Year 11 CCEA Statistics: Interdisciplinary Integrated Problem Training | Year 11 CCEA 统计：跨学科综合题型训练

📚 Year 11 CCEA Statistics: Interdisciplinary Integrated Problem Training | Year 11 CCEA 统计：跨学科综合题型训练

In CCEA Year 11 Statistics, exam success depends on your ability to apply statistical methods in unfamiliar, cross-disciplinary contexts. This article brings together real-world scenarios from biology, business, geography, physics and more. Each section models the kind of integrated problem you might meet in the exam and shows you how to tackle it step by step, strengthening both your calculation skills and your interpretative thinking.

在 CCEA 11 年级统计考试中，成功取决于你在不熟悉的跨学科情境中应用统计方法的能力。本文汇集了来自生物学、商业、地理学、物理学等学科的真实情境。每个小节都模拟了考试中可能出现的综合题型，并逐步向你展示如何解决问题，从而强化你的计算技能与解读能力。

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1. Scatter Graphs and Correlation in Biology | 生物学中的散点图与相关性

In a biology experiment, students measured environmental temperature and the body temperature of a common frog to investigate thermoregulation. The data are recorded below.

在一项生物学实验中，学生测量了环境温度和某种常见青蛙的体温，以研究体温调节。数据记录如下。

Environmental temp (°C)	5	10	15	20	25	30
Frog body temp (°C)	7	12	16	21	26	29

Plot a scatter graph with environmental temperature on the horizontal axis and frog body temperature on the vertical axis. The points lie almost in a straight line, suggesting a very strong positive correlation.

绘制散点图，将环境温度放在横轴，青蛙体温放在纵轴。各点几乎落在一条直线上，表明存在极强的正相关关系。

To quantify the strength, calculate Pearson’s product-moment correlation coefficient r. First find the sums: Σx = 105, Σy = 111, Σxy = 2485, Σx² = 2275, Σy² = 2287, with n = 6. Substitute into the formula.

为了量化强度，计算皮尔逊积矩相关系数 r。先求和：Σx = 105, Σy = 111, Σxy = 2485, Σx² = 2275, Σy² = 2287, n = 6。代入公式。

r = [n Σxy − (Σx)(Σy)] / √[ (n Σx² − (Σx)²)(n Σy² − (Σy)²) ]

This gives r = (6×2485 − 105×111) / √[(6×2275 − 105²)(6×2287 − 111²)] ≈ 0.998. The value is extremely close to +1, confirming a near-perfect positive linear association.

代入得 r = (6×2485 − 105×111) / √[(6×2275 − 105²)(6×2287 − 111²)] ≈ 0.998。该值非常接近 +1，证实了近乎完美的正线性关联。

Biologically, the frog’s body temperature rises almost linearly with the environment, illustrating its ectothermic nature. Such a scatter plot can help ecologists predict an amphibian’s activity level under different weather conditions.

从生物学角度看，青蛙的体温几乎随环境温度线性升高，体现了其变温特性。这种散点图有助于生态学家预测蛙类在不同天气条件下的活动水平。

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2. Mean and Range in Business Sales | 商业销售中的均值与极差

A small bakery recorded its weekly sales (in £100s) over six weeks: 10, 12, 9, 15, 11, 13. The owner wants a simple summary to plan inventory.

一家小面包店记录了六周的销售额（单位：£100）：10, 12, 9, 15, 11, 13。店主希望用简单的统计汇总来规划库存。

Calculate the mean: (10+12+9+15+11+13) / 6 = 70 / 6 ≈ 11.67 (£1167). The range is 15−9 = 6 (£600).

计算均值：(10+12+9+15+11+13) / 6 = 70 / 6 ≈ 11.67（£1167）。极差为 15−9 = 6（£600）。

The mean gives a typical weekly sales level, useful for ordering raw materials. The range shows the variation caused by special events or holidays. However, the range is easily distorted by one extreme week; if Week 4 had been an outlier of 25, the range would jump to 16, masking the usual stability.

均值反映了典型周销售水平，有助于订购原材料。极差显示了由特殊活动或假期引起的波动。然而，极差易受极端值影响；如果第 4 周是一个异常值 25，极差将跃升至 16，掩盖了通常的稳定性。

In a business context, managers often combine the mean with the interquartile range for a more robust description of financial performance. This initial analysis shows typical earnings but highlights the need to explore why Week 4 was unusually high.

在商业情境中，管理者常将均值与四分位距结合，以更稳健地描述财务业绩。这一初步分析显示了典型收入，但也凸显出探究第 4 周为何异常高的必要性。

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3. Probability Trees in Genetics | 遗传学中的概率树图

In Mendel’s pea plant experiment, flower colour is determined by a single gene with allele P (purple, dominant) and p (white, recessive). A heterozygous plant (Pp) is self-pollinated. Draw a probability tree to find the chance of purple flowers in the offspring.

在孟德尔的豌豆实验中，花色由一个基因决定，等位基因 P（紫色，显性）和 p（白色，隐性）。一棵杂合子植株（Pp）自花授粉。绘制概率树图，求子代开紫花的概率。

The parent produces gametes P and p, each with probability ½. The tree has two branches from each parent leading to combinations: PP (½ × ½ = ¼), Pp (½ × ½ = ¼), pP (½ × ½ = ¼), pp (½ × ½ = ¼).

亲本产生配子 P 和 p，概率均为 ½。从每个亲本出发的两条分支产生组合：PP (½ × ½ = ¼), Pp (½ × ½ = ¼), pP (½ × ½ = ¼), pp (½ × ½ = ¼)。

Genotypes PP, Pp and pP all produce purple flowers, while only pp gives white. Therefore, P(purple) = ¼ + ¼ + ¼ = ¾ or 0.75.

基因型 PP、Pp 和 pP 都产生紫花，只有 pp 产生白花。因此 P(紫色) = ¼ + ¼ + ¼ = ¾ 或 0.75。

Extending the tree to two generations allows a geneticist to predict the ratio of phenotypes in a large population. This use of probability trees is a standard tool in introductory genetics and shows how statistics bridges mathematics and biology.

将树图延伸至两代，可让遗传学家预测大种群中的表型比例。概率树图是入门遗传学的标准工具，展示了统计如何连接数学与生物学。

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4. Cumulative Frequency and Box Plots in Geography | 地理学中的累积频数与箱线图

An economic geographer compiles the Gross National Income (GNI) per capita (in thousands of US dollars) for 15 countries: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 25, 30. She wants to summarise the distribution.

一位经济地理学家整理了 15 个国家的人均国民总收入（千美元）：1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 25, 30。她想汇总其分布。

Group the data into class intervals 0−5, 5−10, etc. The cumulative frequency table shows how many countries fall below each upper bound. Plotting cumulative frequency against the upper class boundary gives an S-shaped curve.

将数据分组为 0−5, 5−10 等区间。累积频数表显示有多少国家低于各个上限。以累积频数对组上限作图，得到 S 形曲线。

From the graph, the median GNI is about 8 (thousand), the lower quartile ≈4.5, and the upper quartile ≈14. The interquartile range (IQR) is 9.5, indicating moderate

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