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The question is as follows: Given that $X$ is a random variable, and $E(X)=0$, $\operatorname{var}(X)=1$, find $k$ such that: $P(|X|>k)<0.01$. The trouble that I'm having is that I do not understand how to use the information that $E(X)=0$ and $\operatorname{var}(X)=1$.

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    Have you heard of Chebyshev's inequality? – 2017-02-02
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    Yes I have, but never thoroughly understood where it should be used, thank you for your response! – 2017-02-02
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    Chebyshev's inequality is used to bound the probability that a random variable deviates from its mean by more than some fixed amount, when the random variable is known *only* to have finite variance. With more information, better estimates are usually possible, but at this level of generality this is all that can be done. – 2017-02-02

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So we know that P(|X| ≥ k) ≤ $1\over k^2$ since $E(x)=0$ and $V(x)=1$. Hint: this is the same as solving for $k$ but we have $${1\over k^2} ≤ 0.01$$ We already know that the end result must be $≤ 0.01$ and by following Chebyshev's inequality, you should be able to solve for $k$ now.

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    Again, thank you very much! – 2017-02-02
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    You're welcome. If you have any more similar questions, please refer to the textbook. It's clear and to the point :) – 2017-02-02
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    Just to be sure, the k should be 10? – 2017-02-02
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    Yes but more specifically k should be greater than or equal to 10. – 2017-02-02
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This is a basic concept of Chebyshev's Inequality. In your case, your expected value is 0 while your variance is 1, which is great because it will help simplify the inequality and this will let you solve for k. Here is the link of the probability textbook I used last semester when I took the course. I'm assuming this is in the discrete case so head to page 313 and this should answer your question. https://math.dartmouth.edu/~prob/prob/prob.pdf

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    I see how it all comes together now, thank you very much! – 2017-02-02
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    Sure. I will provide a brief explanation on another post just to make sure you still have any lingering questions. – 2017-02-02