Nilai Mean dan Proporsi

• Pencipta
  Topik Pertanyaan
• Tanggal 26-10-2020 jam 07:39 #7469
  Haerunisa Ashari

  Peserta

  Suatu sampel random sebanyak 40 ruang kelas diambil dari Kampus A STEI yang mempunyai luas rata-rata populasi atau mean (μ) 36,4 meter persegi (m2) dan simpangan baku atau standard deviation (σ) 5,2 m2. Hitunglah probabilitas luas rata-rata ruang kelas sampel tersebut :

  (1) Antara 36 m2 dan 38 m2.

  (2) 36,4 m2 sampai 40 m2.

  (3) Paling luas 37,5 m2.

  (4) Dari 37 m2 hingga 40 m2.

  (5) Paling-paling 36 m2.

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  Topik Pertanyaan

Jawaban :

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  Jawaban
• Tanggal 26-10-2020 jam 19:09 #7478
  StatistikaA

  Peserta

  Diketahui \(\mu = 36\text{,}4\) , dan \(\sigma = 5\text{,}2\)
  Pertanyaan diatas dapat diselesaikan menggunakan Tabel Z
  \[\begin{aligned}Z&=\frac{X-\mu}{\sigma }\end{aligned}\]
  1. \[\begin{aligned}P(36<X<38)&=P(X<38)-P(X<36)\\&=P\left(Z<\frac{38-36\text{,}4}{5\text{,}2}\right)-P\left(Z<\frac{36-36\text{,}4}{5\text{,}2}\right)\\&=P(Z<0\text{,}31)-P(Z<-0\text{,}07)\\&=0\text{,}6217-0\text{,}4721\\&=0\text{,}1496\end{aligned}\]
  2.\[\begin{aligned}P(36\text{,}4<X<40)&=P(X<40)-P(X<36\text{,}4)\\&=P\left(Z<\frac{40-36\text{,}4}{5\text{,}2}\right)-P\left(Z<\frac{36\text{,}4-36\text{,}4}{5\text{,}2}\right)\\&=P(Z<0\text{,}69)-P(Z<0)\\&=0\text{,}7549-0\text{,}5\\&=0\text{,}2549\end{aligned}\]
  3.\[\begin{aligned}P(X<37\text{,}5)&=P\left(Z<\frac{37\text{,}5-36\text{,}4}{5\text{,}2}\right)\\&=P(Z<0\text{,}21)\\&=0\text{,}5832\end{aligned}\]
  4. \[\begin{aligned}P(37<X<40)&=P(X<40)-P(X<37)\\&=P\left(Z<\frac{40-36\text{,}4}{5\text{,}2}\right)-P\left(Z<\frac{37-36\text{,}4}{5\text{,}2}\right)\\&=P(Z<0\text{,}69)-P(Z<0\text{,}12)\\&=0\text{,}7549-0\text{,}5478\\&=0\text{,}2071\end{aligned}\]
  5.\[\begin{aligned}P(X<36)&=P\left(Z<\frac{36-36\text{,}4}{5\text{,}2}\right)\\&=P(Z<-0\text{,}07)\\&=0\text{,}4721\end{aligned}\]

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