Well, Chebyshev tells us: If we are k standard deviations away from mean (smaller or bigger than mean), no more than 1/k2of the values will lie be farther than k standard deviations away from mean.
In your problem, 110 = 50 + 4(15), so we are talking about 4 standard deviations from the mean (k = 4).
Then, no more than 1/42 = 1/16 of our data points can be more than 4 standard deviations away from the mean (50). So the rest (at least 15/16) of the data points are within 4 st. dev. In percents this is 15/16 • 100 = 93.75%
But note that we just found that at least 93.75% of our data must be 4 st. dev. smaller OR bigger than our mean of 50. Since we just want the bigger values (50 to 110), our answer is half that, or __ 46.875% __
This problem does not need empirical rule (and you cannot use it unless the data are assumed to be normal), but the empirical rule states that approx. 68% of data are within 1 standard deviation of mean, 95% of data are within 2 standard deviations of mean, and 99.7% of data lie within 3 standard deviations of mean.
In this data set (if it is normal) with a mean of 50 and st. dev. of 15, that means that
- 68% of data are in the range 50 ± 15, or (35, 65)
- 95% of data are in the range 50 ± 30, or (20, 80)
- 99.7% of data are in range 50 ± 45, or (5, 95)