問題の解答
検索用コード（LaTeX）
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% 例題I5.1.3：平均値の値（One More）★
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右の表は，ある家庭の10日間における1日ごとの水道使用量（整数）の記録である．
(1) このデータの平均値の最小値と最大値を求めよ．
(2) 10日間の水道使用量の平均は453（L）であり，各日の水道使用量は$x,710,620,565,525,450,430,360,315,270$(L)であった．$x$の値を求めよ．
\begin{tabular}{|c||c|}
\hline 使用量の階級（L）&日数\\
\hline 200以上300未満&2\\
\hline 300〜400&2\\
\hline 400〜500&2\\
\hline 500〜600&2\\
\hline 600〜700&1\\
\hline 700〜800&1\\
\hline 計&10\\
\hline
\end{tabular}
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% 例題I5.1.3の解答（One More）★
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(1) データの平均値が最小となるのは，データのそれぞれの値が各階級の最小の値となるときであるから，
\[
\frac{1}{10}(200\times 2+300\times 2+400\times 2+500\times 2+600\times 1+700\times 1)=410
\]
よって，平均値の最小値は，$410(\mathrm{L})$
データの平均値が最大となるのは，データのそれぞれの値が各階級の最大の値となるときであるから，
\[
\frac{1}{10}(299\times 2+399\times 2+499\times 2+599\times 2+699\times 1+799\times 1)=509
\]
よって，平均値の最大値は，$509(\mathrm{L})$
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% 例題I5.1.3の別解
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データの平均値が最大となるのは，データのそれぞれの値が最小の値より99Lだけ大きいときであるから，平均使用量も99L高くなり，$410+99=509(\mathrm{L})$
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% 例題I5.1.3の解答（One More）★
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(2) 合計使用量を考えると，
\[
x+710+620+565+525+450+430+360+315+270=453\times 10
\]
したがって，$x+4245=4530$
よって，$x=285$
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% 問題I5.1.3：平均値の値（One More）★
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右の表は，ある家庭の10日間における1日ごとの電気使用量（整数）の記録である．
(1) このデータの平均値の最小値と最大値を求めよ．
(2) 10日間の電気使用量の平均は12（kWh）であり，各日の電気使用量は$x,18,17,13,12,10,10,8,7,6$(kWh)であった．$x$の値を求めよ．
\begin{tabular}{|c||c|}
\hline 使用量の階級（kWh）&日数\\
\hline 5以上10未満&3\\
\hline 10〜15&4\\
\hline 15〜20&3\\
\hline 計&10\\
\hline
\end{tabular}
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% 問題I5.1.3の解答（One More）★
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(1) データの平均値が最小となるのは，データのそれぞれの値が各階級の最小の値となるときであるから，$\frac{1}{10}(5\times 3+10\times 4+15\times 3)=10$
よって，平均値の最小値は，$10(\mathrm{kWh})$
データの平均値が最大となるのは，データのそれぞれの値が各階級の最大の値となるときであるから，$\frac{1}{10}(9\times 3+14\times 4+19\times 3)=14$
よって，平均値の最大値は，$14(\mathrm{kWh})$
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% 問題I5.1.3の別解
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データの平均値が最大となるのは，データのそれぞれの値が最小の値より4kWhだけ大きいときであるから，平均使用量も4kWh高くなり，$10+4=14(\mathrm{kWh})$
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% 問題I5.1.3の解答（One More）★
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(2) 合計使用量を考えると，
\[
x+18+17+13+12+10+10+8+7+6=12\times 10
\]
したがって，$x+101=120$
よって，$x=19$
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