Final answer:
This question deals with Chebyshev's inequality and asks to demonstrate that for a random variable X with mean μ and variance σ^2, the probability that X deviates from its mean by at least k times its standard deviation is at most 1/k^2.
Step-by-step explanation:
The student is asking to show that for a random variable X with finite mean μ and variance σ2, the probability that the absolute value of the deviation of X from its mean exceeds k times the standard deviation is at most 1/k2. This statement can be formulated as PX - μ ≤ 1/k2. This is a consequence of Chebyshev's inequality which states that for any k > 0, the inequality is true for a random variable with finite variance.
To give an intuitive explanation using an example, let's assume X is the amount of money a student has, which is approximately exponentially distributed with a certain mean μ. If we choose a particular multiplier k as our threshold, the probability that a student's amount of money deviates from the mean by more than that multiplier times the standard deviation is limited by 1 divided by k squared. If k is large, implying we're considering a very large deviation, the probability becomes very small.