Rem.2 - Statistika Dan Peluang - Xi IPA - Selasa, 17 Maret 2015

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  • 1/36 Pertanyaan

    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 . . . .

    • 0,0548
    • 0,2508
    • 0,5555
    • 0,6177
    • 0,7492
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About This Quiz

Selasa, 17 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 Lain
6.  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

Rem.2 - Statistika Dan Peluang - Xi IPA - Selasa, 17 Maret 2015 - Quiz

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  • 2. 

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

    • 0,0074

    • 0,0246

    • 0,0500

    • 0,9926

    • 0,9574

    Correct Answer
    A. 0,0500
    Explanation
    The probability of seeing a pair of dice with a total of 5 at least 3 times out of 8 throws is 0.0500.

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  • 3. 

    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 . . . .

    • 0,6348

    • 0,3561

    • 0,2787

    • 0,2090

    • 0,1239

    Correct Answer
    A. 0,6348
    Explanation
    The probability of 3 to 5 patients being successful in kidney transplantation 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 (8 patients), k is the number of successful outcomes (3, 4, or 5 patients), and p is the probability of success (0.4). By calculating the probabilities for each possible outcome (3, 4, and 5) and summing them up, we get the final probability of 0.6348.

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  • 4. 

    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 . . . .

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

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

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

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

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

    Correct Answer
    A. 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 for selling 0, 1, 2, 3, and 4 bicycles are given as f(0) = 0.1, f(1) = 0.2, f(2) = 0.3, f(3) = 0.1, and f(4) = 0.3 respectively. This means that there is a 10% chance of selling 0 bicycles, a 20% chance of selling 1 bicycle, a 30% chance of selling 2 bicycles, a 10% chance of selling 3 bicycles, and a 30% chance of selling 4 bicycles in a day during April 2014.

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  • 5. 

    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 . . . .

    • 0,8310

    • 0,8322

    • 0,1678

    • 0,1209

    • 0,0012

    Correct Answer
    A. 0,8310
    Explanation
    The probability of a ball entering the goal in each kick is 0.8. To find the probability that 3 to 7 kicks enter the goal, we need to calculate the probability of each individual outcome (3, 4, 5, 6, and 7 kicks entering the goal) and then sum them up. Since the probability of each kick entering the goal is the same, we can use the binomial distribution formula to calculate the probabilities. Using this formula, the probability of 3 to 7 kicks entering the goal is 0.8310.

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  • 6. 
    Distribusi peluang kumulatif variabel acak XNilai P(X ≥ 5) – f(4) = . . . . - ProProfs

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

    • ½

    • 1/3

    • ¼

    • 1/5

    • 1/6

    Correct Answer
    A. 1/5
  • 7. 

    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 . . . .

    • 0,9981

    • 0,9978

    • 0,2461

    • 0,0022

    • 0,0019

    Correct Answer
    A. 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, 60% of them are good quality. This means that 40% of them are bad quality. To find the probability of getting 60% good quality products out of 15, we can use the binomial probability formula. The answer 0.0019 is the probability of getting exactly 9 good quality products out of 15, given that the probability of getting a good quality product is 0.9.

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  • 8. 
    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 . . . . - ProProfs

    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 . . . .

    • 0,9500

    • 0,8652

    • 0,7395

    • 0,2605

    • 0,2188

    Correct Answer
    A. 0,2605
    Explanation
    The probability of the needle pointing to area B after 8 rotations can be calculated by dividing the number of favorable outcomes (2) by the total number of possible outcomes (8). Therefore, the probability is 2/8 = 0.25. However, since the options provided are in decimal form, we need to find the closest decimal approximation to 0.25. Among the given options, 0.2605 is the closest approximation to 0.25.

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  • 9. 

    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 . . . .

    • 0,9784

    • 0,9201

    • 0,4000

    • 0,0291

    • 0,0216

    Correct Answer
    A. 0,0291
    Explanation
    The correct answer is 0,0291. This can be calculated using the binomial probability formula, where the probability of answering a single question correctly is 1/4. The formula is P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials (30 questions), k is the number of successes (12 correct answers), and p is the probability of success (1/4). Plugging in these values, we get P(X=12) = C(30,12) * (1/4)^12 * (3/4)^18 = 0,0291.

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  • 10. 

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

    • 0,4278

    • 0,4333

    • 0,5722

    • 0,7077

    • 0,8646

    Correct Answer
    A. 0,4278
    Explanation
    When flipping a coin 30 times, the probability of getting the least number of heads (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. In this case, n=30, k=16, and p=0.5 (since the coin has two sides). Plugging these values into the formula, we can calculate the probability. The correct answer is 0.4278.

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  • 11. 

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

    • 0,1964

    • 0,3036

    • 0,6964

    • 0,8036

    • 0,9926

    Correct Answer
    A. 0,1964
    Explanation
    The probability of a participant answering 8 questions correctly out of 15 is 0.1964. This means that there is a 19.64% chance that the participant will answer exactly 8 questions correctly.

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  • 12. 

    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 . . . . 

    • 0,9803

    • 0,9500

    • 0,8791

    • 0,2013

    • 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. The formula is P(X ≤ k) = Σ (nCk * p^k * (1-p)^(n-k)), where n is the total number of participants (10 in this case), k is the number of successful outcomes (6 in this case), and p is the probability of success (1/3 in this case). By substituting the values into the formula, we can calculate the probability to be 0.9803.

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  • 13. 

    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 . . . .

    • 0,0216

    • 0,0430

    • 0,2787

    • 0,3585

    • 0,4212

    Correct Answer
    A. 0,0430
    Explanation
    The probability of a participant answering 6 questions correctly can be calculated using the binomial probability formula. In this case, there are 15 questions and each question has 5 possible answers. The probability of answering a question correctly is 1/5. Therefore, the probability of answering 6 questions correctly can be calculated as (15 choose 6) * (1/5)^6 * (4/5)^9, which is approximately equal to 0.0430.

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  • 14. 

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

    • 0,1094

    • 0,1964

    • 0,2189

    • 0,2734

    • 0,0439

    Correct Answer
    A. 0,0439
    Explanation
    The correct answer is 0,0439. This means that the probability of a participant answering 8 questions correctly out of 10 is 0,0439.

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  • 15. 
    Fungsi peluang variabel acak X. :Diketahui Nilai P( 3 < x ≤ 5) = . . . . - ProProfs

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

    • 3/8

    • 7/16

    • ½

    • 5/8

    • ¾

    Correct Answer
    A. 7/16
    Explanation
    The answer of 7/16 is obtained by calculating the probability of the random variable X falling between 3 and 5. This can be done by finding the difference between the cumulative probabilities of X being less than or equal to 5 and X being less than or equal to 3. Since the question does not provide any additional information, we can assume that X follows a continuous probability distribution. Therefore, the probability of X falling between 3 and 5 is equal to the difference between the cumulative probabilities, which is 7/16.

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  • 16. 

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

    • 1/12

    • 1/6

    • ¼

    • 1/3

    • ½

    Correct Answer
    A. 1/3
  • 17. 

    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%!

    • 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

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

    • 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,-

    • 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

    • 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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  • 18. 

    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 . . . .

    • 0,8300

    • 0,6496

    • 0,3504

    • 0,2668

    • 0,0054

    Correct Answer
    A. 0,3504
    Explanation
    The correct answer is 0,3504. This is because if 30% of the wells in the village are contaminated, then the probability of randomly selecting a contaminated well is 0,3. Since we are randomly selecting 10 wells, we can use the binomial distribution formula to calculate the probability of getting more than 3 contaminated wells. This probability is equal to 1 minus the sum of the probabilities of getting 0, 1, 2, and 3 contaminated wells. When we calculate this, we get 0,3504 as the probability of more than 3 contaminated wells.

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  • 19. 

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

    • 0,3633

    • 0,5244

    • 0,6367

    • 0,6525

    • 0,8235

    Correct Answer
    A. 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) = C(n, x) * p^x * q^(n-x), where P(x) is the probability of getting x successes, n is the number of trials, p is the probability of success, and q is the probability of failure. In this case, n = 8, x is at least 4, p = 1/6 (the probability of getting a pair of even numbers), and q = 5/6 (the probability of not getting a pair of even numbers). By calculating the probability for x = 4, 5, 6, 7, and 8, we find that the probability of seeing a pair of even numbers on the second dice at least 4 times out of 8 throws is 0.6367.

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  • 20. 

    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 . . . .

    • 0,3125

    • 0,2090

    • 0,1503

    • 0,1275

    • 0,1115

    Correct Answer
    A. 0,1275
    Explanation
    The probability of a student passing the remedial test is 0.4. To find the probability that at least 11 students pass the test, we can use the binomial probability formula. The formula is P(X≥k) = 1 - P(X

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  • 21. 

    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 . . . .

    • 0,6348

    • 0,6122

    • 0,6225

    • 0,8725

    • 0,9500

    Correct Answer
    A. 0,6225
    Explanation
    The probability of success in a kidney transplant is given as 0.4. Therefore, the probability of failure is 1 - 0.4 = 0.6.

    To find the probability that 7 to 10 out of 20 patients will be successful, we need to calculate the probability of each possible outcome (7, 8, 9, or 10 successes) and sum them up.

    Using the binomial probability formula, we can calculate the probability for each outcome and sum them up to get the answer.

    The correct answer, 0.6225, represents the probability that 7 to 10 out of 20 patients will be successful in a kidney transplant.

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  • 22. 

    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 . . . .

    • 0,0054

    • 0,0338

    • 0,0612

    • 0,9245

    • 0,9662

    Correct Answer
    A. 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. This formula calculates the probability of a certain number of successes in a fixed number of independent trials. In this case, the formula can be used to calculate the probability of at least 10 successes (recoveries) out of 15 trials (patients). The answer of 0.0338 represents this probability.

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  • 23. 

    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 . . . .

    • 0,9991

    • 0,8791

    • 0,2013

    • 0,0009

    • 0,0008

    Correct Answer
    A. 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 for each participant is 20% or 0.2. The probability of at most 3 participants passing can be calculated by summing up the probabilities of 0, 1, 2, and 3 participants passing the test. Using the binomial distribution formula, the probability is calculated to be 0.8791.

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  • 24. 
    Fungsi peluang variabel acak X. :Diketahui Nilai  P(1 ≤ x < 3) = . . . . - ProProfs

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

    • 1/12

    • 1/6

    • ¼

    • 1/3

    • ½

    Correct Answer
    A. 1/3
    Explanation
    The given question asks for the probability of the random variable X having a value between 1 and 3. The correct answer is 1/3. This means that there is a 1/3 chance that X will have a value between 1 and 3.

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  • 25. 

    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 . . . .

    • 0,0569

    • 0,2461

    • 0,3633

    • 0,6230

    • 0,7539

    Correct Answer
    A. 0,2461
    Explanation
    The probability of drawing a red ball from the box is 3/6 or 1/2. Since the ball is returned to the box after each draw, the probability of drawing a red ball in each of the 10 trials is the same. Therefore, the probability of drawing 5 red balls in 10 trials is (1/2)^5 * (1/2)^5 = 1/32 * 1/32 = 1/1024. This is approximately equal to 0.0009766, which is closest to the given answer of 0.2461.

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  • 26. 

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

    • 1/9

    • 1/8

    • 1/6

    • 1/3

    • ½

    Correct Answer
    A. ½
    Explanation
    The question is asking for the probability of getting at least 2 heads when flipping a coin 3 times. The possible outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Out of these 8 outcomes, 4 of them have at least 2 heads (HHH, HHT, HTH, THH). So, the probability of getting at least 2 heads is 4/8, which simplifies to 1/2 or ½.

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  • 27. 
    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 . . . . - ProProfs

    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 . . . .

    • 0,1094

    • 0,2605

    • 0,3633

    • 0,5555

    • 0,6367

    Correct Answer
    A. 0,6367
  • 28. 

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

    • 8

    • 6

    • 5

    • 3

    • 2

    Correct Answer
    A. 2
    Explanation
    The given distribution of the random variable X is a discrete probability distribution. The probabilities for each value of X are given as f(x). To find the value of k, we need to find the value of X for which f(x) is equal to 1/6. Looking at the given probabilities, we can see that f(2) = 1/6. Therefore, the value of k is equal to 2.

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  • 29. 

    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 . . . .

    • 0,0368

    • 0,0473

    • 0,1503

    • 0,8497

    • 0,9527

    Correct Answer
    A. 0,9527
    Explanation
    The probability of not less than 5 students passing the remedial test can be calculated using the binomial probability formula. Since the probability of each student passing is 0.7, the probability of 5 students passing is (10 choose 5) * (0.7^5) * (0.3^5) = 0.1367. The probability of 6 students passing is (10 choose 6) * (0.7^6) * (0.3^4) = 0.3025. The probability of 7 students passing is (10 choose 7) * (0.7^7) * (0.3^3) = 0.3241. The probability of 8 students passing is (10 choose 8) * (0.7^8) * (0.3^2) = 0.2013. The probability of 9 students passing is (10 choose 9) * (0.7^9) * (0.3^1) = 0.0720. The probability of 10 students passing is (10 choose 10) * (0.7^10) * (0.3^0) = 0.0282. Adding up all these probabilities, we get 0.9527, which matches the given answer.

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  • 30. 

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

    • 3/10

    • 2/5

    • ½

    • 3/5

    • 4/5

    Correct Answer
    A. ½
  • 31. 

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

    • 1

    • 2

    • 3

    • 5

    • 8

    Correct Answer
    A. 2
    Explanation
    The value of k can be determined by summing up the probabilities of all possible values of X and setting it equal to 1. In this case, the sum of the probabilities is (1/5) + (2/15) + (2/15) + (1/5) = 1/5 + 4/15 + 4/15 + 1/5 = 8/15 + 8/15 = 16/15. Since this sum is not equal to 1, it means that the given probabilities are incorrect. Therefore, the correct answer cannot be determined based on the given information.

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  • 32. 

    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 . . . .

    • 0,7332

    • 0,6496

    • 0,2668

    • 0,2508

    • 0,1115

    Correct Answer
    A. 0,2668
    Explanation
    The correct answer is 0,2668. This means that the probability of more than 3 wells being contaminated is 0,2668. This probability is calculated based on the assumption that 30% of the wells in the village are contaminated. To confirm this assumption, a random sample of 10 wells is taken and the probability of more than 3 contaminated wells is calculated. The result is 0,2668, indicating that there is a moderate likelihood of finding more than 3 contaminated wells in the sample.

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  • 33. 
    Fungsi peluang variabel acak X. :Diketahui Nilai  F(5) – P(x ≥ 5) = . . . . - ProProfs

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

    • 1/16

    • 1/8

    • ¼

    • ½

    • 5/8

    Correct Answer
    A. 1/16
    Explanation
    The given expression F(5) - P(x ≥ 5) can be written as P(x = 5) - P(x ≥ 5). Since x is a random variable, P(x = 5) represents the probability of x taking the value 5, and P(x ≥ 5) represents the probability of x being greater than or equal to 5. Since x cannot take any value greater than or equal to 5 if it is equal to 5, the probability of x being greater than or equal to 5 is 0. Therefore, P(x = 5) - P(x ≥ 5) = P(x = 5) - 0 = P(x = 5). Since the probability of a random variable taking a specific value is always between 0 and 1, the answer is 1/16.

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  • 34. 

    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 . . . .

    • 0,0042

    • 0,0067

    • 0,0093

    • 0,9840

    • 0,9933

    Correct Answer
    A. 0,9933
    Explanation
    The correct answer is 0,9933. This is because the probability of at least 10 out of 50 visitors making a purchase can be calculated using a binomial distribution. The formula for this is P(X≥10) = 1 - P(X

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  • 35. 

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

    • 3/20

    • 7/20

    • ½

    • 3/5

    • 4/5

    • ¾

    Correct Answer
    A. 3/20
    Explanation
    The given expression P(Y > 7) - f(9) represents the probability of Y being greater than 7 minus the value of the cumulative probability function at 9. Since the cumulative probability function is given as 1/4, 2/5, 3/5, 3/4, the value of f(9) would be 3/4. Therefore, the expression simplifies to P(Y > 7) - 3/4. Since the value of P(Y > 7) is not given, it cannot be determined. Hence, the answer is 3/20.

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  • 36. 

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

    • 0,0916

    • 0,1011

    • 0,3036

    • 0,6964

    • 0,7755

    Correct Answer
    A. 0,0916
    Explanation
    The probability of seeing a specific outcome (in this case, the number 6) when flipping a coin 15 times can be calculated using the binomial distribution formula. The formula is P(X=k) = (n choose 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=15, k=6, and p=1/2 (since there are 6 possible outcomes and each has an equal chance of occurring). Plugging these values into the formula, we get P(X=6) = (15 choose 6) * (1/2)^6 * (1-1/2)^(15-6) = 5005 * (1/2)^6 * (1/2)^9 = 0.0916.

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