【数学I】例題5.1.7：分散と標準偏差（One More）★★

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% 例題I5.1.7：分散と標準偏差（One More）★★
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下の表はA，Bの2つの倉庫で1日あたりの荷物の搬入量(単位：箱)を10日間調査した結果である．

\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|}
\hline 日&1&2&3&4&5&6&7&8&9&10\\
\hline A倉庫の搬入量(箱)&4&6&5&3&7&5&6&5&4&5\\
\hline B倉庫の搬入量(箱)&3&5&4&3&4&5&4&3&5&4\\
\hline
\end{tabular}

(1) A倉庫，B倉庫それぞれの搬入量の平均値$\overline{a},\overline{b},$分散$s_a{}^2,s_b{}^2$,標準偏差$s_a$,$s_b$を求めよ．ただし，$\sqrt{2}=1.41,\sqrt{3}=1.73,\sqrt{5}=2.24$とし，標準偏差は小数第2位を四捨五入して答えよ．

(2) (1)から，A倉庫，B倉庫の2つの倉庫の搬入量の散らばりはどちらが大きいか．

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% 例題I5.1.7の解答（One More）★★
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(1)

\begin{align*}
\overline{a}=&\frac{1}{10}(4+6+5+3+7+5+6+5+4+5)=5\text{(箱)}\\
{s_a}^2=&\frac{1}{10}\{(4-5)^2+(6-5)^2+(5-5)^2+(3-5)^2+(7-5)^2\\
&+(5-5)^2+(6-5)^2+(5-5)^2+(4-5)^2+(5-5)^2\}\\
=&1.2\\
s_a=&\sqrt{1.2}=\frac{\sqrt{120}}{10}=\frac{\sqrt{2}\times\sqrt{3}\times\sqrt{5}}{5}\fallingdotseq 1.1\text{(箱)}\\
\overline{b}=&\frac{1}{10}(3+5+4+3+4+5+4+3+5+4)=4\text{(箱)}\\
s_b{}^2=&\frac{1}{10}\{(3-4)^2+(5-4)^2+(4-4)^2+(3-4)^2+(4-4)^2\\
&+(5-4)^2+(4-4)^2+(3-4)^2+(5-4)^2+(4-4)^2\}\\
=&0.6\\
s_b=&\sqrt{0.6}=\frac{\sqrt{60}}{10}=\frac{\sqrt{3}\times\sqrt{5}}{5}\fallingdotseq 0.8\text{(箱)}
\end{align*}

(2) $s_a>s_b$より，A倉庫の方が搬入量の散らばりが大きい．

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% 問題I5.1.7：分散と標準偏差（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

下の表はA，Bの2つの倉庫で1日あたりの荷物の搬入量(単位：箱)を10日間調査した結果である．

\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|}
\hline 日&1&2&3&4&5&6&7&8&9&10\\
\hline A倉庫の搬入量(箱)&8&6&6&7&7&8&8&7&6&7\\
\hline B倉庫の搬入量(箱)&5&4&5&6&4&5&5&7&4&5\\
\hline
\end{tabular}

(1) A倉庫，B倉庫それぞれの搬入量の平均値$\overline{a},\overline{b},$分散$s_a{}^2,s_b{}^2$,標準偏差$s_a$,$s_b$を求めよ．ただし，$\sqrt{3}=1.73,\sqrt{5}=2.24$とし，標準偏差は小数第2位を四捨五入して答えよ．

(2) (1)から，A倉庫，B倉庫の2つの倉庫の搬入量の散らばりはどちらが大きいか．

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% 問題I5.1.7の解答（One More）★★
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(1)

\begin{align*}
\overline{a}=&\frac{1}{10}(8+6+6+7+7+8+8+7+6+7)=7\text{(箱)}\\
{s_a}^2=&\frac{1}{10}\{(8-7)^2+(6-7)^2+(6-7)^2+(7-7)^2+(7-7)^2\\
&+(8-7)^2+(8-7)^2+(7-7)^2+(6-7)^2+(7-7)^2\}\\
=&0.6\\
s_a=&\sqrt{0.6}=\frac{\sqrt{60}}{10}=\frac{\sqrt{3}\times\sqrt{5}}{5}\fallingdotseq 0.8\text{(箱)}\\
\overline{b}=&\frac{1}{10}(5+4+5+6+4+5+5+7+4+5)=5\text{(箱)}\\
s_b{}^2=&\frac{1}{10}\{(5-5)^2+(4-5)^2+(5-5)^2+(6-5)^2+(4-5)^2\\
&+(5-5)^2+(5-5)^2+(7-5)^2+(4-5)^2+(5-5)^2\}\\
=&0.8\\
s_b=&\sqrt{0.8}=\frac{\sqrt{80}}{10}=\frac{\sqrt{16}\times\sqrt{5}}{10}\fallingdotseq 0.9\text{(箱)}
\end{align*}

(2) $s_b>s_a$より，B倉庫の方が搬入量の散らばりが大きい．

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% 例題I5.1.7の側注：分散と標準偏差（One More）★★
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% 側注1
平均値は整数である．

% 側注2
$s^2=(x^2\text{の平均値})-(x\text{の平均値})^2$を用いて求めてもよい．

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% 問題I5.1.7の側注：分散と標準偏差（One More）★★
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% 側注1
平均値は整数である．

% 側注2
$s^2=(x^2\text{の平均値})-(x\text{の平均値})^2$を用いて求めてもよい．

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% 例題I5.1.7の考え方・補足など：分散と標準偏差（One More）★★
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% 考え方
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(1) 変量$x$のデータが$n$個で，その値が，$x_1,x_2,\ldots,x_n$であるとき，平均値を$\overline{x}$とする．このとき，分散$s^2=\frac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+\cdots+(x_n-\overline{x})^2}{n}\cdots(\mathrm{i})$と標準偏差$s=\sqrt{\frac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+\cdots+(x_n-\overline{x})^2}{n}}$を求める．

なお，分散は，$s^2=(x^2\text{の平均値})-(x\text{の平均値})^2\cdots(\mathrm{ii})$により求めることもできる．$x$の平均値が整数ではないときは，(i)ではなく(ii)を用いた方が計算が楽になることが多い．

(2) 標準偏差（分散）が大きいほど，データの散らばりも大きくなることを利用する．