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Quiz Chapter 5: Probability Distribution

10 questions · Form 5 Additional Mathematics Bab 5: Probability Distribution

Question 1 of 10Score: 0

Which of the following is NOT a condition for a binomial distribution?

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. Which of the following is NOT a condition for a binomial distribution?

  1. The trials are dependent on each other
  2. There are a fixed number of trials
  3. The probability of success remains constant
  4. Each trial has only two possible outcomes
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Answer: A

A key condition of a binomial distribution is that all trials must be strictly independent.

2. If X ~ B(6, 0.3), calculate P(X = 2).

  1. 0.3241
  2. 0.1852
  3. 0.2430
  4. 0.4211
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Answer: A

P(X = 2) = ⁶C₂ (0.3)² (0.7)⁴ = 15 × 0.09 × 0.2401 = 0.324135 ≈ 0.3241.

3. If X ~ B(5, 0.4), find P(X ≥ 1).

  1. 0.9222
  2. 0.0778
  3. 0.6723
  4. 0.8208
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Answer: A

P(X ≥ 1) = 1 - P(X = 0) = 1 - ⁵C₀ (0.4)⁰ (0.6)⁵ = 1 - 1(1)(0.07776) = 1 - 0.07776 = 0.92224 ≈ 0.9222.

4. A discrete random variable X has probability distribution P(X = x) = kx for x = 1, 2, 3, 4. Find the value of k.

  1. 110
  2. 14
  3. 15
  4. 120
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Answer: A

Sum of probabilities = 1 => k(1) + k(2) + k(3) + k(4) = 1 => 10k = 1 => k = 110.

5. The marks of a group of students follow a normal distribution with a mean of 60 and a standard deviation of 12. If 15.87% of the students scored more than M marks, find M.

  1. 72
  2. 68
  3. 84
  4. 75
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Answer: A

P(X > M) = 0.1587 => P(Z > M - 6012) = 0.1587. From tables, Z = 1.00. M - 6012 = 1.00 => M = 72.

6. If X ~ B(n, p) where mean = 15 and standard deviation = 3, find the probability of success p.

  1. 0.4
  2. 0.6
  3. 0.2
  4. 0.8
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Answer: A

Mean np = 15. Standard deviation √(npq) = 3 => npq = 9. Substituting np = 15 gives 15q = 9 => q = 0.6. Thus, p = 1 - 0.6 = 0.4.

7. In a town, the masses of residents are normally distributed with mean 65 kg and variance 25 kg². Find the percentage of residents with masses between 60 kg and 70 kg.

  1. 68.26%
  2. 95.44%
  3. 50.00%
  4. 34.13%
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Answer: A

μ = 65, σ = √25 = 5. For X = 60, Z1 = 60 - 655 = -1.0. For X = 70, Z2 = 70 - 655 = 1.0. P(-1.0 < Z < 1.0) = 1 - 2(0.1587) = 0.6826 = 68.26%.

8. A random variable X is binomially distributed with X ~ B(100, 0.2). Find the mean and variance of X.

  1. Mean = 20, Variance = 16
  2. Mean = 20, Variance = 4
  3. Mean = 80, Variance = 16
  4. Mean = 10, Variance = 16
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Answer: A

Mean = np = 100 × 0.2 = 20. Variance = npq = 100 × 0.2 × 0.8 = 16.

9. In a fair die roll repeated 180 times, what is the standard deviation of obtaining the number '6'?

  1. 5
  2. 25
  3. 30
  4. 6
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Answer: A

n = 180, p = 16, q = 56. Variance = npq = 180 × (16) × (56) = 25. Standard deviation σ = √25 = 5.

10. Find P(Z > -2.0) for Z ~ N(0, 1).

  1. 0.9772
  2. 0.0228
  3. 0.4772
  4. 0.9544
Show answer

Answer: A

P(Z > -2.0) = P(Z < 2.0) = 1 - P(Z > 2.0) = 1 - 0.0228 = 0.9772.

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