4 Biomes: Mathematical Modelling for Edexcel A-Level | 四大生物群落：Edexcel A-Level 数学建模

📚 4 Biomes: Mathematical Modelling for Edexcel A-Level | 四大生物群落：Edexcel A-Level 数学建模

In Edexcel A-Level Mathematics, real-world data questions often use climate or geographic contexts. The four major biomes – tropical rainforest, desert, temperate forest, and tundra – provide a rich setting for applying statistics, probability, and calculus.

在 Edexcel A-Level 数学中，真实数据题常以气候或地理为背景。四大生物群落——热带雨林、沙漠、温带森林和苔原——为统计、概率和微积分的应用提供了丰富的场景。

1. Defining the Data Set | 定义数据集

The table below summarises typical mean annual temperature T (°C), annual precipitation P (mm), and approximate latitude L (degrees) for the four biomes.

下表总结了四个生物群落的典型年平均气温 T（°C）、年降水量 P（mm）和近似纬度 L（度）。

Biome	T (°C)	P (mm)	L (°)
Tropical rainforest	27	2000	0
Desert	22	200	25
Temperate forest	10	900	45
Tundra	-8	250	65

These values are generalised for exam practice. You should always identify the variables, units, and whether the data are continuous or discrete before starting a statistical calculation.

这些数值是为考试练习而概括的。在开始统计计算之前，你应始终明确变量、单位以及数据是连续型还是离散型。

2. Scatter Diagrams and Correlation | 散点图与相关性

A scatter diagram of temperature T against latitude L shows a clear negative association: as latitude increases, mean annual temperature tends to decrease.

以纬度 L 为横轴、气温 T 为纵轴绘制的散点图显示出明显的负相关：随着纬度增加，年平均气温趋于下降。

You should describe correlation in terms of direction, strength, and form. In this case, the relationship is negative, strong, and approximately linear.

你应从方向、强度和形式三方面描述相关性。在此情况下，该关系是负向、较强且近似线性的。

Positive correlation: as one variable increases, the other tends to increase. 正相关：一个变量增加，另一个也趋于增加。
Negative correlation: as one variable increases, the other tends to decrease. 负相关：一个变量增加，另一个趋于减少。
No correlation: no clear linear pattern. 无相关：没有明显的线性模式。

Scatter diagrams can reveal outliers, which may be anomalies or recording errors. A single unusual point can heavily influence the correlation coefficient.

散点图可以揭示异常值，这些异常值可能是异常情况或记录错误。一个不寻常的点会严重影响相关系数。

3. Product Moment Correlation Coefficient | 乘积矩相关系数

The product moment correlation coefficient (PMCC), denoted by r, measures the strength and direction of a linear relationship.

乘积矩相关系数（PMCC）记作 r，用于衡量线性关系的强度和方向。

r = Sxy / √(Sxx × Syy)

Here Sxy = Σ(x − x̄)(y − ȳ), Sxx = Σ(x − x̄)², and Syy = Σ(y − ȳ)². The value of r always lies between −1 and 1.

这里 Sxy = Σ(x − x̄)(y − ȳ)，Sxx = Σ(x − x̄)²，Syy = Σ(y − ȳ)²。r 的值始终在 −1 到 1 之间。

For the four biome points, r is approximately −0.98, suggesting a very strong negative linear correlation between temperature and latitude.

对于四个生物群落的点，r 约为 −0.98，表明气温与纬度之间存在极强的负线性相关。

A value close to −1 does not prove causation; it only indicates a statistical association. Latitude is a proxy for solar energy input, so a physical explanation also exists.

接近 −1 的值并不能证明因果关系；它只表明统计上的关联。纬度是太阳能量输入的代理变量，因此也存在物理解释。

4. Least Squares Linear Regression | 最小二乘线性回归

When the scatter diagram suggests a linear relationship, we can model it using the least squares regression line.

当散点图显示线性关系时，我们可以使用最小二乘回归线进行建模。

T = a + bL

The gradient b is calculated as b = Sxy / Sxx, and the intercept is a = (ΣT)/n − b(ΣL)/n. This line minimises the sum of the squares of the vertical distances from the points to the line.

斜率 b 的计算公式为 b = Sxy / Sxx，截距为 a = (ΣT)/n − b(ΣL)/n。该直线使各点到直线的垂直距离平方和最小。

Using the given data, the regression equation is approximately T = 28.5 − 0.55L. This model predicts that for every 1° increase in latitude, temperature falls by about 0.55°C.

使用给定数据，回归方程约为 T = 28.5 − 0.55L。该模型预测，纬度每增加 1°，气温下降约 0.55°C。

Regression lines can be used for interpolation within the data range, but extrapolation beyond the data should be treated with caution.

回归线可用于数据范围内的插值，但对数据范围外的外推应谨慎处理。

5. Exponential and Logarithmic Models | 指数与对数模型

Precipitation does not vary linearly with latitude. A better model may be an exponential decay: precipitation decreases rapidly from the equatorial rainforest belt towards the poles.

降水量并不随纬度线性变化。更好的模型可能是指数衰减：降水从赤道雨林带向两极迅速减少。

P = 2500 e^(−0.04L)

Here e is the mathematical constant approximately equal to 2.71828. The negative exponent gives a decreasing curve.

这里 e 是数学常数，约等于 2.71828。负指数产生递减曲线。

Exponential models are common in Edexcel questions involving population growth, radioactive decay, and cooling. The parameter 0.04 is the decay rate per degree of latitude.

指数模型在 Edexcel 题目中很常见，涉及人口增长、放射性衰变和冷却等。参数 0.04 是每个纬度单位的衰减率。

To test whether an exponential model is appropriate, you can plot ln P against L. If the data follow P = 2500 e^(−0.04L), then ln P = ln 2500 − 0.04L, giving a straight line with gradient −0.04.

要检验指数模型是否合适，可以绘制 ln P 对 L 的图像。如果数据符合 P = 2500 e^(−0.04L)，则 ln P = ln 2500 − 0.04L，得到斜率为 −0.04 的直线。

Logarithms are essential for transforming nonlinear relationships into linear form, which can then be analysed using regression techniques.

对数对于将非线性关系转换为线性形式至关重要，转换后可利用回归技术进行分析。

6. Probability and the Normal Distribution | 概率与正态分布

Daily temperatures within a biome often follow an approximately normal distribution. For a tropical rainforest, suppose the daily mean temperature X is normally distributed with mean μ = 27°C and standard deviation σ = 3°C.

生物群落内的日气温通常近似服从正态分布。假设在热带雨林中，日平均气温 X 服从均值为 μ = 27°C、标准差为 σ = 3°C 的正态分布。

X ~ N(27, 3²)

This notation states that X follows a normal distribution with mean 27 and variance 9. The standard deviation is the square root of the variance.

该记号表示 X 服从均值为 27、方差为 9 的正态分布。标准差是方差的平方根。

To find probabilities, you must standardise using Z = (X − μ) / σ, where Z is the standard normal variable with mean 0 and standard deviation 1.

要求概率，必须使用 Z = (X − μ) / σ 进行标准化，其中 Z 是均值为 0、标准差为 1 的标准正态变量。

7. Extreme Weather Probability | 极端天气概率

Calculate the probability that a randomly selected day in the tropical rainforest model has a temperature above 35°C.

计算在热带雨林模型中随机选择一天的日平均气温超过 35°C 的概率。

P(X > 35) = P(Z > (35 − 27) / 3) = P(Z > 2.6667)

Using statistical tables or a calculator, P(Z > 2.6667) ≈ 0.00383. So such extreme heat is very rare in this biome model, occurring on about 0.383% of days.

使用统计表或计算器，P(Z > 2.6667) ≈ 0.00383。因此在此生物群落模型中，这种极端高温非常罕见，约发生在 0.383% 的日子里。

In a desert biome, daily temperature variation is larger. If σ = 5°C and μ = 22°C, then P(X > 35) becomes P(Z > 2.6) ≈ 0.00466, still less than 0.5%.

在沙漠生物群落中，日温度变化更大。如果 σ = 5°C 且 μ = 22°C，则 P(X > 35) 变为 P(Z > 2.6) ≈ 0.00466，仍然低于 0.5%。

Extreme event probabilities allow scientists to assess heat stress, drought risk, and ecological vulnerability using a mathematical framework.

极端事件概率使科学家能够使用数学框架评估热压力、干旱风险和生态脆弱性。

8. Hypothesis Testing for the Mean | 均值的假设检验

A researcher claims that the mean annual temperature in a temperate forest is 10°C. A sample of 30 sites gives a mean of 8.4°C. Assuming the population standard deviation is σ = 2°C, test at the 1% significance level.

一位研究者声称温带森林的年平均气温为 10°C。一个包含 30 个地点的样本给出的均值为 8.4°C。假设总体标准差为 σ = 2°C，在 1% 显著性水平下进行检验。

H₀: μ = 10, H₁: μ < 10

The test statistic is Z = (x̄ − μ) / (σ / √n) = (8.4 − 10) / (2 / √30) ≈ −4.38.

检验统计量为 Z = (x̄ − μ) / (σ / √n) = (8.4 − 10) / (2 / √30) ≈ −4.38。

The critical value for a one-tailed test at the 1% level is −2.326. Since −4.38 < −2.326, we reject H₀.

在 1% 显著性水平下，单尾检验的临界值为 −2.326。由于 −4.38 < −2.326，我们拒绝 H₀。

There is sufficient evidence at the 1% level to suggest that the true mean temperature is less than 10°C. Hypothesis testing requires a clear conclusion written in the context of the problem.

在 1% 显著性水平下，有充分证据表明真实平均气温低于 10°C。假设检验需要在问题背景下写出清晰的结论。

9. Calculus: Modelling Rate of Change | 微积分：变化率建模

Seasonal temperature within a temperate forest can be modelled by a trigonometric function over months t after January.

温带森林的季节性温度可用以 1 月后月份 t 为变量的三角函数建模。

T(t) = 10 + 12 sin(π(t − 4) / 6)

Differentiating gives the rate of change of temperature with respect to time.

对其求导可得到温度随时间的变化率。

dT/dt = 12 × (π / 6) cos(π(t − 4) / 6) = 2π cos(π(t − 4) / 6)

The maximum rate of warming occurs when cos(π(t − 4) / 6) = 1, giving dT/dt = 2π ≈ 6.28°C per month.

当 cos(π(t − 4) / 6) = 1 时，升温速率最大，得到 dT/dt = 2π ≈ 每月 6.28°C。

Calculus is also used to find maximum and minimum points, to analyse concavity, and to model accumulation of rainfall over time.

微积分还用于求最大值点和最小值点、分析凹凸性以及建立降雨量随时间的累积模型。

10. Numerical Methods: Solving Exponential Equations | 数值方法：求解指数方程

Suppose you need to find the latitude at which precipitation falls to half of its equatorial maximum. Solve

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4 Biomes: Mathematical Modelling for Edexcel A-Level | 四大生物群落：Edexcel A-Level 数学建模