How would I use Chebyshev's inequality for this problem?

Question

A probability distribution has a mean of 50 and a standard deviation of $2$. Use Chebyshev's inequality to find the minimum probability that an outcome is between $42$ and $58$.

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How would I use Chebyshev's inequality for this problem? Any similar questions, feel free to link it in the comment section, as well as any questions. Thanks

Answer 1

A big hint:

$$42 < X < 58$$

is the same as

$$|X-50|<8$$

Answer 2

Hint:

Let $X$ be the r.v. the problem is referencing. Then $$ \Bbb{P}( 42 \leq X \leq 58 ) = \Bbb{P}(|X-50| \leq 8). $$ Now, use Chebechev's inequality here.