Answer
$94\%$
Work Step by Step
Chebyshev’s theorem The proportion of values from a data set that will fall within k
standard deviations of the mean will be at least $1−1/k^2$, where k is a number greater
than 1 (k is not necessarily an integer).
33. In a distribution of 160 values with a mean of 72, at
least 120 fall within the interval 67–77.
$120/160=0.75$ Let $1−1/k^2=0.75$ we have $k=2$ so $2s=5,s=2.5$
Approximately hat percentage of values should fall in the interval 62–82?
$k=\frac{82-72}{2.5}=4$ and $1-1/4^2=0.94$
