Cari rataanya terlebih dahulu :

$\begin{aligned}\overline{x} & =\frac{3+6+4+7+5}{5}\\
& =\frac{25}{5}\\
& =5
\end{aligned}
$

$\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}=\left(3-5\right)^{2}+\left(6-5\right)^{2}$$+(4-5)^{2}+(7-5)^{2}$$+(5-5)^{2}$

$=4+1+1+4+0$

$=10$

$\begin{aligned}S & =\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}}{n}}\\
& =\sqrt{\frac{10}{5}}\\
& =\sqrt{2}
\end{aligned}
$

Jadi simpangan bakunya adalah $\sqrt{2}.$

$\begin{aligned}\overline{x} & =\frac{10+9+7+8+6}{5}\\
& =\frac{40}{5}\\
& =8
\end{aligned}
$

$\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}=\left(10-8\right)^{2}+\left(9-8\right)^{2}$$+(7-8)^{2}+(8-8)^{2}$$+(6-8)^{2}$

$=4+1+1+0+4$

$=10$

$\begin{aligned}S & =\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}}{n}}\\
& =\sqrt{\frac{10}{5}}\\
& =\sqrt{2}.
\end{aligned}
$

$\begin{aligned}\bar{x} & =\frac{10+6+5+7+8+7+9+12}{8}\\
& =\frac{64}{8}\\
& =8
\end{aligned}
$

$\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}=\left(10-8\right)^{2}+\left(6-8\right)^{2}+(5-8)^{2}$$+(7-8)^{2}+(8-8)^{2}$$+\left(7-8\right)^{2}+\left(9-8\right)^{2}$$+\left(12-8\right)^{2}$

$=4+4+9+1+0+1+1+16$

$=36$

$\begin{aligned}S & =\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}}{n}}\\
& =\sqrt{\frac{36}{8}}\\
& =\sqrt{\frac{9}{2}}\\
& =\frac{3}{2}\sqrt{2}
\end{aligned}
$

Jadi simpangan bakunya adalah $\frac{3}{2}\sqrt{2}.$

Cari dulu rata-rata nya :

$\begin{aligned}\overline{x} & =\frac{2+3+4+5+6+6+7+8+9+10}{10}\\
& =\frac{60}{10}\\
& =6
\end{aligned}
$

$\sum\left(x_{i}-\overline{x}\right)^{2}=\left(2-6\right)^{2}+(3-6)^{2}$$+\left(4-6\right)^{2}+\left(5-6\right)^{2}$$+(6-6)^{2}+(6-6)^{2}$$+(7-6)^{2}+(8-6)^{2}$$+(9-6)^{2}+(10-6)^{2}$

$\sum\left(x_{i}-\overline{x}\right)^{2}=16+9+4+1$$+0+0+1+4+9+16$

$\sum\left(x_{i}-\overline{x}\right)^{2}=60$

Simpangan baku

$\begin{aligned}(S) & =\sqrt{\frac{\sum\left(x_{i}-\overline{x}\right)^{2}}{n}}\\
& =\sqrt{\frac{60}{10}}\\
& =\sqrt{6}.
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{4+6+7+5+4+8+6+8}{8}\\
& =\frac{48}{8}\\
& =6
\end{aligned}
$

$\sum\left(x_{i}-\overline{x}\right)^{2}=\left(4-6\right)^{2}+(6-6)^{2}$$+(7-6)^{2}+\left(5-6\right)^{2}$$+(4-6)^{2}+(8-6)^{2}$$+(6-6)^{2}+\left(8-6\right)^{2}$

$=4+0+1+1+4+4+0+4$

$=18$

Simpangan baku

$\begin{aligned}(S) & =\sqrt{\frac{\sum\left(x_{i}-\overline{x}\right)^{2}}{n}}\\
& =\sqrt{\frac{18}{8}}\\
& =\sqrt{2,25}\\
& =1,5.
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{4+5+8+8+4+6+9+4+7+5}{10}\\
& =\frac{60}{6}\\
& =6
\end{aligned}
$

$\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}=\left(4-6\right)^{2}+\left(5-6\right)^{2}$$+(8-6)^{2}+(8-6)^{2}$$+(4-6)^{2}+\left(6-6\right)^{2}$$+\left(9-6\right)^{2}+\left(4-6\right)^{2}+\left(7-6\right)^{2}$$+\left(5-6\right)^{2}$

$=4+1+4+4$$+4+0+9$$+4+1+1$

$=37$

$\begin{aligned}S & =\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}}{n}}\\
& =\sqrt{\frac{32}{10}}\\
& =\sqrt{3,2}\\
& =1,79.
\end{aligned}
$

Rata- rata dari dari data $a,\, a+2,\, a$$+4,\, a+6,\, a+8:$

$\begin{aligned}\overline{x} & =\frac{a+a+2+a+4+a+6+a+8}{5}\\
& =\frac{5a+20}{5}\\
& =a+4
\end{aligned}
$

$S^{2}=\frac{1}{n}\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}$

$S^{2}=\frac{1}{5}\{\left(a-(a+4)\right)^{2}$$+\left(a+2-(a+4)\right)^{2}$$+(a+4-(a+4))^{2}$$+\left(a+6-(a+4)\right)^{2}$$+\left(a+8-(a+4)\right)^{2}\}$

$\begin{aligned}S^{2} & =\frac{1}{5}\left(16+4+0+4+16\right)\\
& =\frac{40}{5}\\
& =8
\end{aligned}
$

$\begin{aligned}S & =\sqrt{8}\\
& =2\sqrt{2}
\end{aligned}
$

Jadi simpangan baku dari data diatas adalah $2\sqrt{2}.$

$\begin{aligned}\overline{x} & =\frac{2+4+4+5+6+6+7+8+9+9}{10}\\
& =\frac{60}{10}\\
& =6
\end{aligned}
$

$s^{2}=\frac{1}{10}\{(2-6)^{2}+(4-6)^{2}$$+(4-6)^{2}+(5-6)^{2}$$+(6-6)^{2}+(6-6)^{2}$$+(7-6)^{2}+(8-6)^{2}$$+(9-6)^{2}+(9-6)^{2}\}$

$=\frac{48}{10}$

$=4,48$

$s=2,19.$

$\begin{aligned}\overline{x} & =\frac{3(2)+4(4)+5(3)+6(1)}{2+6+4+2}\\
& =\frac{43}{10}\\
& =4,3
\end{aligned}
$

$S^{2}=\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}$

$=\frac{2\left(3-4,3\right)^{2}+4\left(4-4,3\right)^{2}+3\left(5-4,3\right)^{2}+1\left(6-4,3\right)^{2}}{10}$

$=\frac{3,38+0,36+1,47+2,89}{10}$

$=0,81$

$S=0,9.$

$\begin{aligned}\overline{x} & =\frac{6(3)+7(10)+8(15)+9(8)+10(4)}{3+10+15+8+4}\\
& =\frac{320}{40}\\
& =8
\end{aligned}
$

$S^{2}=\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}$

$=\frac{3(6-8)^{2}+10(7-8)^{2}+15(8-8)^{2}+8(9-8)^{2}+4(10-8)^{2}}{40}$

$=\frac{12+10+0+8+16}{40}$

$=\frac{46}{40}$

$=1,15$

$\begin{aligned}S & =\sqrt{1,15}\\
& =1,07
\end{aligned}
$

Jadi simpangan baku dari data diatas adalah $1,07.$

Cari rata-rata terlebih dahulu :

$\begin{aligned}\overline{x} & =\frac{5(2)+6(5)+7(12)+8(7)+9(4)}{2+5+12+7+4}\\
& =\frac{10+30+84+56+36}{30}\\
& =\frac{216}{30}\\
& =7,2
\end{aligned}
$

$S^{2}=\frac{2\left(5-7,2\right)^{2}+5(6-7,2)^{2}+12(7-7,2)^{2}+7(8-7,2)^{2}+4(9-7,2)^{2}}{2+5+12+7+4}$

$\begin{aligned}S^{2} & =\frac{9,68+7,2+0,48+4,48+12,96}{30}\\
& =\frac{34,8}{30}\\
& =1,16
\end{aligned}
$

$\begin{aligned}S & =\sqrt{1,16}\\
& =1,08.
\end{aligned}
$

Jadi simpangan bakunya adalah $1,08.$

$\begin{aligned}S^{2} & =\frac{\sum_{i=1}^{n}\left(x_{i}\right)^{2}}{n}-\left\{ \frac{\sum_{i=1}^{n}x_{i}}{n}\right\} ^{2}\\
& =\frac{131}{5}-\left\{ \frac{25}{5}\right\} ^{2}\\
& =26,2-25\\
& =1,2.
\end{aligned}
$

Jika terjadi perubahan pada setiap data, perubahannya dapat terlihat pada tabel berikut :

Dari data diatas bisa diperoleh :

Rata-rata baru diperoleh $=13,9$

Simpangan baku baru diperoleh $=5,7$

Quartil bawah diperoleh $=2,8.$

Misalkan data awalnya adalah $x_{1},x_{2},x_{3},…,x_{40}$

$\bar{x}=\frac{x_{1}+x_{2}+x_{3}….+x_{40}}{40}$

Setelah setiap data dikalikan $2$ dan dikurangin $10,$ maka menjadi :

$\frac{2x_{1}-10+2x_{2}-10+2x_{3}-10….+2x_{40}-10}{40}$$=\frac{2(x_{1}+x_{2}+x_{3}….+x_{40})-10.40}{40}$

$=2\overline{x}-10$

$=64-10$

$=54$

Dengan menggunakan alternatif diperoleh :

Jika terjadi perubahan pada setiap data, perubahannya dapat terlihat pada tabel berikut :

Jika terjadi perubahan pada setiap data, perubahannya dapat terlihat pada tabel berikut :

Jadi yang memenuhi adalah $C.$