Here's a problem I am stuck on. The problem goes as follows:
Suppose the distribution of scores of a test has mean 100 and standard deviation 16. Calculate an upper bound for the probability $P\{X>148\space or \space X<52\}$.
So here is my progress
By the addivity axiom, $P\{X>148\space or \space X<52\}=P\{X>148\}+P\{X<52\}$. Can I use Chebyshev's Inequality on both probabilities or do I use the one-sided Chebyshev Inequality? Or do I use the corollary from the one-sided chebyshev inequality (stated below)?
$P\{X\ge \mu+a\} \le \frac{\sigma^2}{\sigma^2+a^2}$ (1)
$P\{X\le \mu-a\} \le \frac{\sigma^2}{\sigma^2+a^2}$ (2)
Being that the problem states $\mu$, $\sigma$ and $a$, I believe I should use the corollary to get an upper bound.
Since the interval is symmetric about the mean, use the Chebyshev Inequality in the version $$\Pr(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}.$$ The only issue is that we should choose for $k$. Since $148$ and $52$ are both $3$ standard deviation units from the mean, we can take $k=3$, and conclude that $\frac{1}{9}$ is an upper bound for our probability.
But since here the random variable (test score) is probably integer valued, we really want the probability that $x\ge 149$ or $x\le 51$. That would give $k=\frac{49}{16}$, and lead to a somewhat improved upper bound of roughly $0.1067$.
Remark: Because $\frac{48}{16}$ is the "nice" number $3$, you are probably intended to treat the grade as if it had continuous distribution, and obtain the bound $\frac{1}{9}$.