【高校数学Ⅰ】4-1-3 分散と標準偏差｜問題集

\(x^2\)の平均値は
\(\displaystyle \bar{x^2}=\frac{1}{10}(11^2+5^2+12^2+17^2+7^2\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +15^2+9^2+16^2+12^2+6^2)\)
\(\displaystyle \ \ \ =\frac{1370}{10}\)
\(\ \ \ =137\)

\(x\)の平均値は
\(\displaystyle \bar{x}=\frac{1}{10}(11+5+12+17+7\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +15+9+16+12+6)\)
\(\displaystyle \ \ =\frac{110}{10}\)
\(\ \ =11\)

分散は
\(s^2=137-11^2\)
\(\ \ \ =16\)

標準偏差は
\(s=\sqrt{16}\)
\(\ \ =4\)

\(x^2\)の平均値は
\(\displaystyle \bar{x^2}=\frac{1}{10}(5^2+3^2+6^2+8^2+5^2\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +8^2+5^2+4^2+6^2+5^2)\)
\(\displaystyle \ \ \ =\frac{325}{10}\)
\(\ \ \ =32.5\)

\(x\)の平均値は
\(\displaystyle \bar{x}=\frac{1}{10}(5+3+6+8+5\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +8+5+4+6+5)\)
\(\displaystyle \ \ =\frac{55}{10}\)
\(\ \ =5.5\)

分散は
\(s^2=32.5-5.5^2\)
\(\ \ \ =2.25\)

標準偏差は
\(s=\sqrt{2.25}\)
\(\ \ =1.5\)

\(x^2\)の平均値は
\(\displaystyle \bar{x^2}=\frac{1}{7}(10^2+8^2+4^2+6^2+9^2+5^2+7^2)\)
\(\displaystyle \ \ \ =\frac{371}{7}\)
\(\ \ \ =53\)

\(x\)の平均値は
\(\displaystyle \bar{x}=\frac{1}{7}(10+8+4+6+9+5+7)\)
\(\displaystyle \ \ =\frac{49}{7}\)
\(\ \ =7\)

分散は
\(s^2=53-7^2\)
\(\ \ \ =4\)

標準偏差は
\(s=\sqrt{4}\)
\(\ \ =2\)

\(x^2\)の平均値は
\(\displaystyle \bar{x^2}=\frac{1}{7}(4^2+2^2+1^2+1^2+4^2+2^2+0^2)\)
\(\displaystyle \ \ \ =\frac{42}{7}\)
\(\ \ \ =6\)

\(x\)の平均値は
\(\displaystyle \bar{x}=\frac{1}{7}(4+2+1+1+4+2+0)\)
\(\displaystyle \ \ =\frac{14}{7}\)
\(\ \ =2\)

分散は
\(s^2=6-2^2\)
\(\ \ \ =2\)

標準偏差は
\(s=\sqrt{2}\)
\(\ \ =1.4\)

\(\displaystyle \ \ 6=\frac{1}{5}(2+3+a+8+12)\)
\(30=a+25\)
\(\ \ a=5\)

\(x^2\)の平均値は
\(\displaystyle \bar{x^2}=\frac{1}{5}(2^2+3^2+5^2+8^2+12^2)\)
\(\displaystyle \ \ \ =\frac{246}{5}\)
\(\ \ \ =49.2\)

分散は
\(s^2=49.2-6^2\)
\(\ \ \ =13.2\)

\(\displaystyle \bar{x}=\frac{1}{20}(8×3+12×8)\)
\(\displaystyle \ \ \ =\frac{120}{20}\)
\(\ \ \ =6\)

\(8\)個のデータの2乗の和を\(a\)とすると、
\(\displaystyle 4=\frac{a}{8}-3^2\)
\(a=104\)

\(12\)個のデータの2乗の和を\(b\)とすると、
\(\displaystyle 9=\frac{b}{12}-8^2\)
\(b=876\)

よって、\(20\)個のデータの分散は
\(\displaystyle s^2=\frac{104+876}{20}-6^2\)
\(\displaystyle \ \ \ =\frac{980}{20}-36\)
\(\ \ \ =13\)