10 questions · Form 5 Additional Mathematics Bab 5: Probability Distribution
Given Z ~ N(0, 1), evaluate P(-1.5 < Z < 1.5).
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. Given Z ~ N(0, 1), evaluate P(-1.5 < Z < 1.5).
Answer: A
P(-1.5 < Z < 1.5) = 1 - 2 × P(Z > 1.5) = 1 - 2(0.0668) = 1 - 0.1336 = 0.8664.
2. If X ~ B(6, 0.3), calculate P(X = 2).
Answer: A
P(X = 2) = ⁶C₂ (0.3)² (0.7)⁴ = 15 × 0.09 × 0.2401 = 0.324135 ≈ 0.3241.
3. Given Z ~ N(0, 1) and P(Z < k) = 0.8413, find k.
Answer: A
P(Z < k) = 0.8413 => P(Z > k) = 1 - 0.8413 = 0.1587. From normal distribution tables, Z = 1.00.
4. 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.
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.
5. A continuous random variable X is normally distributed with mean 50 and standard deviation 5. Find the Z-score when X = 62.5.
Answer: A
Z = (X - μ) / σ = 62.5 - 505 = 12.55 = 2.5.
6. For a standard normal variable Z ~ N(0, 1), if P(Z > k) = 0.1587, find the value of k.
Answer: A
From the standard normal distribution table, P(Z > 1.00) = 0.1587. Thus, k = 1.00.
7. 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.
Answer: A
P(X > M) = 0.1587 => P(Z > M - 6012) = 0.1587. From tables, Z = 1.00. M - 6012 = 1.00 => M = 72.
8. If X ~ B(4, 0.5), calculate P(X = 2).
Answer: A
P(X = 2) = ⁴C₂ (0.5)² (0.5)² = 6 × 0.25 × 0.25 = 0.375.
9. If X ~ B(n, p) where mean = 15 and standard deviation = 3, find the probability of success p.
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.
10. If X ~ B(5, 0.4), find P(X ≥ 1).
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.