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Chebyshevs Theorem: Minimum Proportion Within k Standard Deviations

Using chebyshevs theorem, what minimum proportion of observations lies within 2 and 3 standard deviations of the mean, and what \(k\) is required to guarantee at least 90% within \(k\sigma\)?

Subject: Statistics Chapter: Numerical Descriptive Measures Topic: Use of Standard Deviation Chebyshev's Theorem Answer included
chebyshevs Chebyshev's theorem Chebyshev's inequality standard deviation k standard deviations minimum proportion empirical rule comparison variance
Accepted answer Answer included

Chebyshevs theorem (Chebyshev’s inequality) stated precisely

In descriptive statistics, chebyshevs commonly refers to Chebyshev’s theorem (also called Chebyshev’s inequality), which gives a guaranteed minimum proportion of observations within a certain number of standard deviations from the mean.

For any distribution with finite mean \(\mu\) and standard deviation \(\sigma>0\), and for any \(k>1\), \[ P\!\left(\,|X-\mu|

The bound does not assume normality; it applies to any distribution with finite variance.

Minimum proportion within 2 and 3 standard deviations

Apply the formula \(1-\frac{1}{k^2}\) for the requested \(k\) values.

Case \(k=2\)

\[ P\!\left(|X-\mu|<2\sigma\right)\ge 1-\frac{1}{2^2} =1-\frac{1}{4} =\frac{3}{4} =0.75. \]

At least 75% of observations lie within \(2\sigma\) of the mean.

Case \(k=3\)

\[ P\!\left(|X-\mu|<3\sigma\right)\ge 1-\frac{1}{3^2} =1-\frac{1}{9} =\frac{8}{9} \approx 0.8889. \]

At least \(\frac{8}{9}\approx 88.89\%\) of observations lie within \(3\sigma\) of the mean.

Finding \(k\) to guarantee at least 90% within \(k\sigma\)

Require the Chebyshev bound to be at least 0.90: \[ 1-\frac{1}{k^2}\ge 0.90. \]

Solve step-by-step: \[ 1-0.90 \ge \frac{1}{k^2} \quad\Longrightarrow\quad 0.10 \ge \frac{1}{k^2} \quad\Longrightarrow\quad k^2 \ge 10 \quad\Longrightarrow\quad k \ge \sqrt{10}\approx 3.162. \]

A value of \(k=\sqrt{10}\) is the smallest threshold that guarantees at least 90% of observations within \(k\sigma\) by Chebyshev’s theorem.

Quick reference table

\(k\) Minimum proportion within \(k\sigma\) Minimum percent
2 \(\displaystyle 1-\frac{1}{2^2}=\frac{3}{4}\) 75%
3 \(\displaystyle 1-\frac{1}{3^2}=\frac{8}{9}\) 88.89%
\(\sqrt{10}\approx 3.162\) \(\displaystyle 1-\frac{1}{10}=\frac{9}{10}\) 90%
4 \(\displaystyle 1-\frac{1}{16}=\frac{15}{16}\) 93.75%

Visualization: the guaranteed interval \(\mu \pm k\sigma\)

The interval from \(\mu-k\sigma\) to \(\mu+k\sigma\) always contains at least \(1-\frac{1}{k^2}\) of the distribution (for \(k>1\)). The shaded center segment represents the guaranteed proportion.

\(\mu-k\sigma\) \(\mu\) \(\mu+k\sigma\) Guaranteed mass at least \(\,1-\frac{1}{k^2}\,\) \((k>1)\) \((k>1)\)
Chebyshev’s theorem guarantees a minimum proportion inside \([\mu-k\sigma,\mu+k\sigma]\) regardless of distribution shape, provided the variance is finite.

Interpretation and a common comparison

  • Chebyshev’s theorem provides a lower bound; the true proportion within \(k\sigma\) is often larger.
  • For approximately normal data, the empirical rule (about 95% within \(2\sigma\)) is typically much stronger than the Chebyshev guarantee (75% within \(2\sigma\)).
  • When distribution shape is unknown or skewed, Chebyshev’s bound remains valid and conservative.
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