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Statistika (3) Ukuran Penyebaran Data

Statistika (3) Ukuran Penyebaran Data

Assessment

Presentation

•

Mathematics

•

12th Grade

•

Medium

Created by

Sepka Sugita

Used 6+ times

FREE Resource

14 Slides • 8 Questions

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Multiple Choice

Rumus menentukan rentang/jangkauan data adalah  R=R= ....

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Xmax⁡−Xmin⁡X_{\max}-X_{\min}  

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Q3−Q1Q_3-Q_1  

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12(Q3−Q1)\frac{1}{2}\left(Q_3-Q_1\right)  

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1n∑n=1n∣xi−x‾∣\frac{1}{n}\sum_{n=1}^n\left|x_i-\overline{x}\right|  

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Multiple Choice

Rumus menentukan rentang antar kuartil/hamparan data adalah  H=H= ....

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Xmax⁡−Xmin⁡X_{\max}-X_{\min}  

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Q3−Q1Q_3-Q_1  

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12(Q3−Q1)\frac{1}{2}\left(Q_3-Q_1\right)  

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1n∑n=1n∣xi−x‾∣\frac{1}{n}\sum_{n=1}^n\left|x_i-\overline{x}\right|  

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Multiple Choice

Rumus menentukan simpangan kuartil data adalah  Qd=Q_d= ....

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Xmax⁡−Xmin⁡X_{\max}-X_{\min}  

2

Q3−Q1Q_3-Q_1  

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12(Q3−Q1)\frac{1}{2}\left(Q_3-Q_1\right)  

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1n∑n=1n∣xi−x‾∣\frac{1}{n}\sum_{n=1}^n\left|x_i-\overline{x}\right|  

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Multiple Choice

Rumus menentukan simpangan rata-rata data adalah  SR=SR= ....

1

1n∑n=1n(xi−x‾)2\sqrt{\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)^2}  

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1n∑n=1n(xi−x‾)\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)  

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1n∑n=1n(xi−x‾)2\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)^2  

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1n∑n=1n∣xi−x‾∣\frac{1}{n}\sum_{n=1}^n\left|x_i-\overline{x}\right|  

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Multiple Choice

Rumus menentukan ragam (varians) data adalah  s2=s^2= ....

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1n∑n=1n(xi−x‾)2\sqrt{\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)^2}  

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1n∑n=1n(xi−x‾)\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)  

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1n∑n=1n(xi−x‾)2\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)^2  

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1n∑n=1n∣xi−x‾∣\frac{1}{n}\sum_{n=1}^n\left|x_i-\overline{x}\right|  

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Multiple Choice

Rumus menentukan simpangan baku (standar deviasi) data adalah  s=s= ....

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1n∑n=1n(xi−x‾)2\sqrt{\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)^2}  

2

1n∑n=1n(xi−x‾)\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)  

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1n∑n=1n(xi−x‾)2\frac{1}{n}\sum_{n=1}^n\left(x_i-\overline{x}\right)^2  

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1n∑n=1n∣xi−x‾∣\frac{1}{n}\sum_{n=1}^n\left|x_i-\overline{x}\right|  

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Multiple Choice

Metode menentukan simpangan baku (standar deviasi) dari data; 2,3,42,3,4 adalah ....

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s2=(2−3)2+(3−3)2+(4−3)23s^2=\frac{\left(2-3\right)^2+\left(3-3\right)^2+\left(4-3\right)^2}{3}  

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s2=(2+3)2−(3+3)2−(4+3)23s^2=\frac{\left(2+3\right)^2-\left(3+3\right)^2-\left(4+3\right)^2}{3}  

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s=(2−3)2+(3−3)2+(4−3)23s=\sqrt{\frac{\left(2-3\right)^2+\left(3-3\right)^2+\left(4-3\right)^2}{3}}  

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s=(2+3)2−(3+3)2−(4+3)23s=\sqrt{\frac{\left(2+3\right)^2-\left(3+3\right)^2-\left(4+3\right)^2}{3}}  

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Fill in the Blanks

Diketahui Varians suatu data adalah 25.

maka Simpangan Baku data tersebut adalah =

Type answer...

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