📚 Case Study Practice for Year 12 OCR Maths | Year 12 OCR 数学：案例分析实战演练
Case studies in A Level Mathematics bring together pure maths, mechanics and statistics in real-world contexts. This article walks you through a complete case study that integrates differentiation and binomial hypothesis testing, helping Year 12 OCR students sharpen their problem-solving skills and exam technique.
A Level 数学中的案例分析将纯数学、力学和统计学融入真实情境。本文通过一个完整的案例演练，融合微分法与二项假设检验，帮助 Year 12 OCR 学生提升解题技巧与应试能力。
1. Introducing the Case: Premium Headphones | 案例介绍：高端耳机
A technology company is launching a new pair of premium headphones. The marketing team has estimated a linear monthly demand function, and the finance department has provided detailed cost figures. In addition, the quality control team has tested a sample from the production line to check the defect rate. You will use differentiation to find the selling price that maximises profit, and then apply a binomial hypothesis test to decide whether the defect rate exceeds the acceptable level.
一家科技公司正推出一款全新高端耳机。市场团队估算了一条线性月需求函数，财务部门也给出了详细成本数据。此外，质控团队从生产线上抽取样本检测缺陷率。你将使用微分法找出利润最大化的售价，再运用二项假设检验判断缺陷率是否超出可接受水平。
2. Demand and Revenue Modelling | 需求与收益建模
Market research suggests that the monthly quantity demanded q (pairs) is related to the price p (£) by the linear equation:
市场调研表明，月需求量 q（副）与价格 p（英镑）满足如下线性方程：
q = 3000 − 50p, p ≥ 0, q ≥ 0
Total revenue R (£) is price multiplied by quantity: R = p × q = p(3000 − 50p) = 3000p − 50p².
总收益 R（英镑）为价格与数量的乘积：R = p × q = p(3000 − 50p) = 3000p − 50p²。
A few price–demand–revenue combinations are shown below to illustrate how revenue first rises and then falls.
下表列出几组价格-需求-收益组合，说明收益先升后降的规律。
| Price, p (£) | Demand, q | Revenue, R (£) |
|---|---|---|
| 10 | 2500 | 25 000 |
| 20 | 2000 | 40 000 |
| 30 | 1500 | 45 000 |
| 40 | 1000 | 40 000 |
| 50 | 500 | 25 000 |
3. Cost Structure | 成本结构
The production cost is £10 per pair, and the fixed monthly costs (rent, salaries, marketing) total £10 000. Hence the total cost C (£) is:
每副耳机的生产成本为 10 英镑，固定月成本（租金、工资、营销）总计 10 000 英镑。因此总成本 C（英镑）为：
C = 10q + 10 000
Substituting the demand function q = 3000 − 50p gives:
代入需求函数 q = 3000 − 50p 得到：
C = 10(3000 − 50p) + 10 000 = 40 000 − 500p
Notice that cost decreases as price rises, because a higher price reduces demand and therefore the variable cost.
请注意，由于提价导致需求减少，变动成本下降，总成本也随之降低。
4. Profit Function and Optimisation | 利润函数与最优化
Monthly profit L (£) is revenue minus cost:
月利润 L（英镑）为收益减去成本：
L = R − C = (3000p − 50p²) − (40 000 − 500p) = −50p² + 3500p − 40 000
This is a quadratic with a negative coefficient of p², so the graph is an inverted parabola with a maximum at the vertex. To find the price that maximises profit, differentiate L with respect to p and set the derivative equal to zero:
这是一个 p² 系数为负的二次函数，图像为开口向下的抛物线，其顶点对应最大利润。为求出最大化利润的价格，对 p 求导并令导数为零：
dL/dp = −100p + 3500 = 0  → p = 35
5. Verifying the Maximum | 验证最大值
The second derivative is d²L/dp² = −100, which is negative, confirming that p = 35 yields a maximum profit. Substituting p = 35 back into the relevant equations gives:
二阶导数 d²L/dp² = −100 为负，确认 p = 35 对应极大值（即最大利润）。将 p = 35 代回相关方程得到：
- Demand: q = 3000 − 50 × 35 = 1250 pairs
- Revenue: R = 3000 × 35 − 50 × 35² = £61 250
- Cost: C = 40 000 − 500 × 35 = £22 500
- Maximum profit: L = 61 250 − 22 500 = £21 250
q = 1250 副，R = £61 250，C = £22 500，最大利润 L = £21 250。
6. Quality Control: Setting Up the Problem | 质量控制：问题设定
The company claims that the proportion of defective headphones is at most 5% (0.05). To test this claim, a random sample of 50 headphones is taken from the production line and inspected. Four headphones are found to be defective. We will perform a hypothesis test at the 5% significance level to determine whether there is sufficient evidence that the true defect rate exceeds 5%.
该公司声称耳机的缺陷率不超过 5%（0.05）。为检验此声称，从生产线随机抽取 50 副耳机进行检测，发现 4 副有缺陷。我们将在 5% 显著性水平下进行假设检验，以判断是否有充分证据表明真实缺陷率超过 5%。
7. Defining the Binomial Model | 定义二项分布模型
Let X be the number of defective headphones in a sample of 50. Under the null hypothesis that the defect probability p = 0.05, X follows a binomial distribution:
设 X 为 50 副样本中缺陷耳机的数量。在原假设（缺陷概率 p = 0.05）下，X 服从二项分布：
X ~ B(50, 0.05)
The observed test statistic is x appropriate because the company is concerned the defect rate might be higher than claimed.
观测到的检验统计量为 x = 4。由于公司关心缺陷率是否高于声称值，应使用单尾检验。
8. Stating the Hypotheses and Significance Level | 陈述假设与显著性水平
Null hypothesis H₀: p = 0.05
Alternative hypothesis H₁: p > 0.05
Significance level: α = 0.05
原假设 H₀：p = 0.05；备择假设 H₁：p > 0.05；显著性水平 α = 0.05。
If the probability of obtaining 4 or more defectives, assuming H₀ is true, is less than 0.05, we will reject H₀.
若在原假设成立的条件下，得到 4 个或更多缺陷品的概率小于 0.05，我们将拒绝 H₀。
9. Calculating the p-value and Making a Decision | 计算 p-值并做出决策
We need P(X ≥ 4) = 1 − P(X ≤ 3). Using the binomial cumulative distribution for n = 50, p = 0.05:
我们需要计算 P(X ≥ 4) = 1 − P(X ≤ 3)。使用 n = 50, p = 0.05 的二项分布累积概率表：
| x | P(X ≤ x) |
|---|---|
| 0 | 0.0769 |
| 1 | 0.2794 |
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