According to Chebyshev\'s theorem, the proportion of values from a data set that is further than 2 standard deviations from the mean is at most: A. 0.50 B. 0.13 C. 1.00 D. 0.25 Solution Let X be a random variable whose distribution has mean m and standard deviation s, then Chebyshev\'s theorem says precisely that P(|X - m| ks) 1/k^2, that is the probability or in your case proportion that is more than k standard deviations away from the mean is at most 1/k^2. Applying this to your case, the proportion that is more than 2 standard deivations away must be at most 1/2^2 = .25. Therefore, the answer is d..