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Quiz Chapter 5: Probability Distribution

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

10 questions · plays in fullscreen

Full Question List & Answer Key

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

1. 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.

2. 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.

3. 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.

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

  1. 0.9772
  2. 0.0228
  3. 0.4772
  4. 0.9544
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Answer: A

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

5. 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.

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. For X ~ N(160, 100), find P(X < 150).

  1. 0.1587
  2. 0.8413
  3. 0.3413
  4. 0.0228
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Answer: A

μ = 160, σ = √100 = 10. Z = 150 - 16010 = -1.0. P(X < 150) = P(Z < -1.0) = P(Z > 1.0) = 0.1587.

8. 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%.

9. 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.

10. In a manufacturing factory, 10% of items produced are defective. If 8 items are selected at random, find the probability that exactly 1 item is defective.

  1. 0.3826
  2. 0.4305
  3. 0.1000
  4. 0.2799
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Answer: A

X ~ B(8, 0.10). P(X = 1) = ⁸C₁ (0.10)¹ (0.90)⁷ = 8 × 0.10 × 0.478297 = 0.382637 ≈ 0.3826.

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