10 questions · Form 5 Additional Mathematics Bab 5: Probability Distribution
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Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. A discrete random variable X has probability distribution P(X = x) = kx for x = 1, 2, 3, 4. Find the value of k.
Answer: A
Sum of probabilities = 1 => k(1) + k(2) + k(3) + k(4) = 1 => 10k = 1 => k = 110.
2. Given that X ~ B(n, p) has mean = 12 and variance = 3. Find the values of n and p.
Answer: A
np = 12 and npq = 3. Substituting np = 12 into npq gives 12q = 3 => q = 0.25. So p = 1 - 0.25 = 0.75. Then n(0.75) = 12 => n = 16.
3. The total area under a standard normal probability distribution curve is equal to:
Answer: A
By definition, the total area under any probability density curve, including the normal curve, is always 1.
4. For X ~ N(160, 100), find P(X < 150).
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.
5. 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.
6. 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.
7. A random variable X is binomially distributed with X ~ B(100, 0.2). Find the mean and variance of X.
Answer: A
Mean = np = 100 × 0.2 = 20. Variance = npq = 100 × 0.2 × 0.8 = 16.
8. 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.
9. Find P(Z > -2.0) for Z ~ N(0, 1).
Answer: A
P(Z > -2.0) = P(Z < 2.0) = 1 - P(Z > 2.0) = 1 - 0.0228 = 0.9772.
10. 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.