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Chebyshev’s inequality is a probability theorem used to characterize the dispersion or spread of data away from the mean. It was developed by a Russian mathematician called Pafnuty Chebyshev. The theorem states that:

*For any set of observations, whether sample or population data and regardless of the shape of the distribution, the percentage of the observations that lie within k* *standard deviations of the mean is at least \(1 – \cfrac {1}{k^2}\) for all \(k > 1\).*

Simply put, this means that you can use the formula to determine the percentage of observations lying inside or outside a given number of standard deviations. Remember that the standard deviation tells us how far values are from the arithmetic mean. For example, two-thirds of the observations fall within one standard deviation on either side of the mean in a normal distribution. However, Chebyshev’s inequality goes slightly against the 68-95-99.7 rule commonly applied to the normal distribution.

$$ P = 1 – \cfrac {1}{k^2} $$

*Where *

*P is the percentage of observations*

*K is the number of standard deviations*

Suppose we wish to find the percentage of observations lying within two standard deviations of the mean:

k = 2

Hence,

$$ \begin{align*} P & = 1 – \cfrac {1}{2^2} \\ & = 0.75 = 75\% \\ \end{align*} $$

QuestionUsing Chebyshev’s inequality, calculate the percentage of observations that would fall outside 3 standard deviations of the mean.

- 11%
- 89%
- 90%
The correct answer is

B.Working:

note that the question asks for the percentage that would fall

outside3 standard deviations. Therefore:$$ P = 1 – \cfrac {1}{3^2} = 89\% $$

But we are interested in 1 – p , i.e., the outside observations.

Therefore,

$$ 1 – 0.89 = 0.11 \text{ or } 11\% $$