| 設有一組數據\({x_1},\;{x_2},\; \cdots \;,\;{x_n}\)，全距為\(R\)，標準差為\(\sigma \)，則\(\sigma \le \frac{1}{2}R\)。 |
不失一般性，設\({x_1} \le {x_2} \le \cdots \le {x_n}\)，平均數\(\mu = \frac{1}{n}({x_1} + {x_2} + \cdots + {x_n})\)
全距\(R = {x_n} – {x_1}\)，標準差\(\sigma = \sqrt {\frac{1}{n}\left( {{{({x_1} – \mu )}^2} + {{({x_2} – \mu )}^2} + \cdots + {{({x_n} – \mu )}^2}} \right)} \)
考慮底下性質：
| 設\(f(x) = {({x_1} – x)^2} + {({x_2} – x)^2} + \cdots + {({x_n} – x)^2}\)，則當\(x = \frac{1}{n}({x_1} + {x_2} + \cdots + {x_n})\)時，\(f(x)\)有最小值。 |
上述性質進一步可表示為：
| 若有一組數據\({x_1},\;{x_2},\; \cdots \;,\;{x_n}\)，其中標準差為\(\sigma \)，則當\(x = \frac{1}{n}({x_1} + {x_2} + \cdots + {x_n}) = \mu \)時，\(f(x) = {({x_1} – x)^2} + {({x_2} – x)^2} + \cdots + {({x_n} – x)^2}\)有最小值為\(f(\mu ) = {({x_1} – \mu )^2} + {({x_2} – \mu )^2} + \cdots + {({x_n} – \mu )^2} = n \cdot {\sigma ^2}\)。 |
設\(k = \frac{{{x_1} + {x_n}}}{2}\)，\(f(x) = {({x_1} – x)^2} + {({x_2} – x)^2} + \cdots + {({x_n} – x)^2}\)，則
\[\left\{ \begin{array}{l}|{x_1} – k|\; = \;|{x_1} – \frac{{{x_1} + {x_n}}}{2}|\; = \;|\frac{{{x_1} – {x_n}}}{2}|\; = \frac{1}{2}R\\|{x_2} – k|\; = \;|{x_2} – \frac{{{x_1} + {x_n}}}{2}|\; \le \;|\frac{{{x_1} – {x_n}}}{2}|\; = \frac{1}{2}R\\\quad \;\; \vdots \\|{x_i} – k|\; = \;|{x_i} – \frac{{{x_1} + {x_n}}}{2}|\; \le \;|\frac{{{x_1} – {x_n}}}{2}|\; = \frac{1}{2}R\\\quad \;\; \vdots \\|{x_n} – k|\; = \;|{x_n} – \frac{{{x_1} + {x_n}}}{2}|\; = \;|\frac{{{x_n} – {x_1}}}{2}|\; = \frac{1}{2}R\end{array} \right.\]
因此\(f(k) = {({x_1} – k)^2} + {({x_2} – k)^2} + \cdots + {({x_n} – k)^2} \le n \cdot {\left( {\frac{1}{2}R} \right)^2}\)
\( \Rightarrow \)\(f(\mu ) \le f(k) \le n \cdot {\left( {\frac{1}{2}R} \right)^2}\)
\( \Rightarrow \)\(n \cdot {\sigma ^2} \le f(k) \le n \cdot {\left( {\frac{1}{2}R} \right)^2}\)
故\(\sigma \le \frac{1}{2}R\)
