The correct answer is D. 3%.
Chebyshev’s inequality states that for any random variable $X$ with mean $\mu$ and standard deviation $\sigma$, the probability that $|X-\mu|\ge k\sigma$ is less than or equal to $\frac{1}{k^2}$. In other words, at least $1-\frac{1}{k^2}$ of the time, $X$ will be within $k$ standard deviations of its mean.
A “Six Sigma” event is an event that occurs when a random variable is more than 6 standard deviations away from its mean. In other words, a “Six Sigma” event is an event that is extremely unlikely to occur.
Using Chebyshev’s inequality, we can calculate the probability of a “Six Sigma” event as follows:
$$P(|X-\mu|\ge 6\sigma) \le \frac{1}{(6\sigma)^2} = \frac{1}{36} = 3\%$$
Therefore, the probability of a “Six Sigma” event is less than 3%.