【数学I】例題5.1.8：データの値の決定（One More）★★

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I5.1.8：データの値の決定（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

右の表は会社Aと会社Bの4日間における商品の販売数のデータである．

(1) 会社Aの販売数の平均値と分散を求めよ．

(2) 会社Bの1日目から4日目までの販売数の平均値は10個，分散は8であるとき，会社Bの販売数$a,b$を求めよ．ただし，$a<b$とする．

\begin{tabular}{|c||c|c|}
\hline 日目&会社A&会社B\\
\hline 1&4&$a$\\
\hline 2&7&14\\
\hline 3&6&10\\
\hline 4&5&$b$\\
\hline
\end{tabular}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I5.1.8の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 会社Aの販売数の平均値は，

\[
\frac{1}{4}(4+7+6+5)=5.5\text{(個)}
\]

会社Aの販売数の分散は，

\[
\frac{1}{4}(4^2+7^2+6^2+5^2)-5.5^2=31.5-30.25=1.25
\]

(2) 平均値が10個であるから，

\[
\frac{1}{4}(a+14+10+b)=10
\]

したがって，$a+b=16\cdots(\mathrm{i})$

分散が8であるから，

\[
\frac{1}{4}\{(a-10)^2+(14-10)^2+(10-10)^2+(b-10)^2\}=8
\]

ゆえに，$(a-10)^2+(b-10)^2=16\cdots(\mathrm{ii})$

(i) より，$b=16-a$を(ii)に代入すると，$a^2-16a+60=0$

よって，$a=6,10$

これを(i)に代入すると，$a<b$より，$a=6,b=10$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I5.1.8：データの値の決定（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

右の表はクラスAとクラスBの4日間における生徒の読書ページ数のデータである．

(1) クラスAの読書ページ数の平均値と分散を求めよ．

(2) クラスBの1日目から4日目までの読書ページ数の平均値は12ページ，分散は30.5であるとき，クラスBの読書ページ数$a,b$を求めよ．ただし，$a<b$とする．

\begin{tabular}{|c||c|c|}
\hline 日目&クラスA&クラスB\\
\hline 1&5&$a$\\
\hline 2&8&16\\
\hline 3&7&12\\
\hline 4&6&$b$\\
\hline
\end{tabular}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I5.1.8の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) クラスAの読書ページ数の平均値は，$\frac{1}{4}(5+8+7+6)=6.5\text{(ページ)}$

クラスAの読書ページ数の分散は，

\[
\frac{1}{4}(5^2+8^2+7^2+6^2)-6.5^2=43.5-42.25=1.25
\]

(2) 平均値が12ページであるから，$\frac{1}{4}(a+16+12+b)=12$

したがって，$a+b=20\cdots(\mathrm{i})$

分散が$30.5$であるから，

\[
\frac{1}{4}\{(a-12)^2+(16-12)^2+(12-12)^2+(b-12)^2\}=30.5
\]

ゆえに，$(a-12)^2+(b-12)^2=106\cdots(\mathrm{ii})$

(i) より，$b=20-a$を(ii)に代入すると，$a^2-20a+51=0$

よって，$a=3,17$

これを(i)に代入すると，$a<b$より，$a=3,b=17$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I5.1.8の側注：データの値の決定（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$s^2=(x^2\text{の平均値})-(x\text{の平均値})^2$を用いて求めている．なお，
\[
\frac{1}{4}\{(4-5.5)^2+(7-5.5)^2+\cdots\}
\]
と偏差を2乗した値を考えて求めてもよいが，計算に手間が掛かる．

% 側注2
$b$を代入して展開し，整理する．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I5.1.8の側注：データの値の決定（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$s^2=(x^2\text{の平均値})-(x\text{の平均値})^2$を用いて求めている．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I5.1.8の考え方・補足など：データの値の決定（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 考え方
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

データ$x_1,x_2,\ldots,x_n$の平均値を$\overline{x}$，分散を$s^2$とすると，

\[
s^2=\overline{x^2}-(\overline{x})^2
\]

つまり，分散は，$s^2=(x^2\text{の平均値})-(x\text{の平均値})^2$により求めることができる．$x$の平均値が整数ではないときは，偏差を2乗した値の平均$s^2=\frac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+\cdots+(x_n-\overline{x})^2}{n}$で求めようとすると，計算に手間が掛かるので注意すること．