A random variable $X$ has $E(X) = 10$ and $SD(x) = 4$. Find upper and lower bounds for $P(0 < X < 40)$
My guess is we need to use Chebyshev's inequality. Since we know $X$ must be between 0 and 40, can we first break it up into $P(X>0)$ and $P(X<40)$? How would I solve this problem?
Best Answer
Lower bound
Since the event in Chebyshev's inequality
$$P(|X-\mu| \ge \epsilon) \le \frac{\sigma^2}{\epsilon^2}\tag{*}\label{*}$$
is symmetric about $\mu = E(X) = 10$, so we consider $P(0 < X < 20) \le P(0 < X < 40)$ and try to find a lower bound for this question. Now we take $\epsilon = 10, \mu = 10, \sigma = 4$ in \eqref{*}.
\begin{aligned} P(|X - 10| \ge 10) &\le \left(\frac{4}{10}\right)^2 \\ P(X \le 0 \text{ or } X \ge 20) &\le \frac{4}{25} \\ P(0 < X < 20) &= P(|X-10|<10) \\ &= 1-P(|X-10|\ge10) \\ &\ge 1 - \frac{4}{25} = \frac{21}{25} \end{aligned}
Hence a lower bound would be
$$\frac{21}{25} \le P(0 < X < 20) \le P(0 < X < 40).$$
Upper bound
Without further information about random variable $X$, we can only conclude that $P(0 < X < 40)$ is bounded above by one by definition of probability due to the existence of a random variable which is distributed near enough to its mean.
Consider a random variable
$$X = \begin{cases} 14 & \text{ with probability } \frac12 \\ 6 & \text{ with probability } \frac12 \end{cases} $$
It's clear that $\mu = E(X) = 10$ and
\begin{align} \sigma =& \sqrt\frac{(14-10)^2+(6-10)^2}{2} \\ =& \sqrt\frac{4^2 + 4^2}{2} = 4. \end{align}
So $X$ satisfies the hypothesis of the question.
Therefore, in this particular case, $P(0 < X < 40) = 1$. In other words, any number less than one won't be an upper bound to the required probability.
Conclusion
In general, we have $$\frac{21}{25} \le P(0 < X < 40) \le 1.$$