I am trying to solve the following problem:
The radius of a circle is a random variable X, with the mean value $\mu_x=10$, and variance $\sigma^2=5$. Compute a lower bound on the probability that the radius lies between 8 and 12 using Chebyshev's inequality.
According to Chebyshev's inequality:
$$P[|X-\mu_X| \le k] > 1-\frac{\sigma^2}{k^2} \tag{1}$$
This is equivalent to:
$$P[\mu_X - k \le X \le \mu_X + k] > 1-\frac{\sigma^2}{k^2} \tag{2}$$
As we need to compute $P[8 \le X \le 12]$, therefore, comparing $P[\mu_X - k \le X \le \mu_X + k]$ and $P[8 \le X \le 12]$,
$$\mu_X - k = 8 \tag{3}$$ $$\mu_X + k = 12 \tag{4}$$
As $\mu_X=10$, therefore, $k=2$. Using these values in (2) produces the following:
$$P[8 \le X \le 12]> 1-\frac{\sigma^2}{k^2} = 1-\frac{5}{4} \tag{5}$$
$$P[8 \le X \le 12]> -\frac{1}{4} \tag{6}$$
This results in a negative value of the bound on the probability. What could be wrong? I would much appreciate any help with this.
As far as I can tell, nothing has gone wrong. Chebyshev's inequality doesn't tell you anything if what you're looking at is within one standard deviation of the mean -- and in this case, it is.
A more illuminating example might be to consider what actually happens. Suppose that $$P(X=10-\sqrt{5})=\frac12=P(X=10+\sqrt{5})$$ The standard deviation of this data is, of course, $\sqrt{5}$, (with variance $5$), but $P[8\le X\le 12]=0.$ Of course, if the distribution were uniform or normal, then the probability will be greater, but this example demonstrates that Chebyshev, by itself, is absolutely useless in this scenario.