【数学I】例題5.1.11：仮平均を利用したデータの平均値，分散（One More）★★

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% 例題I5.1.11：仮平均を利用したデータの平均値，分散（One More）★★
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次のような変量$x$のデータがある．

\[
650,720,690,680,710,730,680,660,690,710
\]

(1) $y=x-700$とおくことにより，変量$x$のデータの平均値$\overline{x}$を求めよ．

(2) $z=\frac{x-700}{10}$とおくことにより，変量$x$のデータの分散を求めよ．

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% 例題I5.1.11の解答（One More）★★
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(1) $y$のデータの平均値を$\overline{y}$とすると，

\[
\overline{y}=\frac{1}{10}\{-50+20-10-20+10+30-20-40-10+10\}=-8
\]

よって，$\overline{x}$は，

\[
\overline{x}=\overline{y}+700=-8+700=692
\]

(2) $z=\frac{x-700}{10}$とおくと，$z,z^2$の値は次のようになる．

\[
\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|}
\hline
x&650&720&690&680&710&730&680&660&690&710&\text{計}\\
\hline
z&-5&2&-1&-2&1&3&-2&-4&-1&1&-8\\
\hline
z^2&25&4&1&4&1&9&4&16&1&1&66\\
\hline
\end{array}
\]

したがって，$z$のデータの分散は，$z$のデータの平均値を$\overline{z}$とすると，

\[
\overline{z^2}-(\overline{z})^2=\frac{66}{10}-(\frac{-8}{10})^2=5.96
\]

よって，$x$のデータの分散は，

\[
10^2\times 5.96=100\times 5.96=596
\]

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% 問題I5.1.11：仮平均を利用したデータの平均値，分散（One More）★★
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次のような変量$x$のデータがある．このとき，次の問いに答えよ．

\[
550,620,590,570,610,630,560,580,590,600
\]

(1) $y=x-600$とおくことにより，変量$x$のデータの平均値$\overline{x}$を求めよ．

(2) $z=\frac{x-600}{10}$とおくことにより，変量$x$のデータの分散を求めよ．

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(1) $y$のデータの平均値を$\overline{y}$とすると，

\[
\overline{y}=\frac{1}{10}\{-50+20-10-30+10+30-40-20-10+0\}=-10
\]

よって，$\overline{x}$は，$\overline{x}=\overline{y}+600=-10+600=590$

(2) $z=\frac{x-600}{10}$とおくと，$z,z^2$の値は次のようになる．

\[
\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|}
\hline
x&550&620&590&570&610&630&560&580&590&600&\text{計}\\
\hline
z&-5&2&-1&-3&1&3&-4&-2&-1&0&-10\\
\hline
z^2&25&4&1&9&1&9&16&4&1&0&70\\
\hline
\end{array}
\]

したがって，$z$のデータの分散は，$z$のデータの平均値を$\overline{z}$とすると，

\[
\overline{z^2}-(\overline{z})^2=\frac{70}{10}-(\frac{-10}{10})^2=7-1=6
\]

よって，$x$のデータの分散は，$10^2\times 6=100\times 6=600$

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% 例題I5.1.11の側注：仮平均を利用したデータの平均値，分散（One More）★★
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% 側注1
$\overline{x}=\frac{1}{10}(650+\cdots+710)$を用いても平均値は求められるが，計算に手間が掛かる．

% 側注2
分散は，$(z^2\text{の平均値})-(z\text{の平均値})^2$で求めることができる．

% 側注3
${s_x}^2=10^2s_z{}^2$

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% 問題I5.1.11の側注：仮平均を利用したデータの平均値，分散（One More）★★
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% 側注1
$\overline{x}=\frac{1}{10}(550+\cdots+600)$を用いても平均値は求められるが，計算に手間が掛かる．

% 側注2
${s_x}^2=10^2s_z{}^2$

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% 例題I5.1.11の考え方・補足など：仮平均を利用したデータの平均値，分散（One More）★★
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% 考え方
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(1) $y$のデータの平均値を$\overline{y}$とすると，$\overline{y}=\overline{x}-700$，すなわち，$\overline{x}=\overline{y}+700$であることを利用する．直接$\overline{x}=\frac{1}{10}(650+720+\cdots+710)$を計算するより，平均値を楽に求めることができる．このような平均値の計算を簡単にするためにとった700のような値のことを仮平均という．仮平均を自分で設定する場合，平均値に近いと予想される値をとるとよい（計算が楽になる）．

(2) $x,z$のデータの分散をそれぞれ$s_x{}^2,s_z{}^2$とすると，$z=\frac{x-700}{10}$より，$s_x{}^2=10^2s_z{}^2$である．先に変量$x$のそれぞれの値に対応する変量$z$の値を求め，$s_z{}^2$を計算する．