6 Fires in Nature | 自然界中的火灾

📚 6 Fires in Nature | 自然界中的火灾

In A-Level Statistics, we often model the number of natural wildfires occurring in a forest over a fixed time period using the Poisson distribution. This scenario appears in Edexcel exam questions where the mean rate of fires is given and we calculate probabilities, test hypotheses, or evaluate the suitability of the model. In this article we explain step by step how to apply the Poisson distribution to fires in nature, covering conditions, calculations, and exam techniques.

在A-Level统计学中，我们经常使用泊松分布来模拟一片森林在固定时间段内发生的自然野火次数。这种情境常出现在Edexcel考试题中，题目会给出火灾的平均发生率，然后要求计算概率、进行假设检验或评估模型的适当性。本文我们将逐步讲解如何将泊松分布应用于自然界中的火灾，涵盖条件、计算以及应试技巧。

1. The Poisson Model for Wildfires | 野火的泊松模型

Wildfires are relatively rare, random, and occur independently in a large area. The number of fires per year can be modelled by a Poisson random variable X with parameter λ, where λ is the mean number of fires per year.

野火相对罕见、随机，且在大面积区域内独立发生。每年火灾次数可以用参数为λ的泊松随机变量X来建模，λ是每年火灾的平均次数。

For example, a national park records an average of 2.4 fires per fire season. We set λ = 2.4. The Poisson distribution then gives the probability of exactly r fires in a season: P(X = r) = (e⁻λ λʳ) / r!.

例如，一个国家公园每个火灾季节平均发生2.4次火灾。我们设λ=2.4。泊松分布给出一个季节内恰好发生r次火灾的概率：P(X=r)=(e⁻λ λʳ)/r!。

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2. Conditions for a Poisson Distribution | 泊松分布的条件

For the Poisson model to be valid, events must occur singly in continuous space or time, at a constant average rate, and independently of each other. When modelling fires in nature, we assume that one fire does not directly cause another in a different location, and that the underlying risk remains constant throughout the year.

要使泊松模型有效，事件必须在连续的空间或时间中单独发生，且平均发生率恒定，并且相互独立。对自然界火灾建模时，我们假定一场火灾不会直接引发另一处火灾，并且全年潜在风险保持不变。

If fires cluster due to lightning storms or drought conditions, the assumption of independence might be violated. In exam questions you may be asked to comment on whether a Poisson model is appropriate, often by looking at the variance-to-mean ratio.

如果由于雷暴或干旱导致火灾聚集，则独立性的假设可能不成立。考试题目可能会要求你评论泊松模型是否合理，通常通过观察方差与均值的比率来判断。

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3. Defining the Random Variable | 定义随机变量

Always define your random variable clearly in an exam. For example: “Let X be the number of fires occurring in the forest in one year.” State the distribution: X ~ Po(λ). This earns method marks and helps you avoid errors.

在考试中务必清晰定义随机变量。例如：“设X为一年内森林中发生的火灾次数”。写出分布：X ~ Po(λ)。这样做不仅能拿到方法分，还有助于避免错误。

If the time interval changes, adjust λ proportionally. For instance, the number of fires in three years follows Po(3λ). This scaling property is crucial for solving problems involving different observation periods.

如果时间区间改变，需按比例调整λ。例如，三年内的火灾次数服从Po(3λ)。这一缩放特性对解决涉及不同观察周期的问题至关重要。

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4. Poisson Probability Formula | 泊松概率公式

The core formula is:

P(X = r) = (e⁻λ × λʳ) / r!    for r = 0, 1, 2, …

核心公式为：

P(X = r) = (e⁻λ × λʳ) / r!    r = 0, 1, 2, …

Here e ≈ 2.71828. λ is the mean rate. Factorial r! grows quickly, so probabilities for large r become very small. In practice we often use cumulative probability tables.

这里e ≈ 2.71828。λ为平均率。阶乘r!增长很快，因此较大r值的概率变得极小。实际上我们常使用累积概率表。

For λ=2.4, we can compute: P(X=0) = e⁻²·⁴ = 0.0907, P(X=1) = e⁻²·⁴ × 2.4 = 0.2177, P(X=6) = (e⁻²·⁴ × 2.4⁶) / 720, which is quite small. You may need to show substitution in working.

对于λ=2.4，可计算出：P(X=0)= e⁻²·⁴ = 0.0907，P(X=1)= e⁻²·⁴ × 2.4 = 0.2177，P(X=6)= (e⁻²·⁴ × 2.4⁶) / 720，这个概率非常小。在解题过程中可能需要写出代入步骤。

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5. Using Cumulative Poisson Tables | 使用累积泊松分布表

Edexcel provides tables of P(X ≤ r) for various λ. To find P(X = r), calculate P(X ≤ r) – P(X ≤ r−1). To find P(X > r) use 1 – P(X ≤ r). Always check which version of the table you have and draw a simple number line if necessary.

Edexcel提供不同λ取值下P(X ≤ r)的表格。要求P(X = r)时，计算P(X ≤ r) – P(X ≤ r−1)。要求P(X > r)时，使用1 – P(X ≤ r)。始终确认所用表格的形式，必要时画一条简单的数轴。

Example: for λ=2.5, find the probability of at most 3 fires. From tables, P(X ≤ 3) = 0.7576. For more than 3 fires: 1 – 0.7576 = 0.2424.

例如：λ=2.5，求最多发生3次火灾的概率。查表得P(X ≤ 3)=0.7576。超过3次火灾的概率：1 – 0.7576 = 0.2424。

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6. Mean and Variance of the Poisson Distribution | 泊松分布的均值与方差

One special property is that the mean and variance are both equal to λ. For fires data, if the sample variance is much larger than the sample mean, the data may be over-dispersed and a Poisson model might not be the best choice. This is frequently tested.

泊松分布的一个特殊性质是均值与方差相等，均等于λ。对于火灾数据，如果样本方差远大于样本均值，则数据可能存在过度离散，泊松模型可能不是最佳选择。这是常考内容。

You may be asked: “Given data for 8 years, mean number of fires is 2.125 and variance is 4.696. Comment on the suitability of a Poisson model.” Since 4.696 > 2.125 substantially, you would conclude the Poisson may not be appropriate because the variance exceeds the mean.

题中可能会问：“根据8年的数据，火灾平均次数为2.125，方差为4.696。评述泊松模型的合理性。”由于4.696明显大于2.125，可得出结论：方差超出均值，泊松模型可能不合适。

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7. Hypothesis Testing for the Mean Number of Fires | 火灾平均次数的假设检验

A typical exam question: “An environmentalist claims the mean number of fires per year has increased from 2.4. Last year there were 5 fires. Test at the 5% significance level.”

典型的考题：“某环保人士声称每年火灾平均次数已从2.4上升。去年发生了5起火灾。在5%显著性水平下检验。”

Set up hypotheses: H₀: λ = 2.4; H₁: λ > 2.4. Under H₀, X ~ Po(2.4). Find P(X ≥ 5) = 1 – P(X ≤ 4). Using tables, P(X ≤ 4) = 0.9042 (approx). So p-value = 0.0958. Since 0.0958 > 0.05, do not reject H₀. There is insufficient evidence of an increase.

建立假设：H₀: λ = 2.4；H₁: λ > 2.4。在H₀下，X ~ Po(2.4)。求P(X ≥ 5) = 1 – P(X ≤ 4)。查表得P(X ≤ 4) ≈ 0.9042。故p值=0.0958。由于0.0958 > 0.05，不拒绝H₀。没有足够证据表明火灾次数增加。

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8. Critical Region Method | 临界区域法

Alternatively, find the critical value c such that P(X ≥ c) ≤ 0.05. For Po(2.4), try c=6: P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.9643 = 0.0357 < 0.05. For c=5: prob is 0.0958 > 0.05. So critical region is X ≥ 6. Since observed value 5 is not in critical region, do not reject H₀.

另一种方法：找到临界值c，使得P(X ≥ c) ≤ 0.05。对于Po(2.4)，尝试c=6：P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.9643 = 0.0357 < 0.05。c=5时概率为0.0958 > 0.05。因此临界区域为X ≥ 6。由于观测值5不在临界区域内，不拒绝H₀。

Always state your conclusion in context: “There is no significant evidence at the 5% level to suggest the mean number of fires per year has risen above 2.4.”

务必联系背景写出结论：“在5%显著性水平下，没有显著证据表明每年火灾平均次数已超过2.4。”

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9. Two-Tailed Tests and Fire Studies | 双尾检验与火灾研究

A researcher might suspect the fire rate has changed, without specifying direction. Then use a two-tailed test. For X ~ Po(2.4) and at significance level 5%, find critical values in both tails with probabilities ≤ 0.025 each. This tests whether λ ≠ 2.4.

研究人员可能怀疑火灾发生率发生了变化，但没有指明方向。此时应使用双尾检验。对于X ~ Po(2.4)，在5%显著性水平下，求两个尾部各概率≤0.025的临界值。这用于检验λ是否不等于2.4。

Using tables, P(X ≤ 0) = 0.0907 > 0.025, so lower tail not reached. Upper tail: X ≥ 6 gives 0.0357, not ≤ 0.025. So no rejection region in typical single observation. Often combined with larger sample sizes or normal approximation.

查表可知，P(X ≤ 0) = 0.0907 > 0.025，未落入下尾。上尾：X ≥ 6 对应的概率为0.0357，不≤0.025。因此单次观测下通常无拒绝域。常结合更大样本量或正态近似。

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10. Fires as a Poisson Process and Approximations | 火灾作为泊松过程与近似

Fires occurring randomly in a forest can be treated as a Poisson process. If λ is large (e.g. λ > 10), the Poisson can be approximated by a normal distribution N(λ, λ). Use continuity correction: P(X ≥ 15) ≈ P(Y > 14.5) where Y ~ N(λ, λ).

森林中随机发生的火灾可以视为一个泊松过程。当λ较大时（如λ > 10），泊松分布可用正态分布N(λ, λ)近似。使用连续性校正：P(X ≥ 15) ≈ P(Y > 14.5)，其中Y ~ N(λ, λ)。

If events are fires over a large area, we might also approximate the Poisson with a binomial when the probability is small and n large, but Poisson is direct.

如果事件是广大区域内的火灾，也可用二项近似泊松，但通常直接使用泊松。

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11. Limitations and Model Checking | 局限性与模型检验

In reality, fire occurrences can be seasonal (dry summers) or influenced by human activity, making the rate non-constant. A Poisson model ignores climate cycles. When analyzing more than one year, data may show over-dispersion. Always check expected vs. observed frequencies using a χ² goodness-of-fit test if required.

实际上，火灾发生可能具有季节性（干燥夏季）或受人类活动影响，使得发生率非常数。泊松模型忽略了气候周期。分析多年数据时可能出现过度离散。如有需要，应用χ²拟合优度检验比较期望频数与观测频数。

In an exam, you could be asked to carry out a goodness-of-fit test for a Poisson distribution to a set of fire frequency data. Remember to combine categories so expected frequencies ≥ 5, and reduce degrees of freedom accordingly.

在考试中，可能要求对一组火灾频率数据进行泊松分布的拟合优度检验。记得合并类别使期望频数≥5，并相应减少自由度。

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12. Summary and Exam Tactics | 总结与应试策略

Key points for Poisson fires questions: define X, state λ, justify the model using constant mean and independence, correctly use probability formulas and tables, interpret p-values in context, and check for over-dispersion. Always write a final conclusion that refers back to the original problem.

泊松火灾题目的关键点：定义X，写明λ，用恒定的均值和独立性论证模型的合理性，正确使用概率公式与表格，在背景中解释p值，并检查过度离散。务必写出指向原问题的最终结论。

When facing a wordy problem, underline the numerical rate and the time interval. Convert to Poisson parameter carefully. Practice numeric skills with e⁻λ and factorial calculations, but rely heavily on provided tables. Manage your time by not recalculating tabled probabilities unnecessarily.

遇到文字冗长的题目时，划出数字速率和时间间隔，谨慎转化为泊松参数。练习运用 e⁻λ 和阶乘的计算能力，但应充分利用提供的表格。无需重新计算表格中已有的概率值以节约时间。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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