GCSE Cambridge Statistics: Interdisciplinary Integrated Question Practice | GCSE剑桥统计:跨学科综合题型训练

📚 GCSE Cambridge Statistics: Interdisciplinary Integrated Question Practice | GCSE剑桥统计:跨学科综合题型训练

Statistics is the language of data, and in the Cambridge GCSE Statistics examination, questions often arise from real-world contexts spanning biology, geography, economics and beyond. This article provides a comprehensive training on interdisciplinary integrated question types, equipping you with strategies to interpret, analyse and evaluate statistical information from diverse subjects.

统计是数据的语言,在剑桥GCSE统计考试中,题目常源自生物学、地理学、经济学等现实背景。本文提供跨学科综合题型的全面训练,助你掌握解读、分析和评价来自不同学科统计信息的策略。

1. Biology & Medicine: Interpreting Histograms and Cumulative Frequency Curves | 生物与医学:解读直方图和累积频率曲线

A medical study records the systolic blood pressure (mmHg) of 80 patients. The results are grouped as follows: 100–110 (12 patients), 110–120 (18), 120–130 (24), 130–140 (16), 140–150 (10). To construct a histogram, we calculate frequency density = frequency ÷ class width. The class width is 10 mmHg for all groups, so frequency densities are 1.2, 1.8, 2.4, 1.6 and 1.0. From the histogram, we can estimate the median using cumulative frequency. The cumulative frequencies are 12, 30, 54, 70, 80. The median position is the 40.5th value, which lies in the 120–130 group. Using linear interpolation, median ≈ 120 + ((40.5 – 30)/24) × 10 = 120 + (10.5/24)×10 ≈ 124.4 mmHg. This application shows how medical researchers use grouped data to summarise patient health indicators.

一项医学研究记录了80名患者的收缩压(mmHg)。数据分组如下:100–110(12人),110–120(18人),120–130(24人),130–140(16人),140–150(10人)。绘制直方图需计算频数密度 = 频数 ÷ 组距。所有组距均为10 mmHg,因此频数密度依次为1.2、1.8、2.4、1.6、1.0。借助累积频率可估计中位数:累积频数为12、30、54、70、80,中位数的位置为第40.5个值,落在120–130组。线性插值得中位数 ≈ 120 + ((40.5 – 30) / 24) × 10 ≈ 124.4 mmHg。该应用展示了医学研究者如何利用分组数据概括患者健康指标。

Another interdisciplinary question may provide cumulative frequency curves for two treatment groups and ask which treatment lowers blood pressure more effectively. By reading off the median and interquartile range from the curves, you can compare central tendency and spread. Always label axes with correct units, and use the graph to justify your conclusion.

另一种跨学科题目可能给出两个治疗组的累积频率曲线,要求判断哪种治疗更有效降低血压。从曲线中读取中位数和四分位距,便可比较集中趋势与离散程度。务必为坐标轴标注正确单位,并依据图形论证结论。


2. Geography: Comparing Distributions with Box Plots | 地理:用箱线图比较分布

A geography field study records the annual rainfall (mm) at two weather stations over 30 years. The five-number summaries are: Station A – min 520, Q1 680, median 790, Q3 880, max 1020; Station B – min 610, Q1 720, median 760, Q3 810, max 950. Box plots can be drawn on the same scale. From the plot, it is clear that Station A has a higher median rainfall but also greater variability (IQR = 200 vs 90). Outliers may be identified using the 1.5 × IQR rule. This type of question tests your ability to construct, interpret and compare statistical diagrams in a geographical context.

一项地理实地研究记录了30年间两个气象站的年降雨量(mm)。五数概括为:A站 – 最小值520,Q1 680,中位数790,Q3 880,最大值1020;B站 – 最小值610,Q1 720,中位数760,Q3 810,最大值950。可在同一尺度上绘制箱线图。从图中明显看出,A站的中位降雨量更高,但变异性也更大(IQR = 200 vs 90)。异常值可使用1.5 × IQR规则识别。此类题目考查在地理情境下构建、解读和比较统计图表的能力。


3. Economics: Index Numbers and Time Series Analysis | 经济学:指数与时间序列分析

The Cambridge Statistics exam often includes questions on weighted index numbers, such as the Retail Price Index. Suppose a basket of three goods has prices and weights: Good A (weight 4) price £2.00 in base year, £2.20 in current year; Good B (weight 5) £3.00 → £3.30; Good C (weight 1) £1.50 → £1.65. The weighted aggregate price index = Σ (weight × (current price / base price) × 100) / Σ weight. Calculations: A: 4 × (2.20/2.00)×100 = 440; B: 5 × 110 = 550; C: 1 × 110 = 110. Total 1100; Σweight = 10. Index = 110. This indicates a 10% overall price increase. Additionally, time series data for sales can be smoothed using moving averages to identify trends and seasonal fluctuations, which is a vital skill for economic forecasting.

剑桥统计考试常涉及加权指数题目,如零售价格指数。假设一篮子商品有三种,价格和权重为:商品A(权重4)基期价格£2.00,现期£2.20;商品B(权重5)£3.00→£3.30;商品C(权重1)£1.50→£1.65

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