Empirical Rule and Chebyshev's Theorem

Question

Consider a data set that has a mean of 44.6 and a standard deviation of 7.6. According to Chebyshev's Theorem, give an interval that has at least 89.99% of the data.

Answer 1

Chebyshev's: P(|x-µ| >= kσ) <= 1/k² so you solve for x given μ = 44.6, 1/k² = .8999, and kσ = 7.6 (1/sqrt(.8999))

Easiest is to solve for x without the absolute value sign. (This will be the upper bound) The lower bound will be μ - (x_UB- μ) = x_LB

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