📚 Comparing UK University Entry Requirements: A Year 7 Statistics Investigation | 英国大学申请要求对照:七年级统计探究
When you think about studying at a UK university, you might imagine high entry requirements. But how can we compare different universities’ demands? In this Year 7 statistics investigation, we will collect data on typical A-level entry requirements (converted to UCAS Tariff points) for a range of universities and use statistical methods to analyse them. By calculating the mean, median, mode, and range, and creating charts, we can discover which universities are the most competitive and whether certain subject areas have higher thresholds. This project shows how statistics can help us understand real-world information and make informed comparisons.
当你设想在英国上大学时,你可能会想到很高的入学要求。但我们如何比较不同大学的要求呢?在这项七年级统计探究中,我们将收集一系列大学典型的A-level入学要求(转换为UCAS关税分数)数据,并使用统计方法进行分析。通过计算平均数、中位数、众数和范围,并绘制图表,我们可以发现哪些大学最具竞争力,以及某些学科领域是否有更高的门槛。这个项目展示了统计学如何帮助我们理解真实世界的信息并进行明智的比较。
1. Data Collection: Gathering University Requirements | 数据收集:收集大学要求
We selected ten well-known UK universities and recorded their typical A-level entry requirements for science-related courses (such as Computer Science or Mathematics) and humanities courses (such as History). To make the requirements comparable, we converted the A-level grades into UCAS Tariff points using the standard conversion: A* = 56, A = 48, B = 40, C = 32, D = 24, E = 16. For example, a requirement of A*AA becomes 56 + 48 + 48 = 152 points. Data was gathered from official university websites and UCAS course listings, ensuring we used the same entry year where possible.
我们选择了十所英国知名大学,并记录了它们对科学类课程(如计算机科学或数学)和人文学科课程(如历史)的典型A-level入学要求。为了使要求具有可比性,我们使用标准换算将A-level成绩转换为UCAS关税分数:A* = 56, A = 48, B = 40, C = 32, D = 24, E = 16。例如,A*AA的要求变为56 + 48 + 48 = 152分。数据来源于大学官网和UCAS课程列表,我们尽可能使用了相同的入学年份。
2. Organising the Data in a Table | 用表格整理数据
We recorded the tariff points for each university in both subject groups. A well-organised table allows us to see all the data clearly and spot patterns. Below is the data we collected.
我们记录了每所大学在两个学科组中的关税分数。一张整理得当的表格能让我们清楚地看到所有数据并发现模式。以下是我们收集的数据。
| University | Science Points | Humanities Points |
|---|---|---|
| Oxford | 152 | 152 |
| Cambridge | 160 | 152 |
| UCL | 152 | 144 |
| Edinburgh | 144 | 128 |
| Manchester | 144 | 128 |
| Bristol | 152 | 144 |
| Warwick | 136 | 136 |
| Leeds | 128 | 128 |
| Sheffield | 136 | 120 |
| Nottingham | 144 | 136 |
3. Drawing a Bar Chart | 绘制条形图
To visualise the Science tariff points, we can draw a bar chart. On the horizontal axis we place the university names, and on the vertical axis the tariff score. Each bar’s height represents the points for that university. From the chart, Cambridge stands out with the tallest bar at 160 points, while Leeds has the shortest at 128. A bar chart makes it easy to compare values at a glance.
为了将科学课程关税分数可视化,我们可以绘制一个条形图。在横轴上放置大学名称,纵轴为关税分数。每个条形的高度代表该大学的分数。从图表中可以看到,剑桥大学的条形最高,为160分,而利兹大学最低,为128分。条形图让人一目了然地比较数值。
4. Calculating the Mean (Average) | 计算平均数
The mean gives us a typical tariff score for science courses. We add all the science points and divide by the number of universities (10). The sum is:
平均数给出了科学课程典型的关税分数。我们将所有科学课程分数相加,再除以大学数量(10)。总和为:
128 + 136 + 136 + 144 + 144 + 144 + 152 + 152 + 152 + 160 = 1448
Then the mean is calculated as:
然后计算平均数为:
Mean = 1448 ÷ 10 = 144.8 points
So, on average, universities in our sample require about 145 UCAS points for science courses. This single number summarises the centre of the data.
因此,平均而言,我们样本中的大学对科学课程要求约145个UCAS分数。这个单一数值概括了数据的中心。
5. Finding the Median | 找出中位数
The median is the middle value when the data is ordered from smallest to largest. First we sort the science points: 128, 136, 136, 144, 144, 144, 152, 152, 152, 160. Because we have ten numbers (an even count), the median is the average of the 5th and 6th values. Both are 144, so the median is 144 points. This tells us that half of the universities require 144 points or less, and half require 144 points or more for science.
中位数是将数据从小到大排序后处于中间位置的数值。我们首先对科学分数排序:128, 136, 136, 144, 144, 144, 152, 152, 152, 160。因为有十个数值(偶数个),中位数是第5和第6个值的平均数。这两个都是144,所以中位数为144分。这告诉我们,一半的大学对科学课程要求144分或更低,另一半要求144分或更高。
6. Identifying the Mode | 确定众数
The mode is the value that appears most frequently. In the science dataset, 144 appears three times and 152 also appears three times. All other values appear once or twice. Therefore there are two modes: 144 and 152. This makes the distribution bimodal, indicating two clusters of entry requirements – one around
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