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CIE A-Level 数学 9709 Paper 6（Probability & Statistics 2）模拟卷｜75 分・1 小时 15 分
练习卷 · CIE A-Level 数学 9709 Paper 6（Probability & Statistics 2）
CIE A-Level 数学 9709 Paper 6（Probability & Statistics 2）模拟卷：9 题 75 分、1 小时 15 分，涵盖泊松分布、连续随机变数、指数分布、正态线性组合、不偏估计与假设检定，附完整答案与评分要点。
卷面资料 年级：A-Level ｜ 科目：数学 ｜ 时限：75 分钟 ｜ 总分：75 分 ｜ 题数：9
甲部 概率分布与连续随机变数（40 分）
全部题目必须作答，须展示步骤并注明所用公式；答案须准确至三位有效数字。
| # | 题目 | 答案与评分要点 |
|---|---|---|
| 1 | A helpline receives calls at random at an average rate of 3.2 calls per hour. The number of calls in any one hour is modelled by a Poisson distribution. (a) Find the probability that exactly 5 calls arrive in a randomly chosen hour. (b) Find the probability that at most 2 calls arrive in an hour. (c) Find the probability that at least 6 calls arrive in an hour. | (a) X ~ Po(3.2), so P(X = 5) = e^(−3.2) × 3.2^5 ÷ 5! = 0.1140. (b) P(X ≤ 2) = e^(−3.2)(1 + 3.2 + 3.2²÷2) = 0.3799. (c) P(X ≥ 6) = 1 − P(X ≤ 5) = 1 − 0.8946 = 0.1054. The Poisson model requires calls to occur singly, independently and at a constant average rate. [3, 3, 2] |
| 2 | A large box contains 400 components. The probability that any one component is defective is 0.012, independently of the others. (a) Write down the exact distribution of the number of defectives and state why a Poisson approximation is appropriate. (b) Use a Poisson approximation to find the probability that a box contains no defective component. (c) Estimate the probability that a box contains at least 5 defective components. | (a) X ~ B(400, 0.012) exactly; n = 400 is large and p = 0.012 is small with np = 4.8, so X is approximately Po(4.8). (b) P(X = 0) = e^(−4.8) = 0.0082. (c) P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.4763 = 0.5237. [2, 3, 3] |
| 3 | A continuous random variable X has probability density function f(x) = k(4 − x) for 0 ≤ x ≤ 4, and f(x) = 0 otherwise. (a) Show that k = 1/8. (b) Find E(X) and Var(X). (c) Find P(X less than 2) and explain whether the median of X is greater than 2. | (a) The area is k∫₀⁴(4 − x)dx = k(16 − 8) = 8k, so k = 1/8. (b) E(X) = (1/8)∫₀⁴(4x − x²)dx = (1/8)(32 − 64/3) = 1.3333; E(X²) = (1/8)(256/3 − 64) = 2.6667, so Var(X) = 0.8889 (8/9). (c) P(X less than 2) = (1/8)(8 − 2) = 0.75, which is less than 0.5, so the median lies above 2 because the density decreases with x. [2, 3, 3] |
| 4 | The waiting time T, in minutes, between consecutive customers at a counter is modelled by an exponential distribution with mean 5 minutes. (a) Find the probability that a customer waits more than 8 minutes. (b) Find the probability that a waiting time is between 2 and 6 minutes. (c) Find the median waiting time. | The density is f(t) = 0.2e^(−0.2t). (a) P(T > 8) = e^(−1.6) = 0.2019. (b) P(2 < T < 6) = e^(−0.4) − e^(−1.2) = 0.3691. (c) Solve 1 − e^(−0.2m) = 0.5, so m = 5 ln 2 = 3.4657 minutes. [3, 2, 3] |
| 5 | The masses of apples of variety A are normally distributed with mean 20 g and standard deviation 3 g. Variety B has masses normally distributed with mean 15 g and standard deviation 2 g. Two apples are chosen independently, one of each variety. (a) Find the probability that their total mass exceeds 40 g. (b) Find the probability that the mass of the A apple is less than the mass of the B apple plus 3 g. (c) Find the probability that 3A − 2B exceeds 35 g. | (a) A + B ~ N(35, 13), sd 3.6056; z = (40 − 35)/3.6056 = 1.3868, so P = 0.0828. (b) A − B ~ N(5, 13); z = (3 − 5)/3.6056 = -0.5547, so P = 0.2895. (c) 3A − 2B ~ N(30, 97), sd 9.8489; z = 0.5077, so P = 0.3058. [3, 3, 2] |
乙部 统计推断与假设检定（35 分）
全部题目必须作答；假设检定须写出 H₀、H₁、检定统计量及结论。
| # | 题目 | 答案与评分要点 |
|---|---|---|
| 6 | A random sample of 10 packets from a production line gave the following masses, in grams: 12.4, 11.8, 12.9, 12.1, 13.0, 12.6, 11.9, 12.3, 12.7, 12.5. (a) Calculate unbiased estimates of the population mean and variance. (b) State the estimate of the standard error of the sample mean if the population variance is known to be 0.16. (c) Hence find a 95% confidence interval for the population mean. | (a) x̄ = 124.2 ÷ 10 = 12.42 g; Σ(x − x̄)² = 1.4560, so s² = 1.4560 ÷ 9 = 0.16178 and s = 0.4022 g. (b) se = 0.4 ÷ √10 = 0.1265. (c) 12.42 ± 1.96(0.1265) = 12.42 ± 0.2479, giving (12.172, 12.668) g. [3, 3, 2] |
| 7 | Over a long period the number of faults reported on a machine averaged 2.0 per hour. After a service, a random sample of 30 one-hour periods contained 45 faults in total. Test at the 5% significance level whether the mean rate has fallen, assuming the number of faults per hour still has a Poisson distribution. State the distribution used and give the p-value. | H₀: λ = 2.0 per hour; H₁: λ < 2.0. Under H₀ the total in 30 hours is Po(60). The p-value is P(X ≤ 45) = 0.0266. The normal approximation with continuity correction gives z = (45 + 0.5 − 60) ÷ √60 = -1.8719 and Φ(z) = 0.0306. Since the p-value is less than 0.05, reject H₀: there is evidence that the fault rate has fallen. [3, 3, 3] |
| 8 | A machine fills bottles with a volume that is normally distributed with variance 36 ml². A sample of 25 bottles has mean volume 52.4 ml. It is claimed that the mean volume is 50 ml. (a) Test the claim against the alternative that the mean exceeds 50 ml, using a 5% significance level. (b) Find the critical region in terms of the sample mean. (c) Find the probability of a Type II error when the true mean is 53 ml. | (a) H₀: μ = 50; H₁: μ > 50. se = 6 ÷ √25 = 1.2; z = (52.4 − 50) ÷ 1.2 = 2.0000. Since 2.0000 exceeds 1.6449, reject H₀. (b) Reject when x̄ exceeds 50 + 1.6449(1.2) = 51.9738 ml. (c) β = P(x̄ less than 51.9738 | μ = 53) = Φ((51.9738 − 53) ÷ 1.2) = Φ(-0.8551) = 0.1962, so the power is 0.8038. [3, 3, 3] |
| 9 | The number of accidents at a junction in a month has a Poisson distribution with mean 50. (a) Use a normal approximation with a continuity correction to estimate the probability of at most 42 accidents in a month. (b) Estimate the probability of between 45 and 55 accidents inclusive. (c) State one condition for the approximation to be reliable and comment on whether it is met here. | X ~ Po(50) has mean 50 and sd √50 = 7.0711. (a) P(X ≤ 42) ≈ Φ((42.5 − 50) ÷ 7.0711) = Φ(-1.0607) = 0.1444. (b) P(45 ≤ X ≤ 55) ≈ Φ(0.7778) − Φ(-0.7778) = 0.5633. (c) The approximation needs λ to be large (in practice at least about 10–15); here λ = 50 and the distribution is only slightly skew, so the approximation is acceptable although the continuity correction is essential. [3, 3, 3] |
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