【数学I】例題5.1.10：変量の変換（One More）★★

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% 例題I5.1.10：変量の変換（One More）★★
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変量$x$のデータの平均値$\overline{x}$が$\overline{x}=30$，分散${s_x}^2=20$であるとする．このとき，次の式によって得られる変量$y$のデータについて，平均値$\overline{y}$，分散${s_y}^2$，標準偏差$s_y$を求めよ．ただし，$\sqrt{5}=2.24$とし，標準偏差は小数第2位を四捨五入して，小数第1位まで求めよ．

(1) $y=x+10$

(2) $y=4x$

(3) $y=-3x+4$

(4) $y=\frac{x-30}{2\sqrt{5}}$

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(1)

\begin{align*}
&\overline{y}=\overline{x}+10=30+10=40\\
&{s_y}^2=1^2\times{s_x}^2=20\\
&s_y=1\times s_x=\sqrt{20}=2\sqrt{5}=4.48\fallingdotseq 4.5
\end{align*}

(2)

\begin{align*}
&\overline{y}=4\overline{x}=4\times 30=120\\
&{s_y}^2=4^2\times{s_x}^2=16\times 20=320\\
&s_y=4\times s_x=4\times\sqrt{20}=8\sqrt{5}=17.92\fallingdotseq 17.9
\end{align*}

(3)

\begin{align*}
&\overline{y}=-3\overline{x}+4=-3\times 30+4=-90+4=-86\\
&{s_y}^2=(-3)^2\times{s_x}^2=9\times 20=180\\
&s_y=|{-3}|\times s_x=3\times\sqrt{20}=6\sqrt{5}=13.44\fallingdotseq 13.4
\end{align*}

(4)

\begin{align*}
&\overline{y}=\frac{\overline{x}-30}{2\sqrt{5}}=\frac{30-30}{2\sqrt{5}}=0\\
&{s_y}^2=\frac{{s_x}^2}{(2\sqrt{5})^2}=\frac{20}{20}=1\\
&s_y=\frac{s_x}{2\sqrt{5}}=\frac{\sqrt{20}}{2\sqrt{5}}=1
\end{align*}

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% 問題I5.1.10：変量の変換（One More）★★
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変量$x$のデータの平均値$\overline{x}$が$\overline{x}=50$，分散${s_x}^2=36$であるとする．このとき，次の式によって得られる変量$y$のデータについて，平均値$\overline{y}$，分散${s_y}^2$，標準偏差$s_y$を求めよ．

(1) $y=x-20$

(2) $y=3x$

(3) $y=-2x+10$

(4) $y=\frac{x-50}{6}$

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(1)

\begin{align*}
&\overline{y}=\overline{x}-20=50-20=30\\
&{s_y}^2=1^2\times{s_x}^2=36\\
&s_y=1\times s_x=\sqrt{36}=6
\end{align*}

(2)

\begin{align*}
&\overline{y}=3\overline{x}=3\times 50=150\\
&{s_y}^2=3^2\times{s_x}^2=9\times 36=324\\
&s_y=3\times s_x=3\times\sqrt{36}=18
\end{align*}

(3)

\begin{align*}
&\overline{y}=-2\overline{x}+10=-2\times 50+10=-100+10=-90\\
&{s_y}^2=(-2)^2\times{s_x}^2=4\times 36=144\\
&s_y=|{-2}|\times s_x=2\times\sqrt{36}=12
\end{align*}

(4)

\begin{align*}
&\overline{y}=\frac{\overline{x}-50}{6}=\frac{50-50}{6}=0\\
&{s_y}^2=\frac{{s_x}^2}{6^2}=\frac{36}{36}=1\\
&s_y=\frac{s_x}{6}=\frac{\sqrt{36}}{6}=1
\end{align*}

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% 例題I5.1.10の側注：変量の変換（One More）★★
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% 側注1
標準偏差$s_y$は，分散${s_y}^2$より求めてもよい．

例：(1)${s_y}^2=20$より，
\[
s_y=\sqrt{20}=2\sqrt{5}
\]

(2) ${s_y}^2=320$より，
\[
s_y=\sqrt{320}=8\sqrt{5}
\]

% 側注2
(4)は変量$x$を平均値が0，標準偏差が1となるように標準化したものである（下の余談を参照）．

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% 問題I5.1.10の側注：変量の変換（One More）★★
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% 側注1
標準偏差$s_y$は，分散${s_y}^2$より求めてもよい．

例：(1)${s_y}^2=36$より，
\[
s_y=\sqrt{36}=6
\]

(2) ${s_y}^2=324$より，
\[
s_y=\sqrt{324}=18
\]

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% 例題I5.1.10の考え方・補足など：変量の変換（One More）★★
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% 考え方
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$a,b$は定数とする．変量$x$のデータから$y=ax+b$によって変量$y$のデータが得られるとき，$x,y$のデータの平均値をそれぞれ$\overline{x},\overline{y}$，分散をそれぞれ$s_x{}^2,s_y{}^2$，標準偏差をそれぞれ$s_x,s_y$とすると，次の表のようにまとめられる．

\begin{tabular}{|c||c|c|c|}
\hline 変量&平均値&分散&標準偏差\\
\hline$x$&$\overline{x}$&$s_x{}^2$&$s_x$\\
\hline$y=ax+b$&$\overline{y}=a\overline{x}+b$&$s_y{}^2=a^2s_x{}^2$&$s_y=|a|s_x$\\
\hline
\end{tabular}

平均値$\overline{y}=a\overline{x}+b$，分散$s_y{}^2=a^2s_x{}^2$，標準偏差$s_y=|a|s_x$を利用して，それぞれの値を求める．

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$c(c\neq 0)$を定数とする．変量の変換$z=\frac{x-x_0}{c}$を考えると，$z=\frac{1}{c}x-\frac{x_0}{c}$であるから，平均値$\overline{z}$は，$\overline{z}=\frac{1}{c}\overline{x}-\frac{x_0}{c}=\frac{\overline{x}-x_0}{c}$となる．また，標準偏差$s_z$は，$s_z=|\frac{1}{c}|s_x$となる．

ここで，$s_x>0$とし，$x_0=\overline{x},c=s_x$とすると，$\overline{z}=\frac{\overline{x}-\overline{x}}{s_x}=0,s_z=|\frac{1}{s_x}|s_x=1$となる．

この変換を標準化といい，変量$z$を，$x$を標準化して得られた変量という．標準化は，複数のデータの平均値や散らばり具合を揃えることに利用される．また，ここでは扱わないが，標準化に関する概念として偏差値がよく知られている．