For ANY distribution with mean $\mu$ and standard deviation $\sigma$, Chebyshev's inequality states that at least $1 - \frac{1}{k^2}$ of the data lies within $k$ standard deviations of the mean.

A dataset has mean = 120 and SD = 8. At minimum, what percentage of data falls between 88 and 152?

(Note: 88 = 120 − 4(8) and 152 = 120 + 4(8))