In a certain distribution, the mean is 50 with a standard deviation of 6 . Use Chebyshev's theorem to tell the probabilitythat a number lies in the following interval. Round your results to the nearest whole percent.
-Between 20 and 80
A) At least $96 \%$
B) At least $94 \%$
C) At least $98 \%$
D) At least $99 \%$

Question

Quizplus · Accepted Answer

Chebyshev's theorem states that at least $1 - \frac{1}{k^2}$ of the data lies within $k$ standard deviations of the mean. Here, $k = \frac{80-50}{6} = \frac{30}{6} = 5$. So, at least $1 - \frac{1}{5^2} = 1 - \frac{1}{25} = 0.96$ or $96\%$ of the data lies between 20 and 80.

In a Certain Distribution, the Mean Is 50 with a Standard