$\begin{aligned}\overline{x} & =\frac{5+7+8+9+11+11+12+13+14}{9}\\
& =\frac{90}{9}\\
& =10
\end{aligned}
$

Rata-rata $=\frac{3+4+4+5+6+7+7+7+8+9}{10}$$=\frac{60}{10}$$=6$

Median $=\frac{6+7}{2}=6,5$

Modus $=7$

Jangkauan $=9-3=6$

Jadi semua pernyataan benar.

$\begin{aligned}\overline{x} & =\frac{x_{1}+x_{2}+x_{3}+….+x_{n}}{n}\\
16 & =\frac{x+x+2+x+4+2x-3+2x}{5}\\
80 & =7x+3\\
7x & =77\\
x & =11
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{\sum_{i=1}^{n}x_{i}}{n}\\
6 & =\frac{4+2+8+5+6+a+9}{7}\\
42 & =34+a\\
a & =8
\end{aligned}
$

Jika modusnya adalah $3,$ maka nilai $x=3$

Data setelah diurutkan :

$2,2,3,3,3,3,4,4,4,5$

Median Jatuh pada :

$\begin{aligned}\frac{x_{5}+x_{6}}{2} & =\frac{3+3}{2}\\
& =3
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{2p+(2p-2)+(p+2)+(p+1)+p}{5}\\
17 & =\frac{7p+1}{5}\\
85 & =7p+1\\
7p & =84\\
p & =12
\end{aligned}
$

Jadi anak tertua $=2p=24.$

$\begin{aligned}\overline{x} & =\frac{\sum f_{i}x_{i}}{\sum f_{i}}\\
\overline{x} & =\frac{480.000}{80}\\
& =6.000
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{n_{1}\overline{x_{1}}+n_{2}\overline{x_{2}}+n_{3}\overline{x_{3}}+n_{4}\overline{x_{4}}}{n_{1}+n_{2}+n_{3}+n_{4}}\\
\overline{x} & =\frac{5×4000+5×2500+10×2000+15×1500}{5+5+10+15}\\
\overline{x} & =\frac{20.000+12500+20000+22500}{5+5+10+15}\\
& =\frac{75000}{35}\\
& =2142,85\\
& \approxeq2143.
\end{aligned}
$

Urutan data dari yang terkecil hingga terbesar sbb :

$-4,-3,0,1,4,4,4,7,9$

Jangkauan $=9-(-4)=13$

$Q_{1}=\frac{-3+0}{2}=-1,5$ dan $Q_{2}=\frac{4+7}{2}=5,5$

Median $=4$

Simpangan kuartil $=\frac{1}{2}\left(Q_{3}-Q_{1}\right)$

$=\frac{1}{2}\left(5,5-(-1,5)\right)$

$=\frac{1}{2}(7)$

$=3,5.$

Modus $=4$

Jadi semua pernyataan benar.

$\begin{aligned}\overline{x} & =\frac{5(4000)+8(2500)+10(2000)+17(1000)}{5+8+10+17}\\
& =\frac{20000+20000+20000+17000}{40}\\
& =\frac{77.000}{40}\\
& =1.925
\end{aligned}
$

Dik : $\overline{x_{1}}=7$ dan $n_{1}=25$

$\overline{x_{2}}=8$ dan $n_{2}?$

$\overline{x_{3}}=7\frac{1}{2}$ dan $n_{3}=5+n_{2}$

$\begin{aligned}n_{2}+n_{3} & =75\\
n_{2}+5+n_{2} & =75
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{n_{1}\overline{x_{1}}+n_{2}\overline{x_{2}}+n_{3}\overline{x_{3}}}{n_{1}+n_{2}+n_{3}}\\
& =\frac{25(7)+35(8)+40(7\frac{1}{2})}{100}\\
& =\frac{175+280+300}{100}\\
& =\frac{755}{100}\\
& =7,55.
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{n_{1}\overline{x_{1}}+n_{2}\overline{x_{2}}}{n_{1}+n_{2}}\\
& =\frac{16\cdot10+17\cdot20}{10+20}\\
& =\frac{500}{30}\\
& =16\frac{2}{3}
\end{aligned}
$

Karena nilai simpangan kuartil adalah $1\frac{1}{2},$

$\begin{aligned}Q_{d} & =\frac{1}{2}\left(Q_{3}-Q_{1}\right)\\
1\frac{1}{2} & =\frac{1}{2}\left(Q_{3}-Q_{1}\right)\\
\left(Q_{3}-Q_{1}\right) & =3
\end{aligned}
$

Nilai median adalah $5\frac{1}{2}$ maka susunan terurut yang mungkin :

$3,a,5,6,7,8$

$\begin{aligned}Me & =\frac{5+6}{2}\\
& =5\frac{1}{2}
\end{aligned}
$

$\begin{aligned}\left(Q_{3}-Q_{1}\right) & =3\\
7-a & =3\\
a & =4
\end{aligned}
$

Data diatas urutkan dari yang terkecil hingga terbesar adalah $3,4,5$$,6,7,8$

Jadi rata-rata data diatas adalah $\frac{3+4+5+6+7+8}{6}=\frac{33}{6}$$=5,5$$=5\frac{1}{2}.$

$x_{0}=\frac{x_{1}+x_{2}+x_{3}…+x_{10}}{10}$

$\overline{x}=\frac{\left(\frac{1}{2}x_{1}+2\right)+\left(\frac{1}{2}x_{2}+4\right)+\left(\frac{1}{2}x_{3}+6\right)+……+\left(\frac{1}{2}x_{10}+20\right)}{10}$

$=\frac{1}{2}\left(\frac{x_{1}+x_{2}+x_{3}…+x_{10}}{10}\right)+\frac{2+4++6+…+20}{10}$

$=\frac{1}{2}x_{0}+\frac{110}{10}=\frac{1}{2}x_{0}+11$

Jadi rata-rata baru adalah $\frac{1}{2}x_{0}+11.$

$\overline{x}$ minimum $\leq\overline{x}\leq\overline{x}$ maksimum

$\frac{3(20)+48}{4}\leq\overline{x}\leq\frac{20+3(48)}{4}$

$\frac{60+48}{4}\leq\overline{x}\leq\frac{20+144}{4}$

$27\leq\overline{x}\leq41$

Pernyataan diatas semua benar.