think I get it!! I got n=94? Thanks so much, I really appreciate the help – sarahusher May 8 '14 at 10:23