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

$\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}$$=2\left(3-4,3\right)^{2}+4\left(4-4,3\right)^{2}$$+3\left(5-4,3\right)^{2}+1\left(6-4,3\right)^{2}$

$=3,38+0,36+1,47+2,89$

$\sum f_{i}=10$

$\begin{aligned}S^{2} & =\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}\\
& =\frac{3,38+0,36+1,47+2,89}{10}\\
& =0,81\\
S & =0,9
\end{aligned}
$

$D_{i}=x_{\frac{i(n+1)}{10}}$

$\begin{aligned}D_{7} & =x_{\frac{7(40+1)}{10}}\\
& =x_{\frac{287}{10}}\\
& =x_{28\frac{7}{10}}
\end{aligned}
$

$D_{7}=x_{28}+\frac{7}{20}(x_{29}-x_{28})$

$\begin{aligned}D_{7} & =8+\frac{7}{20}\left(8-8\right)\\
& =8.
\end{aligned}
$

$\begin{aligned}\frac{i\cdot n}{10} & =\frac{6.40}{10}\\
& =\frac{240}{10}\\
& =24
\end{aligned}
$

kelas desil ke enam berada pada kelas 53 – 57

$\begin{aligned}D_{6} & =tb+p\left(\frac{\frac{6n}{10}-fk_{6}}{f_{6}}\right)\\
& =52,5+5\left(\frac{\frac{6×40}{10}-17}{12}\right)\\
& =52,5+5\left(\frac{24-17}{12}\right)\\
& =52,5+2,9\\
& =55,4.
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{\sum f_{i}\cdot x_{i}}{\sum f_{i}}\\
& =\frac{4784}{32}\\
& =149,5
\end{aligned}
$

$\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}=1\left(132-149,5\right)^{2}+2(137-149,5)^{2}$$+5(142-149,5)^{2}+6\left(147-149,5\right)^{2}$$+10(152-149,5)^{2}$$+6\left(157-149,5\right)^{2}$$+2\left(162-149,5\right)^{2}$

$=306,25+312,5+281,25$$+37,5+62,5+337,5+312,5$

$=1650$

$\sum f_{i}=32$

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

$S^{2}=\frac{1650}{32}$

$=51,56$

$\begin{aligned}S & =\sqrt{51,56}\\
& =7,18
\end{aligned}
$

Jadi simpangan bakunya adalah $7,18.$

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}
$

$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}$$=9,68+7,2+0,48$$+4,48+12,96$

$=34,8$

$2+5+12+7+4=30$

$\begin{aligned}S^{2} & =\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}\overline{x} & =\frac{\sum f_{i}.x_{i}}{\sum f}\\
& =\frac{614}{18}\\
& =34,11
\end{aligned}
$

$\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}=4\left(28-34,11\right)^{2}$$+8\left(33-34,11\right)^{2}$$+4\left(38-34,11\right)^{2}$$+2\left(43-34,11\right)^{2}$

$=29,16+42,32+319,74+302,58$

$=693,8$

$\sum f_{i}=18$

$\begin{aligned}S^{2} & =\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}\\
& =\frac{693,8}{18}\\
& =\left(4,581\right)^{2}
\end{aligned}
$

$\begin{aligned}S & =\sqrt{20,99}\\
& =4,581
\end{aligned}
$

Jadi simpangan baku dari data diatas adalah $4,581.$

$\overline{x}=\frac{\sum f_{i}.x_{i}}{\sum f_{i}}=\frac{1641}{30}=54,7$

$\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}=3\left(48-54,7\right)^{2}$$+6\left(51-54,7\right)^{2}$$+8\left(54-54,7\right)^{2}$$+7\left(57-54,7\right)^{2}$$+6\left(60-54,7\right)^{2}$

$=134,67+82,14$$+3,92+37,03$$+168,54$

$=426,3$

$\sum f_{i}=30$

$\begin{aligned}S^{2} & =\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}\\
& =\frac{426,3}{30}\\
& =14,21
\end{aligned}
$

$\begin{aligned}S & =\sqrt{14,21}\\
& =3,77
\end{aligned}
$

Jadi simpangan bakunya adalah $3,77.$

$\begin{aligned}\overline{x} & =\frac{33(4)+38(7)+43(9)+48(10)}{4+7+9+10}\\
& =\frac{1265}{30}\\
& =42,17
\end{aligned}
$

$\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}=4\left(33-42,17\right)^{2}$$+7\left(38-42,17\right)^{2}$$+9\left(43-42,17\right)^{2}$$+10\left(48-42,17\right)^{2}$

$=336,36+121,72$$+6,20+339,89$

$\sum f_{i}=30$

$\begin{aligned}S^{2} & =\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}\\
& =\frac{804,17}{30}\\
& =26,81
\end{aligned}
$

$\begin{aligned}S & =\sqrt{26,81}\\
& =5,18.
\end{aligned}
$

$2+4+12+a+11+8+1=50$

$a+38=50$

$a=12$

$\overline{x}=\frac{4(2)+5(4)+6(12)+7(12)+8(11)+9(8)+10(1)}{2+4+12+12+11+8+1}$

$=\frac{354}{50}$

$=7,08$

$\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}=2(4-7,08)^{2}$$+4\left(5-7,08\right)^{2}$$+12\left(6-7,08\right)^{2}$$+12\left(7-7,08\right)^{2}$$+11\left(8-7,08\right)^{2}$$+8(9-7,08)^{2}$$+1\left(10-7,08\right)^{2}$

$\sum f_{i}=50$

$\begin{aligned}S^{2} & =\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}\\
& =2,8798
\end{aligned}
$

$\begin{aligned}S & =\sqrt{2,8798}\\
& =1,697.
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{\sum f_{i}x_{i}}{\sum f_{i}}\\
& =\frac{7905}{50}\\
& =158,1
\end{aligned}
$

$\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}=7\left(147-158,1\right)^{2}$$+10\left(152-158,1\right)^{2}$$+13\left(157-158,1\right)^{2}$$+9\left(162-158,1\right)^{2}$$+7\left(167-158,1\right)^{2}$$+4\left(172-158,1\right)^{2}$

$=862,47+372,1$$+15,73+136,89$$+554,47+772,84$

$=2714,5$

$\sum f_{i}=40$

$\begin{aligned}S^{2} & =\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}\\
& =\frac{2714,5}{50}\\
& =54,29
\end{aligned}
$

$\begin{aligned}S & =\sqrt{54,29}\\
& =7,37
\end{aligned}
$

Jadi simpangan baku dari data diatas adalah $7,37.$

$\begin{aligned}S^{2} & =\frac{\sum fx_{i}^{2}}{n}-\left(\frac{\sum fx_{i}}{n}\right)^{2}\\
& =\frac{1124}{20}-\left(\frac{140}{20}\right)^{2}\\
& =56,2-49\\
& =7,2
\end{aligned}
$

Jadi ragamnya adalah $7,2.$

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

berdasarkan soal diatas kita dapat peroleh :

Jadi yang memenuhi adalah C.

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

Berdasarkan data diatas dapat diperoleh :

$\begin{aligned}3+4+x+7+y+5+4 & =36\\
x+y+23 & =36\\
x+y & =13
\end{aligned}
$

$x=13-y$….(1)

Rata-rata dari $36$ siswa adalah $6$ sehingga :

$\begin{aligned}\overline{x} & =\frac{3.3+4.4+5x+6.7+7y+8.5+9.4}{36}\\
& =\frac{9+16+5x+42+7y+40+36}{36}\\
6 & =\frac{143+5x+7y}{36}\\
216 & =143+5x+7y
\end{aligned}
$

$5x+7y=73…..(2)$

Substitusikan pers (1) ke pers (2) sehingga diperoleh :

$\begin{aligned}5\left(13-y\right)+7y & =73\\
65-5y+7y & =73\\
2y & =73-65\\
& =8\\
y & =4
\end{aligned}
$

Substitusikan nilai y = 4 ke pers (1) sehingga diperoleh nilai $x=9$

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

$=27+16+9+0+4+20+36$

$=112$

$\sum f_{i}=36$

$\begin{aligned}S^{2} & =\frac{\sum f_{i}\left(x_{i}-\overline{x}\right)^{2}}{\sum f_{i}}\\
& =\frac{112}{36}\\
& =3,11
\end{aligned}
$

$\begin{aligned}S & =\sqrt{3,11}\\
& =1,76
\end{aligned}
$

Jadi simpangan bakunya adalah $1,76.$