Problem 20.6
This problem uses the opposite signs greater/less than or equal to signs used in Chebyshev's Inequality, so whereas Chebyshev's equation says P(|X-E[X]|>c)< Var[X]/(c^2), it gives you P(|X-E[X]|<a)> .99. Does this mean that you would switch in c for a and make c=.01 so that you can switch the signs back to Chebysev's original equation? Is this how you would approach this problem?