Question

Suppose that quiz scores in a beginning statistics class have a mean of 7.5 with a standard deviation of 0.4. Using Chebyshev's Theorem, state the range in which at least 88.9% of the data will reside. Please do not round your answers.

Answer 1

Chebyshev's theorem says that the probability that a random variable differs from its mean by some multiple of the standard deviation is bounded below as

$P\left(|X - \mu| \le k\sigma\right) \ge \dfrac1{k^2}$

Solve for k :

$0.889 = \dfrac1{k^2} \implies k^2 = \dfrac1{0.889} \implies k = \dfrac1{\sqrt{0.889}}$

Then the theorem says

$P\left(|X - \mu| \le \dfrac{\sigma}{\sqrt{0.889}}\right) \ge 0.889$

which is to say that 88.9% of the data will fall in the interval

$|X - 7.5| \le \dfrac{0.4}{\sqrt{0.889}} \\\\ -\dfrac{0.4}{\sqrt{0.889}} \le X - 7.5 \le \dfrac{0.4}{\sqrt{0.889}} \\\\ \boxed{7.5 - \dfrac{0.4}{\sqrt{0.889}} \le X \le 7.5 + \dfrac{0.4}{\sqrt{0.889}}}$

or roughly corresponding to scores between 7.076 and 7.924.