I'm trying to solve a problem using Chebychev^' s Inequality:

"Suppose that X is a random variable with mean and variance both equal to 20. What can be said about P(0<X<40)?"


Restating the original equation:


Rewriting to match Chebychev's Inequality

P(|X-20|≤20)≤1- 1/k^2 ;k>0

Find k: P(|X-20|≤(kσ=20))≤1-1/k^2

P(|X-20|≤(k√20=20))≤1- 1/k^2

P(|X-20|≤(k=√20))≤1- 1/k^2



The probability that X falls between 0 and 40 is less than or equal to 0.95.

Now, I know that my answer is wrong. I am either sloppy at inequalities, or I don't understand how to use them correctly.

In this case, the correct answer should be


Can anyone help me understand where my misunderstanding is?

Here's an image that may be easier to read: screen capture of work.


1 Answer 1


This line is false: $P(|X-20|≤20)≤1- 1/k^2 ;k>0$.

You started with: $A=P(|X-20|\geq20)≤ 1/k^2=B$.

To get to the next line, you need to change the direction of the inequality because you multiply both side by a negative constant ($-1$):

$$A\leq B \Rightarrow 1-A \geq 1-B$$

  • $\begingroup$ I see it now. Ugh. $\endgroup$
    – Jred
    Commented Jul 23, 2022 at 15:53

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