# Rem.1 - Statistika Dan Peluang - Xi IPA - Senin, 16 Maret 2015

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Pertanyaan: 36 | Attempts: 1,019  Settings  Senin, 16 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

• 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

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 (0.5 for a true/false question), and q is the probability of failure (1 - p). Plugging in the values, we get P(8) = (15C8)(0.5^8)(0.5^7) = 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

E. 0,0439
Explanation
The given answer, 0.0439, represents the probability of a participant answering 8 out of 10 true-false questions correctly. This can be calculated using the binomial probability formula, where n is the number of trials (10 in this case), p is the probability of success (0.5 since it is a true-false question), and k is the number of successful trials (8 in this case). Plugging in these values into the formula, we get the probability of 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

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 each participant passing the test is 20% or 0.2, and the sample size is 10. By calculating the cumulative probability of 0, 1, 2, and 3 participants passing the test, we find that the probability of at most 3 participants passing the test is 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

A. 0,9803
Explanation
The probability of at most 6 out of 10 participants passing the test can be calculated using the binomial distribution formula. Since the probability of each participant passing the test is 1/3, the probability of a participant failing the test is 2/3. Using the binomial distribution formula, the probability can be calculated as P(X ≤ 6) = Σ (10Ck) * (1/3)^k * (2/3)^(10-k), where k ranges from 0 to 6. After calculating the probabilities for each value of k and summing them up, we find that the probability is approximately 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

C. 0,0500
Explanation
The probability of getting a pair of dice with a sum of 5 or more at least 3 times in 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 = 56. Plugging these values into the formula, we get P(X = 3) = 56 * (1/6)^3 * (5/6)^5 = 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

C. 0,6367
Explanation
The probability of getting 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) = (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. In this case, n = 8, k = 4, p = 1/6 (since there are 3 even numbers out of 6 possible outcomes on a dice), and (nCk) = 8C4 = 70. Plugging these values into the formula, we get P(X=4) = (8C4) * (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

B. 0,2461
Explanation
The probability of drawing a red ball from the box is 3/6 or 1/2. Since the balls are replaced after each draw, the probability of drawing a red ball in each of the 10 trials is independent. Therefore, the probability of drawing 5 red balls in 10 trials 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 in each trial. Plugging in the values, we get P(X=5) = (10C5) * (1/2)^5 * (1-1/2)^(10-5) = 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

E. 0,0019
Explanation
The given question states that 90% of the goods produced by a production machine are of good quality. Therefore, the probability of a good quality item being produced is 0.9. Out of 15 items produced, the probability of 60% (9 items) being of good quality is calculated by using the binomial distribution 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 successes, p is the probability of success, and (nCk) is the combination formula. Substituting the values, we get P(X=9) = (15C9) * 0.9^9 * (1-0.9)^(15-9) = 0.0019. Therefore, the correct answer is 0.0019.

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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

D. 0,2605
Explanation
The probability of the needle pointing to area B after 8 rotations can be calculated using the concept of probability. Since the board is divided into 4 equal areas, the probability of the needle pointing to area B on any given rotation is 1/4. Therefore, the probability of the needle pointing to area B exactly 2 times after 8 rotations can be calculated using the binomial probability formula. The correct answer, 0.2605, is the result of this calculation.

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• 10.

• A.

0,1094

• B.

0,2605

• C.

0,3633

• D.

0,5555

• E.

0,6367

E. 0,6367
• 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

D. 0,0291
Explanation
The correct answer is 0,0291. This answer represents the probability of a participant answering 12 questions correctly out of 30 multiple-choice questions, where each question has four answer choices.

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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

B. 0,0430
Explanation
The probability of a participant answering 6 questions correctly out of 15 multiple-choice questions, where each question has 5 answer choices, can be calculated using the binomial probability formula. The formula is P(X=k) = (nCk) * (p^k) * (q^(n-k)), where n is the total number of trials, k is the number of successful outcomes, p is the probability of success on a single trial, and q is the probability of failure on a single trial. In this case, n = 15, k = 6, p = 1/5, and q = 4/5. Plugging these values into the formula, we get P(X=6) = (15C6) * ((1/5)^6) * ((4/5)^9) = 0.0430. Therefore, the correct answer is 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

D. 0,6177
Explanation
The probability of drawing at most 6 green balls can be calculated by finding the probability of drawing 0, 1, 2, 3, 4, 5, or 6 green balls and adding them together. To find the probability of drawing a specific number of green balls, we can use the binomial probability formula. The formula is P(X=k) = (nCk) * (p^k) * (q^(n-k)), where n is the total number of draws, k is the number of successful outcomes (drawing green balls), p is the probability of a successful outcome (drawing a green ball), q is the probability of a failure (drawing a red ball), and (nCk) is the number of combinations of n items taken k at a time. By calculating the probabilities for each case and adding them together, we find that the probability of drawing at most 6 green balls is 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

A. 0,4278
Explanation
When flipping a coin 30 times, the probability of getting the minimum number of heads (which is 16) can be calculated 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, and p is the probability of success on a single trial. In this case, n=30, k=16, and p=0.5 (since the coin has two sides). Plugging in these values into the formula, we can calculate the probability to be approximately 0.4278.

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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

A. 0,0916
Explanation
The probability of seeing a specific side of a coin when it is tossed is always 1/2. In this case, the coin is tossed 15 times. The probability of seeing a specific side (in this case, the number side) 6 times in 15 tosses can be calculated using the binomial probability formula. Using this formula, the probability is calculated to be 0.0916.

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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

D. 0,1275
Explanation
The question asks for the probability that at least 11 students pass the remedial test. To solve this, we can use the binomial probability formula. The probability of exactly k successes in n independent Bernoulli trials is given by the formula P(X=k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes, and p is the probability of success in a single trial. In this case, n = 20, k ≥ 11, and p = 0.4. We need to calculate the sum of probabilities for k = 11, 12, 13, ..., 20. By doing this calculation, we find that the probability of at least 11 students passing the remedial test is 0.1275.

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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

E. 0,9527
Explanation
The probability of a student passing the remedial test is 0.7. To find the probability of at least 5 students passing, we can use the binomial probability formula. The formula is P(X ≥ k) = 1 - P(X < k-1), where X is the number of students passing and k is the minimum number of students passing. In this case, k = 5. Using the formula, we can calculate P(X ≥ 5) = 1 - P(X < 4). Plugging in the values, P(X ≥ 5) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)). Calculating the probabilities for each value of X and subtracting from 1, we get 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

A. 0,6348
Explanation
The probability of a patient successfully undergoing a kidney transplant is 0.4. If there are 8 patients undergoing the transplant, the probability of 3 to 5 patients being successful can be calculated using the binomial probability formula. The formula is P(x) = (nCx) * (p^x) * (q^(n-x)), where n is the number of trials, x is the number of successful outcomes, p is the probability of success, and q is the probability of failure (1-p). In this case, n = 8, p = 0.4, and q = 0.6. By calculating P(3) + P(4) + P(5), we get the answer 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

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 probability formula. The answer 0.6225 is likely obtained by summing the probabilities of 7, 8, 9, and 10 successful transplants out of 20, using the binomial probability 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

E. 0,9933
Explanation
The probability of at least 10 out of 50 visitors making a purchase 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 visitors making a purchase and k is the minimum number of visitors making a purchase. In this case, k = 10. By plugging in the values into the formula, we can calculate the probability to be 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

A. 0,8310
Explanation
The probability of a ball entering the goal in each kick is 0.8. To find the probability of 3 to 7 kicks entering the goal, we need to calculate the probability of 3, 4, 5, 6, and 7 kicks entering the goal and add them together. Since each kick is independent, we can use the binomial probability formula. The probability of k successes in n independent trials, where the probability of success is p, is given by the formula: P(X = k) = (nCk) * (p^k) * ((1-p)^(n-k)). In this case, n = 8, p = 0.8, and we need to calculate P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7). After performing the calculations, we get the answer 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 lebih dari 3 sumur yang tercemar adalah . . . .

• A.

0,7332

• B.

0,6496

• C.

0,2668

• D.

0,2508

• E.

0,1115

C. 0,2668
Explanation
The correct answer is 0,2668. This means that the probability of more than 3 wells being contaminated, given that the estimated contamination rate is 30%, is 0.2668. This can be calculated by using probability theory and statistical methods to analyze the random sampling of 10 wells.

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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

C. 0,3504
Explanation
Based on the given information, it is estimated that 30% of the wells in the village are contaminated. To verify this estimation, a random sample of 10 wells is taken. The question asks for the probability that more than 3 wells are contaminated if the estimation is correct. To find this probability, we can use the binomial distribution formula. The probability of having more than 3 contaminated wells can be calculated by summing the probabilities of having 4, 5, 6, 7, 8, 9, or 10 contaminated wells. When the calculations are done, the probability is found to be 0.3504, which matches the given answer.

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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

B. 0,0338
Explanation
The probability of a patient recovering from the flu is 40%. If 15 people are known to have contracted the disease, the probability of at least 10 people recovering can be calculated using the binomial probability 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, k = 10. By substituting the values into the formula, we can calculate the probability to be 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

E. ½
Explanation
The question is asking for the probability of getting 2 or more heads when a coin is flipped 3 times. Since there are 2 possible outcomes (heads or tails) for each flip, there are a total of 2^3 = 8 possible outcomes when the coin is flipped 3 times. Out of these 8 outcomes, there are 4 outcomes where 2 or more heads appear: HHT, HTH, THH, and HHH. Therefore, the probability of getting 2 or more heads is 4/8, which simplifies to 1/2 or 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

B. 7/16
Explanation
The given question asks for the probability of the random variable X falling between 3 and 5. The correct answer is 7/16, which means that there is a 7/16 chance that X will be greater than 3 and less than or equal to 5.

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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

A. 1/16
Explanation
The given expression is F(5) - P(x ≥ 5) = . . . . This implies that we need to find the difference between the cumulative probability of X being less than or equal to 5 and the probability of X being greater than or equal to 5. The answer is 1/16, which means that the cumulative probability of X being less than or equal to 5 is 1/16 greater than the probability of X being greater than or equal to 5.

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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

D. 1/3
Explanation
The correct answer is 1/3 because the probability is asking for the likelihood that the random variable X takes on a value between 1 and 3 (inclusive). Since there are three possible values (1, 2, and 3) and each value has an equal chance of occurring, the probability of X being between 1 and 3 is 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

D. 1/3
Explanation
The given equation is P(x ≥ 2) - F(3) = . . . . This equation 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 of 1/3 suggests that the probability of X being greater than or equal to 2 is equal to the CDF of X at 3. This implies that the random variable X has a continuous probability distribution, such as a uniform distribution, where the probability is evenly distributed over a range. Therefore, the probability of X being greater than or equal to 2 is equal to the probability of X being at 3, which is 1/3.

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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

C. ½
Explanation
The given question is asking for the value of P(6 ≤ Y < 9), which represents the probability that the random variable Y takes on a value between 6 (inclusive) and 9 (exclusive). The correct answer is ½, which means that there is a 50% chance that Y falls within this range.

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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

A. 3/20
Explanation
The given question is asking for the value of P(Y > 7) - f(9), where Y is a random variable with a cumulative probability distribution function given. To find P(Y > 7), we need to subtract the cumulative probability at 7 from 1 (since the cumulative probability gives the probability up to that value). From the given cumulative probability distribution, we can see that F(7) = 3/5. Next, we need to find f(9), which represents the probability of Y being exactly 9. From the given distribution, we can see that f(9) = 1/4. Finally, subtracting f(9) from P(Y > 7) gives us 3/5 - 1/4 = 12/20 - 5/20 = 7/20. Therefore, the correct answer is 7/20.

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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

D. 1/5
Explanation
The given question is asking for the cumulative probability distribution of a random variable X, specifically the probability of X being greater than or equal to 5. The answer is 1/5 because it is equal to the difference between the cumulative probability of X being greater than or equal to 4 and the cumulative probability of X being greater than or equal to 5. In other words, it represents the additional probability of X being 5 when X is already greater than or equal to 4.

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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

B. 2
Explanation
The correct answer is 2 because the probability distribution given shows that the probability of X being equal to 2 is k/15. Since the sum of all probabilities in a probability distribution must equal 1, we can set up the equation 1/5 + k/15 + 2/15 + k/5 = 1. By solving this equation, we find that k = 2. Therefore, the value of k is 2.

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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

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 bikes sold in a day during April 2014. The probabilities are assigned to each possible value of X. For example, f(0) = 0.1 means that there is a 10% chance that no bikes will be sold in a day. Similarly, f(1) = 0.2 means that there is a 20% chance that one bike will be sold in a day. The probabilities assigned to each value of X add up to 1, representing the total probability.

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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

E. 2
Explanation
The given distribution of the random variable X is discrete. The probability function f(x) is given as 1/3k/9(2k+1)/18(1/6). To find the value of k, we need to solve the equation 1/3k/9(2k+1)/18(1/6) = 1. Simplifying this equation, we get k/9(2k+1)/18 = 1. Cross-multiplying, we get 2k^2 + k - 9 = 0. Factoring this quadratic equation, we get (2k - 3)(k + 3) = 0. Solving for k, we find two possible values: k = 3/2 or k = -3. However, since k cannot be negative, the value of k is 3/2, which is not listed in the answer choices. Therefore, the correct answer is not available.

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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 Back to top