📚 Exam-style practice: Paper 3 | 考试风格练习：卷三
This article provides a set of exam-style questions and detailed solutions tailored for Edexcel A-Level Mathematics Paper 3, covering Statistics and Mechanics. Each question is designed to reflect the style and difficulty of the actual examination, with step-by-step bilingual explanations to strengthen understanding and exam technique.
本文提供一套针对爱德思 A-Level 数学卷三（统计与力学）的考试风格题目及详细解答。每道题目均按照真实考试的题型与难度设计，附带逐步双语讲解，以加深理解并提升应试技巧。
1. Statistical Distribution and Expectation | 统计分布与期望
A discrete random variable X has the following probability distribution:
离散随机变量 X 的概率分布如下：
| x | 1 | 2 | 3 | 4 | 5 |
| P(X=x) | 1/10 | 1/5 | 1/5 | k | 2k |
Find the value of k, E(X), Var(X) and P(2 < X ≤ 5).
求 k 的值、E(X)、Var(X) 以及 P(2 < X ≤ 5)。
Solution | 解答
Step 1: Use Σ P(X=x) = 1.
步骤1：利用概率之和为1。
1/10 + 1/5 + 1/5 + k + 2k = 1
ons with denominator 10: 0.1 + 0.2 + 0.2 + 3k = 0.5 + 3k = 1, so 3k = 0.5, giving k = 1/6.
将分数通分为分母10：0.1 + 0.2 + 0.2 + 3k = 0.5 + 3k = 1，故 3k = 0.5，解得 k = 1/6。
Step 2: Compute E(X) = Σ x·P(X=x).
步骤2：计算 E(X) = Σ x·P(X=x)。
E(X) = 1×(1/10) + 2×(1/5) + 3×(1/5) + 4×(1/6) + 5×(2/6)
+ 0.4 + 0.6 + 4/6 + 10/6 = 1.1 + 14/6 = 1.1 + 7/3 ≈ 1.1 + 2.3333 = 3.4333 (exact: 103/30).
= 0.1 + 0.4 + 0.6 + 4/6 + 10/6 = 1.1 + 14/6 = 1.1 + 7/3 ≈ 1.1 + 2.3333 = 3.4333（精确值：103/30）。
Step 3: Find E(X²).
步骤3：求 E(X²)。
E(X²) = 1²×(1/10) + 2²×(1/5) + 3²×(1/5) + 4²×(1/6) + 5²×(2/6)
= 0.1 + 4/5 + 9/5 + 16/6 + 50/6 = 0.1 + 0.8 + 1.8 + 66/6 = 2.7 + 11 = 13.7.
Then Var(X) = E(X²) – [E(X)]² = 13.7 – (103/30)². Compute (103/30)² ≈ 11.8011, hence Var(X) ≈ 13.7 – 11.8011 = 1.8989 (exact: 137/10 – 10609/900 = simple form would be 611/900).
则 Var(X) = E(X²) – [E(X)]² = 13.7 – (103/30)²。计算 (103/30)² ≈ 11.8011，故 Var(X) ≈ 13.7 – 11.8011 = 1.8989（精确值：137/10 – 10609/900，化简为 611/900）。
Step 4: P(2 < X ≤ 5) = P(X=3) + P(X=4) + P(X=5) = 1/5 + 1/6 + 2/6 = 0.2 + 0.5 = 0.7.
步骤4：P(2 < X ≤ 5) = P(X=3)+P(X=4)+P(X=5) = 1/5 + 1/6 + 2/6 = 0.2 + 0.5 = 0.7。
2. Binomial Distribution | 二项分布
A biased coin is tossed 8 times. The probability of heads on each toss is 0.3. Find the probability of obtaining exactly 3 heads, and the probability of at least 6 heads.
抛掷一枚不均匀硬币8次。每次正面朝上的概率为0.3。求恰好出现3次正面的概率，以及至少6次正面的概率。
Solution | 解答
The number of heads X ~ B(8, 0.3). P(X=r) = ₈Cᵣ (0.3)ʳ (0.7)⁸⁻ʳ.
正面次数 X ~ B(8, 0.3)。P(X=r) = ₈Cᵣ (0.3)ʳ (0.7)⁸⁻ʳ。
, so P(X=3) = 56 × 0.3³ × 0.7⁵ = 56 × 0.027 × 0.16807 ≈ 56 × 0.00453789 = 0.2541 (3 s.f.).
当 r=3：₈C₃ = 56，故 P(X=3) = 56 × 0.3³ × 0.7⁵ = 56 × 0.027 × 0.16807 ≈ 56 × 0.00453789 = 0.2541（保留三位有效数字）。
For at least 6 heads: P(X≥6) = P(6) + P(7) + P(8).
至少6次正面：P(X≥6) = P(6)+P(7)+P(8)。
P(6) = ₈C₆ × 0.3⁶ × 0.7² = 28 × 0.000729 × 0.49 = 28 × 0.00035721 ≈ 0.01000.
P(7) = ₈C₇ × 0.3⁷ × 0.7 = 8 × 0.0002187 × 0.7 = 8 × 0.00015309 ≈ 0.001225.
P(8) = ₈C₈ × 0.3⁸ = 1 × 0.00006561 ≈ 0.0000656.
Sum ≈ 0.01129 (3 s.f.).
总和约为 0.0113（三位有效数字）。
3. Normal Distribution | 正态分布
The heights of adult males in a population are normally distributed with mean 175 cm and standard deviation 6 cm. Find the proportion of males taller than 182 cm. Also, find the symmetric interval around the mean that contains 95% of heights.
某群体成年男性身高服从正态分布，均值为 175 cm，标准差为 6 cm。求身高高于 182 cm 的比例，并求以均值为中心的包含 95% 身高的对称区间。
Solution | 解答
Let H ~ N(175, 6²). For 182 cm, standardise: Z = (182 – 175)/6 = 7/6 ≈ 1.1667.
设 H ~ N(175, 6²)。对于 182 cm，标准化：Z = (182 – 175)/6 = 7/6 ≈ 1.1667。
P(H > 182) = P(Z > 1.1667) = 1 – Φ(1.1667). From tables, Φ(1.17) ≈ 0.8790. So probability ≈ 0.1210.
P(H > 182) = P(Z > 1.1667) = 1 – Φ(1.1667)。查表得 Φ(1.17) ≈ 0.8790，故概率约为 0.1210。
For 95% symmetric interval: z-value for middle 95% is 1.96. So limits: μ ± 1.96σ = 175 ± 1.96×6 = [163.24 cm, 186.76 cm].
对于 95% 对称区间：中间 95% 对应的 z 值为 1.96。故区间为 μ ± 1.96σ = 175 ± 1.96×6 = [163.24 cm, 186.76 cm]。
4. Hypothesis Testing with Binomial | 二项分布假设检验
A seed supplier claims that the germination rate is at least 85%. In a trial, 18 out of 25 seeds germinate. Test this claim at the 5% significance level, stating your hypotheses, test statistic, critical region and conclusion clearly.
某种子供应商声称发芽率不低于 85%。在一次试验中，25 粒种子有 18 粒发芽。在 5% 显著性水平下检验该声明，并明确陈述假设、检验统计量、拒绝域及结论。
Solution | 解答
et p be the population germination proportion. H₀: p = 0.85, H₁: p < 0.85 (one-tailed).
设 p 为总体发芽比例。H₀: p = 0.85，H₁: p < 0.85（单侧）。
Under H₀, number germinating X ~ ). Test statistic: X=18. Lower tail critical region: smallest r such that P(X ≤ r) ≤ 0.05.
在 H₀ 下，发芽数 X ~ B(25, 0.85)。检验统计量：X=18。下尾拒绝域：寻找最小的 r 使得 P(X ≤ r) ≤ 0.05。
P(X ≤ 17) ≈ 0.013; P(X ≤ 18) ≈ 0.044; P(X ≤ 19) ≈ 0.111. So the critical region is X ≤ 18.
计算累积概率：P(X ≤ 15) 非常小；P(X ≤ 16) ≈ 0.003；P(X ≤ 17) ≈ 0.013；P(X ≤ 18) ≈ 0.044；P(X ≤ 19) ≈ 0.111。故拒绝域为 X ≤ 18。
Since the observed value 18 falls in the critical region, we reject H₀. There is sufficient evidence at the 5% level to suggest that the germination rate is less than 85%.
由于观测值 18 落在拒绝域内，拒绝 H₀。在 5% 显著性水平下有充分证据表明发芽率低于 85%。
5. Kinematics: SUVAT Equations | 运动学：SUVAT 方程
A cyclist accelerates uniformly from rest along a straight road, reaching a speed of 12 m s⁻¹ after 8 seconds. Find the acceleration and the distance covered during this period. The cyclist then maintains this speed for 20 seconds and finally decelerates uniformly to rest in 5 seconds. Calculate the total distance travelled.
一名自行车骑手从静止开始沿直线匀加速，8 秒后速度达到 12 m s⁻¹。求加速度及此期间行驶的距离。随后该骑手保持该速度匀速行驶 20 秒，最后在 5 秒内匀减速至静止。计算总路程。
Solution | 解答
g v = u + at: 12 = 0 + a×8 → a = 1.5 m s⁻². Distance s₁ = ut + ½at² = 0 + ½×1.5×64 = 48 m.
阶段一：初速度 u=0，末速度 v=12，时间 t=8。由 v = u + at 得：12 = 0 + a×8 → a = 1.5 m s⁻²。行驶距离 s₁ = ut + ½at² = 0 + ½×1.5×64 = 48 m。
Phase 2: constant speed 12 m s⁻¹ for 20 s → s₂ = 12×20 = 240 m.
阶段二：以 12 m s⁻¹ 匀速行驶 20 秒 → s₂ = 12×20 = 240 m。
Phase 3: deceleration to rest, u=12, v=0, t=5. From v = u + at: 0 = 12 + a×5 → a = –2.4 m s⁻². s₃ = ut + ½at² = 12×5 + ½×(–2.4)×25 = 60 – 30 = 30 m.
阶段三：匀减速至静止，u=12，v=0，t=5。由 v = u + at：0 = 12 + a×5 → a = –2.4 m s⁻²。s₃ = ut + ½at² = 12×5 + ½×(–2.4)×25 = 60 – 30 = 30 m。
Total distance = 48 + 240 + 30 = 318 m.
总路程 = 48 + 240 + 30 = 318 m。
6. Newton's Second Law and Friction | 牛顿第二定律与摩擦力
A block of mass 4 kg is pulled along a rough horizontal table by a rope inclined at 25° above the horizontal. The tension in the rope is 20 N, and the coefficient of friction between the block and the table is 0.3. Determine the acceleration of the block.
质量 4 kg 的方块放在粗糙水平桌面上，被一根与水平方向成 25° 向上的绳子拉动。绳中拉力为 20 N，方块与桌面间的动摩擦因数为
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