Suppose a random variable X has mean μX = 6.3, and variance V(X) = 2.2. According to Chebyshev's Theorem, what is the upper bound for the probability that the distance between the value of X and μ will be at least 4.8?
I don't even know where to start. But If I can see the steps to complete this problem, it would help out tremendously with the others.
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