Rem.3 - Statistika Dan Peluang - Xi IPA - Rabu, 18 Maret 2015

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Rem.3 - Statistika Dan Peluang - Xi IPA - Rabu, 18 Maret 2015 - Quiz

Rabu, 18 Maret 2015 - Pukul 19.00 s.d. 19.30 wib.1. Berdo'alah sebelum mulai mengerjakan.2. Tuliskan NAMA, KELAS, dan NOMOR ABSEN sebelum mulai mengerjakan.3. Kerjakan seluruh soal secara berurutan.4. Kerjakan sendiri soal ini tanpa bantuan orang lain.5. Kerjakan Soal-Soal Berikut Dengan Jujur dan Tanpa Bantuan Orang Lain6. Pilihlah salah satu jawaban yang paling benar.7. Dilarang mengerjakan soal-soal ini lebih dari satu kali dan apabila mengerjakan soal-soal ini lebih dari satu kali, maka nilai yang diambil adalah nilai yang terkecil.8. Apabila telah menyelesaikan semua soal harap mengunduh SERTIFIKAT hasil Ulangan.I n g a t ! .... Tuhan Maha Mengetahui             ⊙ Kerjakan soal-soal ini dengan jujur.⊙ Kejujuran adalah satu keharusan.⊙ Jujur itu ketulusan.⊙ Jujur itu gambaran hati.⊙ Jujur itu berkata dan berbuat apa adanya.⊙ Jujur itu kebenaran.⊙ Jujur itu sifat mulia.⊙ Jujur itu lawan dari dusta.⊙ Jujur itu lawan dari kebohongan.⊙ Dengan kejujuran kau kan di percaya.⊙ Dengan kejujuran keberkahan berlimpah.⊙ Dengan kejujuran kemunafikan lari darimu.⊙ Dengan kejujuran manusia mencintaimu.⊙ Dengan kejujuran Yang Maha Kuasa menyertaimu


Questions and Answers
  • 1. 

    Dalam suatu tes peserta diminta mengerjakan 15 soal pilihan benar-salah. Peluang seorang peserta tes menjawab dengan benar 8 soal adalah . . . . 

    • A.

      0,1964

    • B.

      0,3036

    • C.

      0,6964

    • D.

      0,8036

    • E.

      0,9926

    Correct Answer
    A. 0,1964
    Explanation
    The probability of a participant answering 8 questions correctly out of 15 can be calculated using the binomial probability formula. The formula is P(x) = (nCx) * p^x * q^(n-x), where n is the total number of trials (15), x is the number of successful trials (8), p is the probability of success in a single trial (0.5 for a true/false question), and q is the probability of failure in a single trial (1 - p). Plugging in the values, we get P(8) = (15C8) * (0.5^8) * (0.5^7) = 3003 * 0.00390625 * 0.0078125 = 0.1964. Therefore, the correct answer is 0.1964.

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  • 2. 

    Dalam suatu tes peserta diminta mengerjakan 10 soal pilihan benar-salah. Peluang seorang peserta tes menjawab dengan benar 8 soal adalah . . . . 

    • A.

      0,1094

    • B.

      0,1964

    • C.

      0,2189

    • D.

      0,2734

    • E.

      0,0439

    Correct Answer
    E. 0,0439
    Explanation
    The correct answer is 0,0439. This can be determined by using the binomial probability formula. The formula is P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful outcomes, p is the probability of success, and C(n,k) is the combination formula. In this case, n = 10 (number of questions), k = 8 (number of correct answers), and p = 0.5 (since it is a true/false question). Plugging in these values, we get P(X=8) = C(10,8) * 0.5^8 * 0.5^2 = 45 * 0.00390625 * 0.25 = 0.0439.

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  • 3. 

    Sebuah perusahaan membutuhkan beberapa karyawan baru. Sebuah tes seleksi karyawan diadakan dan dari seluruh peserta tes hanya 20% yang lulus. Jika dari para peserta tes tersebut diambil sampel secara acak sebanyak 10 peserta, peluang paling banyak terdapat 3 peserta lulus tes adalah . . . .

    • A.

      0,9991

    • B.

      0,8791

    • C.

      0,2013

    • D.

      0,0009

    • E.

      0,0008

    Correct Answer
    B. 0,8791
    Explanation
    The probability of at most 3 participants passing the test can be calculated using the binomial distribution formula. In this case, the probability of passing the test is 20% or 0.2, and the sample size is 10. The formula is P(X ≤ 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3). By plugging in the values, we can calculate the probability to be 0.8791.

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  • 4. 

    Sebuah perusahaan X membutuhkan beberapa pegawai baru. Sebuah tes seleksi pegawai diadakan dan dari seluruh peserta tes hanya 1/3 yang lulus. Jika dari para peserta tes tersebut diambil sampel secara acak sebanyak 10 peserta, peluang paling banyak terdapat 6 peserta lulus tes adalah . . . . 

    • A.

      0,9803

    • B.

      0,9500

    • C.

      0,8791

    • D.

      0,2013

    • E.

      0,0569

    Correct Answer
    A. 0,9803
    Explanation
    The probability of at most 6 participants passing the test can be calculated using the binomial distribution formula. In this case, the probability of a participant passing the test is 1/3, and the sample size is 10. To find the probability of at most 6 participants passing the test, we need to calculate the probability of 0, 1, 2, 3, 4, 5, and 6 participants passing the test and sum them up. By using this formula, the probability is found to be 0.9803.

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  • 5. 

    Dua dadu dilambungkan bersama – sama sebanyak 8 kali. Peluang terlihat pasangan mata dadu berjumlah 5 paling sedikit 3 kali adalah . . . .

    • A.

      0,0074

    • B.

      0,0246

    • C.

      0,0500

    • D.

      0,9926

    • E.

      0,9574

    Correct Answer
    C. 0,0500
    Explanation
    The probability of getting a pair of dice with a total of 5 or more at least 3 times out of 8 throws can be calculated using the binomial probability formula. The formula is P(X = k) = (nCk) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful outcomes, p is the probability of success, and (nCk) is the combination formula. In this case, n = 8, k = 3, p = 1/6 (since there are 6 possible outcomes for each dice throw), and (nCk) = 8C3. Plugging in these values into the formula, we can calculate the probability to be approximately 0.0500.

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  • 6. 

    Dua dadu dilambungkan bersama – sama sebanyak 8 kali. Peluang terlihat pasangan mata dadu dari dadu kedua bilangan genap paling sedikit 4 kali adalah . . . .

    • A.

      0,3633

    • B.

      0,5244

    • C.

      0,6367

    • D.

      0,6525

    • E.

      0,8235

    Correct Answer
    C. 0,6367
    Explanation
    The probability of seeing a pair of even numbers on the second dice at least 4 times out of 8 throws can be calculated using the binomial distribution. The formula for the binomial distribution is P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful outcomes, p is the probability of success, and C(n,k) is the number of combinations. In this case, n=8, k=4, p=1/6 (since there are 6 possible outcomes for each dice throw), and C(8,4) = 70. Plugging in these values, we get P(X=4) = 70 * (1/6)^4 * (5/6)^4 = 0.6367. Therefore, the correct answer is 0.6367.

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  • 7. 

    Sebuah kotak berisi bola merah, kuning dan biru dengan perbandingan 3 : 2 : 1. Dari kotak diambil 1 bola secara acak kemudian dikembalikan. Jika percobaan dilakukan 10 kali, peluang terambil bola 5 bola merah adalah . . . .

    • A.

      0,0569

    • B.

      0,2461

    • C.

      0,3633

    • D.

      0,6230

    • E.

      0,7539

    Correct Answer
    B. 0,2461
    Explanation
    The probability of drawing a red ball from the box is 3/6 or 1/2. Since each draw is independent, the probability of drawing a red ball 5 times in 10 draws can be calculated using the binomial probability formula. Plugging in the values, the probability is calculated as (10 choose 5) * (1/2)^5 * (1/2)^5 = 252 * 1/32 * 1/32 = 0.2461.

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  • 8. 

    Sembilan puluh persen dari sejumlah barang yang dihasilkan sebuah mesin produksi berkualitas baik. Dari 15 barang yang dihasilkan mesin produksi, peluang barang yang dihasilkan 60% barang berkualitas baik adalah . . . .

    • A.

      0,9981

    • B.

      0,9978

    • C.

      0,2461

    • D.

      0,0022

    • E.

      0,0019

    Correct Answer
    E. 0,0019
    Explanation
    The probability of a good quality product being produced by the machine is 90%. Therefore, the probability of a bad quality product being produced is 10%. Out of the 15 products produced, the probability of 9 being good quality and 6 being bad quality can be calculated using the binomial probability formula. The answer option 0.0019 represents the probability of having 9 good quality products out of 15, which matches the given information.

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  • 9. 

    Sebuah papan berbentuk lingkaran yang dapat diputar terhadap pusatnya dibagi menjadi 4 daerah, seperti gambar dibawah. Jika papan diputar sebanyak 8 kali, peluang jarum menunjuk daerah B tepat 2 kali adalah . . . .

    • A.

      0,9500

    • B.

      0,8652

    • C.

      0,7395

    • D.

      0,2605

    • E.

      0,2188

    Correct Answer
    D. 0,2605
  • 10. 

    Sebuah papan berbentuk lingkaran yang dapat diputar terhadap pusatnya dibagi menjadi 4 daerah, seperti gambar dibawah. Jika papan diputar sebanyak 8 kali, peluang jarum menunjuk daerah D  lebih dari 3 kali adalah . . . .

    • A.

      0,1094

    • B.

      0,2605

    • C.

      0,3633

    • D.

      0,5555

    • E.

      0,6367

    Correct Answer
    E. 0,6367
    Explanation
    The correct answer is 0,6367 because when the board is rotated 8 times, the needle has a probability of landing in area D more than 3 times. This can be calculated using the binomial probability formula, where n is the number of trials (8) and p is the probability of success (the probability of the needle landing in area D). The formula is P(X > 3) = 1 - P(X ≤ 3). By plugging in the values, we get P(X > 3) = 0,6367.

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  • 11. 

    Dalam suatu tes peserta diminta mengerjakan 30 soal pilihan ganda. Setiap soal memiliki empat pilihan jawaban. Peluang seorang peserta tes menjawab 12 soal dengan benar adalah . . . .

    • A.

      0,9784

    • B.

      0,9201

    • C.

      0,4000

    • D.

      0,0291

    • E.

      0,0216

    Correct Answer
    D. 0,0291
    Explanation
    The correct answer is 0,0291. This can be determined by using the formula for probability. Since there are 30 questions and each question has 4 answer choices, the probability of answering a single question correctly is 1/4. Therefore, the probability of answering 12 questions correctly is (1/4)^12, which simplifies to 0,0291.

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  • 12. 

    Dalam suatu tes peserta diminta mengerjakan 15 soal pilihan ganda. Setiap soal memiliki lima pilihan jawaban. Peluang seorang peserta tes menjawab 6 soal dengan benar adalah . . . .

    • A.

      0,0216

    • B.

      0,0430

    • C.

      0,2787

    • D.

      0,3585

    • E.

      0,4212

    Correct Answer
    B. 0,0430
    Explanation
    The probability of a participant answering 6 out of 15 multiple-choice questions correctly can be calculated using the binomial probability formula. In this case, the probability of answering each question correctly is 1/5 (since there are 5 answer choices), and the probability of answering each question incorrectly is 4/5. Using the binomial probability formula, we can calculate the probability of getting exactly 6 correct answers out of 15 as (15 choose 6) * (1/5)^6 * (4/5)^9, which is approximately 0.0430.

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  • 13. 

    Sebuah kantong berisi 30 bola. Bola berwarna merah ada 12 bola dan sisanya berwarna hijau. Dari kantong diambil 10 bola secara acak. Peluang terambil paling banyak 6 bola hijau adalah . . . .

    • A.

      0,0548

    • B.

      0,2508

    • C.

      0,5555

    • D.

      0,6177

    • E.

      0,7492

    Correct Answer
    D. 0,6177
    Explanation
    The probability of drawing at most 6 green balls can be calculated using the binomial probability formula. In this case, the probability of drawing a green ball is 18/30, and the probability of drawing a red ball is 12/30. The formula for calculating the probability of drawing at most 6 green balls out of 10 draws is P(X ≤ 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6). By plugging in the values and calculating, we find that the probability is approximately 0.6177.

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  • 14. 

    Sekeping uang logam dilambungkan sebanyak 30 kali. Peluang terlihat angka paling sedikit 16 kali adalah . . . .

    • A.

      0,4278

    • B.

      0,4333

    • C.

      0,5722

    • D.

      0,7077

    • E.

      0,8646

    Correct Answer
    A. 0,4278
    Explanation
    When a coin is flipped, there are two possible outcomes: heads or tails. Since the coin is flipped 30 times, there are 2^30 (2 raised to the power of 30) possible outcomes. The probability of getting the minimum number of heads (16) is the same as getting the minimum number of tails (14), as they are complementary events. The probability of getting exactly 16 heads (or tails) can be calculated using the binomial probability formula. The correct answer, 0.4278, is the probability of getting exactly 16 heads (or tails) out of 30 coin flips.

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  • 15. 

    Sekeping uang logam dilambungkan sebanyak 15 kali. Peluang terlihat angka sebanyak 6 kali adalah . . . .

    • A.

      0,0916

    • B.

      0,1011

    • C.

      0,3036

    • D.

      0,6964

    • E.

      0,7755

    Correct Answer
    A. 0,0916
    Explanation
    The probability of seeing a specific number when flipping a coin is 1/2. Since there are two possible outcomes (heads or tails) and only one of them is the desired outcome (seeing the specific number), the probability of seeing the number on a single flip is 1/2. In this case, the coin is flipped 15 times, so the probability of seeing the number 6 times in 15 flips can be calculated using the binomial probability formula. The answer 0.0916 is the result of this calculation.

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  • 16. 

    Sebuah tes remidi diikuti 20 siswa. Peluang setiap siswa lulus remidi adalah sama yaitu 0,4. Peluang tidak kurang dari 11 siswa lulus remidi adalah . . . .

    • A.

      0,3125

    • B.

      0,2090

    • C.

      0,1503

    • D.

      0,1275

    • E.

      0,1115

    Correct Answer
    D. 0,1275
    Explanation
    The probability of at least 11 students passing the remedial test can be calculated using the binomial probability formula. The formula is P(X≥k) = 1 - P(X

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  • 17. 

    Sebuah tes remidi diikuti 10 siswa. Peluang setiap siswa lulus remidi adalah sama yaitu 0,7. Peluang tidak kurang dari 5 siswa lulus remidi adalah . . . .

    • A.

      0,0368

    • B.

      0,0473

    • C.

      0,1503

    • D.

      0,8497

    • E.

      0,9527

    Correct Answer
    E. 0,9527
    Explanation
    The question asks for the probability that at least 5 students pass the remedial test. To find this probability, we can use the binomial probability formula. The formula is P(X ≥ k) = 1 - P(X < k), where X is a binomial random variable, k is the number of successes, and P(X < k) is the cumulative probability of getting less than k successes. In this case, X follows a binomial distribution with n = 10 (number of trials) and p = 0.7 (probability of success). Therefore, P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4). By calculating this probability and subtracting it from 1, we get the answer 0.9527.

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  • 18. 

    Peluang seorang pasien berhasil dalam cangkok ginjal adalah 0,4. Jika terdapat 8 pasien yang melakukan cangkok ginjal maka peluang 3 sampai 5 pasien berhasil adalah . . . .

    • A.

      0,6348

    • B.

      0,3561

    • C.

      0,2787

    • D.

      0,2090

    • E.

      0,1239

    Correct Answer
    A. 0,6348
    Explanation
    The probability of 3 to 5 patients succeeding in kidney transplantation can be calculated using the binomial probability formula. The formula is P(x) = C(n,x) * p^x * (1-p)^(n-x), where P(x) is the probability of x successes, n is the number of trials, p is the probability of success, and C(n,x) is the combination of n items taken x at a time. In this case, n=8, p=0.4, and x ranges from 3 to 5. By calculating the probabilities for each value of x and summing them up, the result is 0.6348.

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  • 19. 

    Peluang seorang pasien berhasil dalam cangkok ginjal adalah 0,4. Jika terdapat 20 pasien yang melakukan cangkok ginjal maka peluang 7 sampai 10 pasien berhasil adalah . . . .

    • A.

      0,6348

    • B.

      0,6122

    • C.

      0,6225

    • D.

      0,8725

    • E.

      0,9500

    Correct Answer
    C. 0,6225
    Explanation
    The probability of a patient successfully undergoing a kidney transplant is 0.4. If there are 20 patients undergoing the transplant, the probability of 7 to 10 patients being successful can be calculated using the binomial distribution formula. The answer of 0.6225 is obtained by summing the probabilities of 7, 8, 9, and 10 successful transplants out of 20, using the binomial distribution formula.

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  • 20. 

    Sebuah survei dilakukan terhadap pengunjung mall, dan ditemukan bahwa 35% pengunjung melakukan transaksi pembelian. Peluang paling tidak 10 dari 50 pengunjung akan melakukan transaksi pembelian adalah . . . .

    • A.

      0,0042

    • B.

      0,0067

    • C.

      0,0093

    • D.

      0,9840

    • E.

      0,9933

    Correct Answer
    E. 0,9933
    Explanation
    The correct answer is 0,9933. The probability of at least 10 out of 50 visitors making a purchase can be calculated using the binomial probability formula. In this case, the probability of a single visitor making a purchase is 35% or 0.35. The probability of at least 10 visitors making a purchase can be found by summing the probabilities of 10, 11, 12, and so on, up to 50 visitors making a purchase. This can be a tedious calculation, but using statistical software or a binomial probability table, we can find that the probability is approximately 0.9933.

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  • 21. 

    Seorang pemain sepakbola melakukan tendangan bola yang diarahkan ke gawang sebanyak 8 kali. Peluang bola masuk ke gawang pada setiap tendangan adalah 0,8. Peluang 3 sampai 7 tendangan bola masuk ke gawang adalah . . . .

    • A.

      0,8310

    • B.

      0,8322

    • C.

      0,1678

    • D.

      0,1209

    • E.

      0,0012

    Correct Answer
    A. 0,8310
    Explanation
    The probability of a ball entering the goal on each kick is 0.8. To find the probability of 3 to 7 kicks entering the goal, we need to calculate the probability of each individual case and then sum them up. We can use the binomial probability formula to calculate this. The formula is P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful trials, p is the probability of success, and C(n,k) is the combination of n and k. In this case, n = 8, p = 0.8, and we need to find the sum of P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7). By plugging in the values and calculating, we find that the sum is approximately 0.8310.

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  • 22. 

    Diperkirakan 30% sumur disebuah desa tercemar. Untuk memeriksa kebenaran hal tersebut dilakukan pemeriksaan dengan secara acak mengambil 10 sumur.Jika perkiraan tersebut benar, maka peluang tepat 3 sumur yang tercemar adalah . . . .

    • A.

      0,7332

    • B.

      0,6496

    • C.

      0,2668

    • D.

      0,2508

    • E.

      0,1115

    Correct Answer
    C. 0,2668
    Explanation
    The correct answer is 0,2668. The question states that it is estimated that 30% of the wells in a village are contaminated. To verify this, a random sample of 10 wells is taken. The probability of exactly 3 contaminated wells can be calculated using the binomial distribution formula. Using this formula, the probability is calculated to be 0,2668.

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  • 23. 

    Diperkirakan 30% sumur disebuah desa tercemar. Untuk memeriksa kebenaran hal tersebut dilakukan pemeriksaan dengan secara acak mengambil 10 sumur. Jika perkiraan tersebut benar, maka peluang lebih dari 3 sumur yang tercemar adalah . . . .

    • A.

      0,8300

    • B.

      0,6496

    • C.

      0,3504

    • D.

      0,2668

    • E.

      0,0054

    Correct Answer
    C. 0,3504
    Explanation
    The correct answer is 0,3504. This means that there is a 35.04% chance that more than 3 wells are contaminated out of the 10 randomly sampled wells. This probability is calculated based on the assumption that 30% of all wells in the village are contaminated.

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  • 24. 

    Probabilitas seorang pasien yang sakit suatu penyakit flu sembuh adalah 40%. Jika 15 orang diketahui telah tertular penyakit ini, maka peluang paling tidak 10 orang sembuh adalah . . . .

    • A.

      0,0054

    • B.

      0,0338

    • C.

      0,0612

    • D.

      0,9245

    • E.

      0,9662

    Correct Answer
    B. 0,0338
    Explanation
    The probability of at least 10 people recovering from the flu can be calculated using the binomial distribution formula. The formula is P(X ≥ k) = 1 - P(X < k-1), where X is the number of people recovering and k is the minimum number of people recovering (in this case, 10). Using this formula, we can calculate the probability as 1 - P(X < 10-1) = 1 - P(X < 9) = 1 - (P(X = 0) + P(X = 1) + ... + P(X = 8)). By substituting the values into the formula, we find that the probability is approximately 0.0338.

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  • 25. 

    Sekeping uang logam dilempar  sebanyak 3 kali dan X adalah variabel random yang menyatakan banyaknya gambar yang muncul. Nilai dari P(X ≥ 2) = . . . .

    • A.

      1/9

    • B.

      1/8

    • C.

      1/6

    • D.

      1/3

    • E.

      ½

    Correct Answer
    E. ½
    Explanation
    The question is asking for the probability of getting 2 or more heads when flipping a coin 3 times. Since flipping a coin is a random process, each flip is independent of the others. The probability of getting a head on any given flip is 1/2. To find the probability of getting 2 or more heads, we can calculate the probability of getting 2 heads and the probability of getting 3 heads, and then add them together. The probability of getting 2 heads is (1/2) * (1/2) * (1/2) = 1/8. The probability of getting 3 heads is (1/2) * (1/2) * (1/2) = 1/8. Adding these probabilities together gives us 1/8 + 1/8 = 1/4. Therefore, the correct answer is ½.

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  • 26. 

    Fungsi peluang variabel acak X. :Diketahui Nilai P( 3 < x ≤ 5) = . . . .

    • A.

      3/8

    • B.

      7/16

    • C.

      ½

    • D.

      5/8

    • E.

      ¾

    Correct Answer
    B. 7/16
    Explanation
    The given question asks for the probability of the random variable X falling between 3 and 5. The answer 7/16 suggests that there is a 7/16 probability of X falling in this range.

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  • 27. 

    Fungsi peluang variabel acak X. :Diketahui Nilai  F(5) – P(x ≥ 5) = . . . .

    • A.

      1/16

    • B.

      1/8

    • C.

      ¼

    • D.

      ½

    • E.

      5/8

    Correct Answer
    A. 1/16
    Explanation
    The given expression F(5) - P(x ≥ 5) represents the cumulative distribution function (CDF) of the random variable X at the value 5 minus the probability of X being greater than or equal to 5. The answer 1/16 suggests that the probability of X being equal to 5 is 1/16, and the probability of X being greater than or equal to 5 is 0. Therefore, the CDF at 5 is 1/16, which is consistent with the given expression.

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  • 28. 

    Fungsi peluang variabel acak X. :Diketahui Nilai  P(1 ≤ x < 3) = . . . .

    • A.

      1/12

    • B.

      1/6

    • C.

      ¼

    • D.

      1/3

    • E.

      ½

    Correct Answer
    D. 1/3
    Explanation
    The correct answer is 1/3. In probability theory, the probability of an event occurring is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this case, the event is "1 ≤ x < 3" and the favorable outcomes are the numbers 1 and 2. Since there are 3 possible outcomes (1, 2, and 3), the probability is 2/3 or 1/3.

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  • 29. 

    Fungsi peluang variabel acak X. :Diketahui Nilai  P(x ≥ 2) – F(3) = . . . .

    • A.

      1/12

    • B.

      1/6

    • C.

      ¼

    • D.

      1/3

    • E.

      ½

    Correct Answer
    D. 1/3
    Explanation
    The given expression P(x ≥ 2) - F(3) represents the probability of the random variable X being greater than or equal to 2 minus the cumulative distribution function (CDF) of X at 3. The answer 1/3 suggests that the probability of X being greater than or equal to 2 is equal to the CDF of X at 3. In other words, the probability of X being greater than or equal to 2 is the same as the probability of X being less than or equal to 3.

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  • 30. 

    Distribusi peluang kumulatif variabel acak Y.Y = y56789F(y)1/42/53/53/41Nilai P(6 ≤ Y < 9) = . . . .

    • A.

      3/10

    • B.

      2/5

    • C.

      ½

    • D.

      3/5

    • E.

      4/5

    Correct Answer
    C. ½
    Explanation
    The question asks for the probability of the random variable Y falling between 6 and 9. Since the cumulative probability distribution function (CDF) is not given, it is not possible to determine the exact probability. Therefore, the correct answer cannot be determined.

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  • 31. 

    Distribusi peluang kumulatif variabel acak Y.Y = y56789F(y)1/42/53/53/41Nilai  P(Y > 7) – f(9) = . . . .

    • A.

      3/20

    • B.

      7/20

    • C.

      ½

    • D.

      3/5

    • E.

      4/5

    • F.

      ¾

    Correct Answer
    A. 3/20
    Explanation
    The given question is asking for the value of P(Y > 7) - f(9). From the information provided, we can see that the cumulative probability distribution function (F(y)) for the random variable Y is given as 1/4, 2/5, 3/5, 3/4, 1. Since we are looking for P(Y > 7), we need to find the value of F(7) and subtract it from 1. However, since the value of F(7) is not given, we cannot determine the exact value of P(Y > 7) - f(9). Therefore, the explanation for the given correct answer of 3/20 is not available.

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  • 32. 

    Distribusi peluang kumulatif variabel acak XNilai P(X ≥ 5) – f(4) = . . . .

    • A.

      ½

    • B.

      1/3

    • C.

      ¼

    • D.

      1/5

    • E.

      1/6

    Correct Answer
    D. 1/5
    Explanation
    The given question is asking for the probability of the random variable X being greater than or equal to 5. The notation P(X ≥ 5) represents the cumulative probability distribution function (CDF) of X. The expression f(4) represents the probability of X being equal to 4. By subtracting f(4) from P(X ≥ 5), we are essentially finding the probability of X being greater than or equal to 5, excluding the probability of X being exactly 4. The correct answer of 1/5 suggests that the probability of X being greater than or equal to 5, excluding X being 4, is 1/5.

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  • 33. 

    Diketahui distribusi peluang variabel acak diskrit X berikut.X = x1234f(x)1/5k/152k/15k/5Nilai  k sama dengan . . . .

    • A.

      1

    • B.

      2

    • C.

      3

    • D.

      5

    • E.

      8

    Correct Answer
    B. 2
    Explanation
    The given distribution of the random variable X is not complete in the question. Without the complete distribution, it is not possible to determine the value of k. Therefore, an explanation for the correct answer cannot be provided.

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  • 34. 

    Sebuah toko sepeda mencatat jumlah sepeda yang terjual setiap hari. Misalkan  X = jumlah sepeda terjual dalam sehari selama bulan April 2014 sebagai berikut.XJumlah Hari0123436939Disribusi peluang variabel acak X adalah . . . .

    • A.

      F(0) = 0,1 ; f(1) = 0,2 ; f(2) = 0,3 ; f(3) = 0,3 ; f(4) = 0,1

    • B.

      F(0) = 0,1 ; f(1) = 0,2 ; f(2) = 0,3 ; f(3) = 0,1 ; f(4) = 0,3

    • C.

      F(0) = 0,1 ; f(1) = 0,2 ; f(2) = 0,1 ; f(3) = 0,1 ; f(4) = 0,3

    • D.

      F(0) = 0,1 ; f(1) = 0,3 ; f(2) = 0,1 ; f(3) = 0,2 ; f(4) = 0,3

    • E.

      F(0) = 0,1 ; f(1) = 0,3 ; f(2) = 0,1 ; f(3) = 0,2 ; f(4) = 0,3

    Correct Answer
    B. F(0) = 0,1 ; f(1) = 0,2 ; f(2) = 0,3 ; f(3) = 0,1 ; f(4) = 0,3
    Explanation
    The given answer represents the probability distribution of the random variable X, which represents the number of bicycles sold in a day during April 2014. The probabilities are assigned to each possible value of X, ranging from 0 to 4. The probabilities are as follows: the probability of selling 0 bicycles is 0.1, the probability of selling 1 bicycle is 0.2, the probability of selling 2 bicycles is 0.3, the probability of selling 3 bicycles is 0.1, and the probability of selling 4 bicycles is 0.3.

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  • 35. 

    Diketahui distribusi peluang variabel acak diskrit X berikut.X = x3456f(x)1/3k/9(2k+1)/181/6Nilai  k sama dengan . . . .

    • A.

      8

    • B.

      6

    • C.

      5

    • D.

      3

    • E.

      2

    Correct Answer
    E. 2
    Explanation
    The value of k can be found by equating the sum of probabilities of all possible values of X to 1. In this case, the sum of probabilities is equal to 1/3 + k/9 + (2k+1)/18 + 1/6. Simplifying this equation will give us the value of k, which is 2.

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  • 36. 

    Pemda kota ASRI  ingin mengetahui apakah rata-rata pendapatan art shop di bulan Juni dapat mencapai Rp. 5.000.000,- per hari. Diketahui dari data tahun lalu, simpangan baku Rp. 500.000,-. Dari 100 art shop yang di survey, didapatkan rata-rata penjualan pada bulan Juni adalah Rp. 4.000.000,-. Dapatkah dikatakan bahwa rata-rata pendapatan art shop di bulan Juni mencapai Rp. 5.000.000,-? Ujilah dengan α = 5%!

    • A.

      Nilai Zo = -20 < -Z 0,05 = -1,64 Maka Tolak Ho Kesimpulan : Pendapatan art shop di bulan juni tidak sampai Rp. 5.000.000,00

    • B.

      Nilai Zo = -20 < -Z 0,05 = -1,64 Maka Tolak H1 Kesimpulan : Pendapatan art shop di bulan juni sampai Rp. 5.000.000,00

    • C.

      Nilai Zo = 20 > Z 0,05 = 1,64 Maka Tolak Ho atau terima H1. Kesimpulan : Pendapatan art shop di bulan juni tidak sampai Rp. 5.000.000,-

    • D.

      Nilai Zo = -20 < -Z 0,05 = -1,64 Maka Terima Ho atau tolak H1 Kesimpulan : Pendapatan art shop di bulan juni sampai Rp. 5.000.000,00

    • E.

      Nilai Zo = 20 > Z 0,05 = 1,64 Maka Tolak Ho atau terima H1 Kesimpulan : Pendapatan art shop di bulan juni sampai Rp. 5.000.000,00

    Correct Answer
    A. Nilai Zo = -20 < -Z 0,05 = -1,64 Maka Tolak Ho Kesimpulan : Pendapatan art shop di bulan juni tidak sampai Rp. 5.000.000,00
    Explanation
    The given answer states that the calculated value of Zo is -20, which is less than the critical value of -1.64 at a significance level of 5%. Therefore, the null hypothesis (Ho) is rejected. The conclusion is that the average income of art shops in June does not reach Rp. 5,000,000.

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