📚 A Case Study on Study Time and Exam Performance | 学习时间与考试成绩案例分析实战演练
Imagine you are a Year 12 student exploring the link between the number of hours spent revising per week and the final exam score in mathematics. You collect data from a random sample of 50 students at your school, recording their weekly study hours (to the nearest half hour) and their percentage scores in the exam. This case study walks you through the entire statistical investigation — from data collection and descriptive statistics to probability models, hypothesis testing, and regression analysis — applying the core concepts of the OCR Year 12 Statistics syllabus.
假设你是一名12年级学生，正在探索每周数学复习时间与期末考试成绩之间的关系。你从学校随机抽取了50名学生作为样本，记录下他们每周的学习时长（精确到半小时）和考试百分制成绩。本案例将带你走完一次完整的统计调查过程——从数据收集、描述性统计，到概率模型、假设检验和回归分析——应用 OCR 12年级统计课程的核心概念。
1. Data Collection and Sampling | 数据收集与抽样
We used a simple random sample of 50 Year 12 students. Each student was asked to report their average weekly study time for mathematics over the term. Their final exam score was obtained from the school records with permission. The sample size (n = 50) is large enough for the Central Limit Theorem to apply, and we assume the data are independent and representative.
我们使用了一个包含50名12年级学生的简单随机样本。要求每名学生报告他们在该学期内每周的数学平均学习时长。他们的期末考试成绩经允许从学校记录中获取。样本量（n = 50）足够大，中心极限定理可以适用，并且我们假定数据是独立且具有代表性的。
2. Descriptive Statistics: Central Tendency and Spread | 描述性统计：集中趋势与离散程度
Let X denote weekly study hours and Y denote the exam score (%). The summary statistics computed from the sample are as follows:
设 X 表示每周学习小时数，Y 表示考试分数（百分比）。由样本计算出的汇总统计量如下：
| Statistic | Study Hours (X) | Exam Score (Y) |
|---|---|---|
| Mean | x̄ = 9.4 h | ȳ = 62.5% |
| Standard deviation | sx = 3.2 h | sy = 12.1% |
| Median | 9.0 h | 63% |
| IQR | 4.5 h | 17% |
The mean study time is 9.4 hours per week, with moderate variability (standard deviation 3.2 h). The exam scores average 62.5%, and the standard deviation of 12.1% indicates a wide spread of performance. The medians are close to the means, suggesting roughly symmetric distributions.
平均学习时间为每周9.4小时，具有中等变异性（标准差3.2小时）。考试平均分为62.5%，标准差12.1%表明成绩分布较广。中位数与均值接近，提示分布大致对称。
3. Visualising Data: Histograms and Box Plots | 数据可视化：直方图与箱线图
We construct histograms for both variables. The histogram of study hours shows a slight right skew, with most students studying between 5 and 14 hours. The exam score histogram is approximately bell-shaped. A side-by-side box plot comparing study hours for students who passed (score ≥ 50%) and those who failed reveals that the pass group has a higher median and smaller IQR.
我们为两个变量分别绘制直方图。学习时长的直方图呈轻微右偏态，多数学生的学习时间在5至14小时之间。考试分数的直方图近似钟形。将及格学生（分数≥50%）与不及格学生的学习时长做并列箱线图，可以看出及格组的学习时间中位数更高，且四分位距更小。
4. Probability Distributions: Modelling Exam Pass Rates | 概率分布：建模考试通过率
Suppose the probability that a randomly chosen student passes the exam (score ≥ 50%) is estimated as p = 0.76 from past records. We can model the number of passes in a sample of 20 students using the binomial distribution B(20, 0.76).
假设根据过往记录，随机抽取一名学生通过考试（分数≥50%）的概率估计为 p = 0.76。我们可以用二项分布 B(20, 0.76) 来对20名学生样本中的通过人数进行建模。
Let R ~ B(20, 0.76). Then P(R = r) = 20Cr × (0.76)r × (0.24)20−r.
设 R ~ B(20, 0.76)，则 P(R = r) = 20Cr × (0.76)r × (0.24)20−r。
5. Binomial Distribution: Probability of a Certain Number of Passes | 二项分布：一定数量学生通过的概率
Calculate the probability that exactly 15 out of 20 students pass the exam. Using the formula:
计算20名学生中恰好有15人通过考试的概率。使用公式：
P(R = 15) = C(20,15) × (0.76)15 × (0.24)5 ≈ 0.202
This means there is about a 20.2% chance of observing exactly 15 passes in a random sample of 20 students if the underlying pass rate is 76%.
这意味着，如果真实的通过率是76%，那么在20名学生的随机样本中，观察到恰好15人通过的概率约为20.2%。
We could also find the probability of at least 17 passes, which may indicate an unusually high-performing group.
我们也可以计算至少17人通过的概率，这可能暗示着一个表现异常优异的群体。
6. Normal Distribution: Approximating Scores | 正态分布：近似成绩分布
The exam scores (Y) are assumed to follow a normal distribution N(μ, σ²) based on the histogram shape. Using the sample estimates, we can approximate Y ~ N(62.5, 12.1²). This allows us to calculate probabilities, such as the proportion of students scoring above 80%.
根据直方图形状，假设考试成绩 Y 遵循正态分布 N(μ, σ²)。利用样本估计，我们可以近似 Y ~ N(62.5, 12.1²)。这使我们能够计算概率，例如得分超过80%的学生比例。
Standardizing: Z = (80 − 62.5) / 12.1 ≈ 1.446. From normal tables, P(Z > 1.446) ≈ 0.074, so about 7.4% of students score above 80%.
标准化：Z = (80 − 62.5) / 12.1 ≈ 1.446。查正态分布表，P(Z > 1.446) ≈ 0.074，因此大约7.4%的学生得分高于80%。
We can also find the score that corresponds to the top 10% — the 90th percentile — by solving z = 1.2816, so the score is μ + z
Published by TutorHao | Year 12 统计 Revision Series | aleveler.com
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