📚 Year 9 OCR Statistics: UK University Admission Requirements Comparison | 英国大学申请要求统计对照
In Year 9 OCR Statistics, one of the most engaging applications of descriptive statistics is to compare university entrance requirements across the UK. By collecting real data on A‑level offers, converting them into UCAS Tariff points and using tools such as box plots, mean, median and interquartile range, GCSE students can see how statistics brings clarity to an important life decision. This article guides you through a complete investigation, from data collection to interpretation, using only the statistical techniques taught at this stage.
在九年级OCR统计课程中，将描述性统计用于比较英国各大学的入学要求，是一项非常吸引人的实践活动。通过收集真实的A‑level录取成绩数据、将其转换为UCAS积分，并运用箱线图、平均数、中位数和四分位距等工具，GCSE学生可以亲眼见证统计学如何为一个重要的人生决策带来清晰的分析。本文从数据收集到结果解读，带领你完成一次完整的统计调查，全程只用到本阶段所学的统计方法。
1. Understanding UCAS Tariff Points | 了解UCAS积分体系
compare offers, we need a common numerical scale. In the UK, UCAS Tariff points convert A‑ equals 56 + 48 + 48 = 152 points. This system allows us to treat otherwise qualitative grades as quantitative data, making statistical analysis possible.
在对录取要求进行比较之前，我们需要一个统一的数值尺度。英国的UCAS积分体系将A‑level等级转换为数字：A*等于56分，A等于48分，B等于40分，C等于32分，D等于24分，E等于16分。因此，A*AA这样的录取条件就相当于56 + 48 + 48 = 152分。这一体系使我们能够将原本的定性等级转化为定量数据，从而进行统计分析。
40 points C = 32 points D = 24 points E = 16 points
A* = 56 分 A = 48 分 B = 40 分 C = 32 分 D = 24 分 E = 16 分
For most university courses, applicants take three A‑ f is 3 × 56 = 168 points. The tariff points act as the raw data for our comparative study.
对于大多数大学课程，申请人会修读三门A‑level，因此最高积分为 3 × 56 = 168 分。这些积分就构成了我们比较研究的原始数据。
2. Data Collection: A‑Level Requirements for a Mathematics Degree | 数据收集：数学专业的A‑level要求
Imagine we are investigating entry requirements for a typical Mathematics BSc at ten different UK universities. We visit their official websites or the UCAS course search and note down the standard A‑level offers for September 2025 entry. The data is shown in the table below. Remember that actual offers may vary by background and context, but these are widely accepted typical grades.
假设我们正在调查英国十所不同大学数学理学学士课程的入学要求。我们浏览各校的官网或UCAS课程搜索页面，记录下2025年9月入学的标准A‑level录取条件。实际录取可能因个人背景和情况而有所不同，但以下数据是普遍认可的典型要求。
| University | Typical A‑Level Offer | UCAS Tariff Points |
|---|---|---|
| Oxford | A*A*A | 160 |
| Cambridge | A*A*A | 160 |
| Imperial College London | A*AA | 152 |
| LSE | A*AA | 152 |
| Warwick | A*AA | 152 |
| UCL | AAA | 144 |
| Manchester | AAA | 144 |
| Edinburgh | AAA | 144 |
| Birmingham | AAB | 136 |
| Leeds | AAB | 136 |
We now have ten data points: 160, 160, 152, 152, 152, 144, 144, 144, 136, 136. This dataset is ready for statistical analysis.
现在我们有了十个数据点：160, 160, 152, 152, 152, 144, 144, 144, 136, 136。这个数据集已经准备好用于统计分析。
3. Organising the Data: A Stem‑and‑Leaf Diagram | 数据整理：茎叶图
A stem‑and‑leaf diagram helps us see the shape of the data while preserving every original value. We choose stems for the hundreds and tens digits, and leaves for the units. For example, 160 is written as 16 | 0. The ordered diagram for our dataset is shown below.
茎叶图能让我们看到数据的分布形状，同时保留每一个原始数值。我们选取百位和十位数字作为茎，个位数字作为叶。例如，160 写作 16 | 0。我们按顺序排列的茎叶图如下。
Stem | Leaf 13 | 6 6 14 | 4 4 4 15 | 2 2 2 16 | 0 0 Key: 13 | 6 means 136 tariff points
茎 | 叶 13 | 6 6 14 | 4 4 4 15 | 2 2 2 16 | 0 0 图例: 13 | 6 表示 136 个积分
From the diagram we can immediately see that the most frequent tariff band is in the 150s, and the distribution has two peaks at 144–152 and at 160. The stem‑and‑leaf also reveals potential gaps: for instance, no university in this sample requires exactly 148 points.
从图中我们可以立即看出，最常见的积分区间集中在150分段，并且数据在144–152和160处呈现两个峰值。茎叶图还揭示了可能存在的空缺：例如，在这个样本中，没有一所大学的录取要求恰好是148分。
4. The Average Entry Requirement: Mean and Median | 平均入学要求：均值与中位数
versities. Adding our ten values gives 1480 points, so the mean is 1480 ÷ 10 = 148 points.
算术平均值通过将所有积分求和，再除以大学的数量得到。十个数据的总和为1480分，因此均值等于 1480 ÷ 10 = 148 分。
Mean = ∑x ÷ n = 1480 ÷ 10 = 148 points
均值 = ∑x ÷ n = 1480 ÷ 10 = 148 分
es. The 5th value is 144, the 6th is 152, so the median is (144 + 152) ÷ 2 = 148 points. Interestingly, the mean and the median are equal in this dataset, indicating the distribution is roughly symmetric.
为了找出中位数，我们先将数据排序，然后确定中间位置。由于共有十个数据（偶数个），中位数落在第5个和第6个值的中间。第5个值是144，第6个是152，因此中位数 = (144 + 152) ÷ 2 = 148 分。有趣的是，该数据集的均值和中位数相等，这表明分布大致是对称的。
5. Measures of Spread: Range and Interquartile Range (IQR) | 离散程度的度量：极差与四分位距
The range is the simplest measure of spread: Maximum – Minimum = 160 – lues. A more robust measure is the interquartile range (IQR), which captures the middle 50% of the data.
极差是最简单的离散度量：最大值减去最小值 = 160 – 136 = 24分。但极差容易受极端值的影响。一个更稳健的度量是四分位距（IQR），它反映了中间50%数据的分散程度。
To calculate the IQR we need Q₁ (lower quartile) and Q₃ (upper quartile). Using the ten ordered values:
要计算IQR，我们需要求出下四分位数Q₁和上四分位数Q₃。利用排序后的十个数据：
Position of Q₁ = (10 + 1) ÷ 4 = 2.75, so Q₁ = 2nd value + 0.75 × (3rd – 2nd) = 136 + 0.75 × (144 – 136) = 136 + 6 = 142.
Position of Q₃ = 3 × (10 + 1) ÷ 4 = 8.25, so Q₃ = 8th value + 0.25 × (9th – 8th) = 152 + 0.25 × (160 – 152) = 152 + 2 = 154.
Q₁的位置 = (10 + 1) ÷ 4 = 2.75，因此 Q₁ = 第2个值 + 0.75 × (第3个 – 第2个) = 136 + 0.75 × (144 – 136) = 136 + 6 = 142。
Q₃的位置 = 3 × (10 + 1) ÷ 4 = 8.25，因此 Q₃ = 第8个值 + 0.25 × (第9个 – 第8个) = 152 + 0.25 × (160 – 152) = 152 + 2 = 154。
IQR = Q₃ – Q₁ = 154 – 142 = 12 points
IQR = Q₃ – Q₁ = 154 – 142 = 12 分
An IQR of 12 points tells us that the central half of universities have tariff requirements clustered within a fairly narrow band around the median.
12分的四分位距告诉我们，中间一半大学的积分要求紧密地聚集在中位数附近的一个较窄区间内。
6. Box Plot: Visual Summary of the Five‑Number Summary | 箱线图：五数概括的可视化
The five‑ Q₁ = 142, median = 148, Q₃ = 154, max = 160. We can sketch a box plot using a simple text‑based representation.
五数概括（最小值、Q₁、中位数、Q₃、最大值）能够极好地概述数据的特征。对于我们的样本：最小值 = 136，Q₁ = 142，中位数 = 148，Q₃ = 154，最大值 = 160。我们可以用简单的文本方式勾勒出箱线图。
Min Q₁ Med Q₃ Max |------[ | ]------| 136 142 148 154 160
最小值 Q₁ 中位数 Q₃ 最大值 |------[ | ]------| 136 142 148
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