$\begin{aligned}\overline{x} & =\frac{\sum F_{i}x_{i}}{\sum F}\\
& =\frac{54\cdot6+55\cdot10+56\cdot14+57\cdot20+58\cdot10}{6+10+14+20+10}\\
& =\frac{324+550+784+1140+580}{60}\\
& =\frac{3.378}{60}\\
& =56,3.
\end{aligned}
$

$\begin{aligned}\overline{x} & =\frac{\sum f_{i}.x_{i}}{x_{i}}\\
& =\frac{330}{20}\\
& =16,5.
\end{aligned}
$

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

$\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}
$

$\begin{aligned}\overline{x} & =\frac{\sum f_{i}x_{i}}{\sum f_{i}}\\
6 & =\frac{4\cdot20+5\cdot40+6\cdot70+8a+10\cdot10}{20+40+70+a+10}\\
& =\frac{80+200+420+8a+100}{140+a}\\
& =\frac{800+8a}{140+a}\\
6(140+a) & =800+8a\\
840+6a & =800+8a\\
2a & =40\\
a & =20
\end{aligned}
$

Misalkan $n_{1}=$ jumlah pegawai pria dan $\overline{x_{1}}=$ rata-rata gaji pegawai pria

$n_{2}=$ jumlah pegawai wanita dan $\overline{x_{2}}=$ rata-rata gaji pegawai wanita

$\overline{x}=\frac{n_{1}\overline{x_{1}}+n_{2}\overline{x_{2}}}{n_{1}+n_{2}}$

$250.000=\frac{260.000n_{1}+210.000n_{2}}{n_{1}+n_{2}}$

$250.000n_{1}+250.000n_{2}$$=260.000n_{1}+210.000n_{2}$

$\begin{aligned}10.000n_{1} & =40.000n_{2}\\
\frac{n_{1}}{n_{2}} & =\frac{40.000}{10.000}\\
& =\frac{4}{1}\\
n_{1}:n_{2} & =4:1
\end{aligned}
$

Misalkan rata-rata ulangan Tono adalah $a$.

Jumlah nilai dari $43$ siswa adalah $43\times56=2408$

Gunakan rumus rata-rata gabungan :

$\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}}\\
55 & =\frac{43×56+25+a}{43+1+1}\\
55\times45 & =2408+25+a\\
2475 & =2433+a\\
a & =42.
\end{aligned}
$

$\overline{x_{l}}=165$ dan banyaknya laki-laki $=n_{l}$

$\overline{x_{p}}=155$ dan bayaknya perempuan = $n_{p}$

$\overline{x}=162$

$\overline{x}=\frac{n_{l}.\overline{x_{l}}+n_{p}.\overline{x_{p}}}{n_{l}+n_{p}}$

$162=\frac{165.n_{l}+155n_{p}}{n_{l}+n_{p}}$

$162n_{l}+162n_{p}=165\cdot n_{l}+155n_{p}$

$3n_{l}=7n_{p}$

$n_{l}:n_{p}=7:3$

$\begin{aligned}\overline{x} & =\frac{\Sigma f\cdot x}{\Sigma f}\\
& =\frac{2(3)+4(4)+8(5)+12(6)+16(7)+4(8)}{46}\\
& =\frac{6+16+40+72+112+40}{46}\\
& =\frac{278}{46}\\
& =6\frac{1}{23}
\end{aligned}
$

Syarat lulus $=\overline{x}-1=5\frac{1}{23}$

Jadi banyaknya siswa yang lulus adalah $12+16+4=32.$

Berikut ini tabel anak usia 16 anak sekarang.

$\begin{aligned}\overline{x} & =\frac{\Sigma f\cdot x}{\Sigma f}\\
& =\frac{144}{16}\\
& =9.
\end{aligned}
$

Gunakan rumus rata-rata gabungan :

Misalkan banyaknya $A=2x,$ $B=3x$ dan $C=3x$

$\begin{aligned}\overline{x} & =\frac{n_{A}.\overline{x_{A}}+n_{B}.\overline{x_{B}}+n_{c}.\overline{x_{B}}}{n_{A}+n_{B}+n_{C}}\\
& =\frac{2x\cdot(8)+3x\cdot(7,8)+3x\cdot(8,2)}{2x+3x+3x}\\
& =\frac{16x+23,4x+24,6x}{8x}\\
& =\frac{64x}{8x}\\
& =8
\end{aligned}
$

Jadi rata-rata keseluruhannya adalah $8.$

$n_{A}=35$ dan $n_{B}=40$

$\overline{x_{B}}=\overline{x_{A}}+5$

$\begin{aligned}\overline{x} & =\frac{n_{A}\cdot\overline{x_{A}}+n_{B}\cdot\overline{x_{B}}}{n_{A}+n_{B}}\\
57\frac{2}{3} & =\frac{35\cdot\overline{x_{A}}+40(\overline{x_{A}}+5)}{35+40}\\
75\cdot\left(\frac{173}{3}\right) & =75\overline{x_{A}}+200
\end{aligned}
$

$\begin{aligned}\overline{x_{A}} & =\frac{4325-200}{75}\\
& =\frac{4125}{75}\\
& =55.
\end{aligned}
$

Misalkan jumlah data kelompok I adalah F1 dan jumlah data kelompok II adalah F2.

$a=$ nilai seorang siswa yang berpindah

$\overline{x}_{1}=\frac{\sum F_{1}}{5}$

$\sum F_{1}=5\cdot63=315$

$\overline{x}_{II}=\frac{\sum F_{II}}{5}$

$\sum F_{II}=58\cdot5=290$

Seorang siswa kelompok I pindah ke kelompok II

$\begin{aligned}\overline{x}_{1}\mbox{ baru} & =\frac{\sum F_{1}-a}{4}\\
65 & =\frac{315-a}{4}\\
315-a & =65\cdot4\\
& =260\\
a & =315-260\\
& =55
\end{aligned}
$

Jadi nilai siswa yang berpindah posisi adalah $55.$

Rata-rata baru kelompok II setelah seorang siswa berpindah dari kelompok I adalah :

$\begin{aligned}\overline{x}_{II}\mbox{ baru} & =\frac{\sum F_{II}+a}{6}\\
& =\frac{290+55}{6}\\
& =\frac{345}{6}\\
& =57,5.
\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\\
2n_{2} & =70\\
n_{2} & =35
\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}
$