Pernyataan berikut yang benar adalah... [tex]\bold{B.~~ \bar{x} < Me < Mo}[/tex]
[tex] \small{\boxed{\begin{array}{c|c|c |c } \bold{Nilai}& \bold{Frekuensi~(fi)} & \bold{xi} & \bold{fi~.~xi}\\\hline 31–41&4& \frac{31+41}{2} = 36 & 144 \\\hline 42–52 & 5 & \frac{42+52}{2}= 47 & 235 \\\hline 53–63 & 6 & \frac{53+63}{2} = 58 &348 \\\hline 64–74 & 9 & \frac{64+74}{2} = 69 & 621 \\\hline 75–85 & 12 & \frac{75+85}{2} = 80 & 960 \\\hline 86–96 & 4 & \frac{86+96}{2}= 91&364 \\\hline \sum & 40& - & 2672 \end{array}}}[/tex]
[tex] \bar{x} = \frac{ \sum fi \: . \: xi}{ \sum fi} = \frac{2672}{40} = 66.8[/tex]
[tex]\boxed{\begin{array}{c|c|c} \bold{Nilai}& \bold{Frekuensi~(f)} & \bold{fk}\\\hline 31–41&4 & 4\\\hline 42–52 & 5 & 9 \\\hline 53–63 & 6 & 15 \\\hline \pink{64–74} & \pink9 & \pink{ 24} \\\hline 75–85 & 12 &36 \\\hline 86–96 & 4 &40\end{array}}[/tex]
Dari tabel diatas diketahui:
[tex]\begin{aligned}Me&= Tb +\left(\frac{\frac{n}{2} - fk_{me}}{f_{me}}\right) \times p \\ &=63.5 + \left( \frac{20 - 15}{9} \right) \times 11 \\&=63.5 + \frac{5}{9} \times 11 \\ &=63.5 + \frac{55}{9} \\ &=63.5 + 6.1 \\&=69.6 \end{aligned}[/tex]
[tex]\boxed{\begin{array}{c|c} \bold{Nilai}& \bold{Frekuensi~(f)} \\\hline 31–41&4\\\hline 42–52 & 5 \\\hline 53–63 & 6 \\\hline {64–74} & 9 \\\hline \pink{75–85} & \pink{12} \\\hline 86–96 & 4 \end{array}}[/tex]
Dari tabel diatas, diketahui:
[tex]\begin{aligned}Mo&= Tb +\left(\frac{d1}{d1 + d2}\right) \times p \\ &=74.5 +\left(\frac{3}{3 + 8}\right) \times 11 \\&=74.5 + \frac{3}{11} \times 11 \\ &=74.5 + 3 \\&=77.5 \end{aligned}[/tex]
Opsi A: [tex]\bar{x} = Me = Mo[/tex]
Opsi B: [tex]\bar{x} < Me < Mo[/tex]
Opsi C: [tex]Mo < Me < \bar{x} [/tex]
Opsi D: [tex]\bar{x} < Mo < Me[/tex]
Opsi E: [tex]Me < \bar{x}< Mo[/tex]
[tex] \\ [/tex]
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[tex]\colorbox{black}{\color{cyan}\#LearnWithPanda}[/tex]
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Pernyataan berikut yang benar adalah... [tex]\bold{B.~~ \bar{x} < Me < Mo}[/tex]
Pembahasan
1. Menghitung Mean
[tex] \small{\boxed{\begin{array}{c|c|c |c } \bold{Nilai}& \bold{Frekuensi~(fi)} & \bold{xi} & \bold{fi~.~xi}\\\hline 31–41&4& \frac{31+41}{2} = 36 & 144 \\\hline 42–52 & 5 & \frac{42+52}{2}= 47 & 235 \\\hline 53–63 & 6 & \frac{53+63}{2} = 58 &348 \\\hline 64–74 & 9 & \frac{64+74}{2} = 69 & 621 \\\hline 75–85 & 12 & \frac{75+85}{2} = 80 & 960 \\\hline 86–96 & 4 & \frac{86+96}{2}= 91&364 \\\hline \sum & 40& - & 2672 \end{array}}}[/tex]
[tex] \bar{x} = \frac{ \sum fi \: . \: xi}{ \sum fi} = \frac{2672}{40} = 66.8[/tex]
2. Menghitung Median
[tex]\boxed{\begin{array}{c|c|c} \bold{Nilai}& \bold{Frekuensi~(f)} & \bold{fk}\\\hline 31–41&4 & 4\\\hline 42–52 & 5 & 9 \\\hline 53–63 & 6 & 15 \\\hline \pink{64–74} & \pink9 & \pink{ 24} \\\hline 75–85 & 12 &36 \\\hline 86–96 & 4 &40\end{array}}[/tex]
Dari tabel diatas diketahui:
[tex]\begin{aligned}Me&= Tb +\left(\frac{\frac{n}{2} - fk_{me}}{f_{me}}\right) \times p \\ &=63.5 + \left( \frac{20 - 15}{9} \right) \times 11 \\&=63.5 + \frac{5}{9} \times 11 \\ &=63.5 + \frac{55}{9} \\ &=63.5 + 6.1 \\&=69.6 \end{aligned}[/tex]
3. Menghitung Modus
[tex]\boxed{\begin{array}{c|c} \bold{Nilai}& \bold{Frekuensi~(f)} \\\hline 31–41&4\\\hline 42–52 & 5 \\\hline 53–63 & 6 \\\hline {64–74} & 9 \\\hline \pink{75–85} & \pink{12} \\\hline 86–96 & 4 \end{array}}[/tex]
Dari tabel diatas, diketahui:
[tex]\begin{aligned}Mo&= Tb +\left(\frac{d1}{d1 + d2}\right) \times p \\ &=74.5 +\left(\frac{3}{3 + 8}\right) \times 11 \\&=74.5 + \frac{3}{11} \times 11 \\ &=74.5 + 3 \\&=77.5 \end{aligned}[/tex]
4. Analisis Jawaban
Opsi A: [tex]\bar{x} = Me = Mo[/tex]
Opsi B: [tex]\bar{x} < Me < Mo[/tex]
Opsi C: [tex]Mo < Me < \bar{x} [/tex]
Opsi D: [tex]\bar{x} < Mo < Me[/tex]
Opsi E: [tex]Me < \bar{x}< Mo[/tex]
[tex] \\ [/tex]
Pelajari Lebih Lanjut
Detail Jawaban
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[tex]\colorbox{black}{\color{cyan}\#LearnWithPanda}[/tex]