The Poisson Distribution | 泊松分布

📚 The Poisson Distribution | 泊松分布

The Poisson distribution is a discrete probability distribution used to model the number of events that occur in a fixed interval of time or space. In Edexcel A Level Mathematics, it appears in Statistics as a tool for modelling random events such as calls to a call centre, flaws in fabric, radioactive emissions, or accidents over a given period. This article covers the definition, conditions, mean and variance, cumulative probabilities, approximation to the binomial distribution, worked examples, and exam tips for the Poisson distribution.

泊松分布是一种离散概率分布，用于建模在固定时间或空间间隔内发生的事件次数。在 Edexcel A Level 数学中，它出现在统计部分，用于对呼叫中心来电、布料瑕疵、放射性发射或某段时间内的事故等随机事件进行建模。本文涵盖泊松分布的定义、使用条件、均值与方差、累积概率、对二项分布的近似、例题以及考试技巧。

1. Definition and Probability Mass Function | 定义与概率质量函数

A random variable X follows a Poisson distribution with parameter λ, written X ~ Po(λ), if X counts the number of events in a fixed interval and has the probability mass function shown below. The parameter λ must be positive and represents the average number of events in that interval.

若随机变量 X 服从参数为 λ 的泊松分布，记作 X ~ Po(λ)，则 X 表示在一个固定区间内事件发生的次数，其概率质量函数如下。参数 λ 必须为正数，表示该区间内事件发生的平均次数。

P(X = x) = e⁻λ λˣ / x!, x = 0, 1, 2, …

Here e is Euler’s number, approximately 2.71828, λ is the average rate, and x must be a non-negative integer. Since the Poisson distribution is discrete, P(X = x) gives a probability rather than a probability density.

其中 e 是欧拉数，约为 2.71828，λ 是平均发生率，x 必须是非负整数。由于泊松分布是离散分布，P(X = x) 给出的是概率而不是概率密度。

2. Conditions for a Poisson Model | 泊松模型的条件

The Poisson distribution is appropriate when the following modelling conditions are met. The events must occur independently, occur one at a time, have a constant average rate, and be random across the given interval.

当满足以下建模条件时，使用泊松分布是合适的。事件必须独立发生，每次只发生一个，平均发生率保持恒定，并在给定的区间内随机出现。

Events occur independently within the interval; one event does not change the probability of another. | 事件在区间内独立发生；一个事件不会改变另一个事件发生的概率。
Events occur singly, not in batches. | 事件单独发生，不能成批发生。
The average rate λ is constant throughout the interval. | 平均发生率 λ 在整个区间内保持恒定。
The probability of more than one event in a very small sub-interval is negligible. | 在一个非常小的子区间内发生一个以上事件的概率可以忽略不计。

If events happen in clusters or the rate changes over time, the Poisson model may not be valid. Edexcel exam questions often ask you to state whether these conditions are realistic in context.

如果事件成簇发生或发生率随时间变化，泊松模型可能不适用。Edexcel 考试题经常要求你说明这些条件在具体情境中是否现实。

3. Mean and Variance | 均值与方差

If X ~ Po(λ), then the expected value and variance are both equal to λ. This is a unique property of the Poisson distribution and provides a quick way to check whether a Poisson model is plausible for a data set.

如果 X ~ Po(λ)，那么期望值和方差都等于 λ。这是泊松分布的一个独特性质，为检验一组数据是否适合用泊松模型提供了一种快速方法。

E(X) = λ and Var(X) = λ

Because the mean equals the variance, a data set with a sample mean much smaller or much larger than its sample variance is unlikely to come from a Poisson distribution. In modelling questions, comparing the sample mean and sample variance can help justify or reject a Poisson model.

由于均值等于方差，若一组数据的样本均值远小于或远大于其样本方差，则该数据不太可能来自泊松分布。在建模题中，比较样本均值与样本方差可以帮助证明使用泊松模型是否合理。

4. Shape and Key Probabilities | 分布形状与关键概率

For small values of λ, the Poisson distribution is right-skewed. As λ increases, the distribution becomes more symmetric and eventually resembles a normal distribution. The shape depends entirely on λ.

当 λ 较小时，泊松分布呈右偏态。随着 λ 增大，分布变得更加对称，最终近似于正态分布。分布形状完全取决于 λ。

The probability of exactly zero events is P(X = 0) = e⁻λ. This result is useful because probabilities for x = 1, 2, 3, … can be built up from it using a recurrence relation.

恰好发生零次事件的概率为 P(X = 0) = e⁻λ。这个结果很有用，因为 x = 1、2、3 等的概率可以通过递推关系从它逐步求得。

5. Using the Probability Formula | 使用概率公式

To calculate Poisson probabilities, substitute the value of λ and the required x into the formula. For example, if X ~ Po(4), then P(X = 6) is calculated as shown below.

要计算泊松概率，请将 λ 的值和所需的 x 代入公式。例如，若 X ~ Po(4)，则 P(X = 6) 的计算如下所示。

P(X = 6) = e⁻⁴ × 4⁶ / 6! = 0.1042 (4 d.p.)

A recurrence relation can make manual calculation faster. Once P(X = 0) = e⁻λ is known, each next probability can be found using the formula below.

使用递推关系可以加快手算速度。一旦已知 P(X = 0) = e⁻λ，即可使用以下公式求出下一个概率。

P(X = x + 1) = λ/(x + 1) × P(X = x)

This relation is especially useful when working without a calculator or when only a basic scientific calculator is allowed. It avoids recalculating large powers and factorials separately.

当不使用计算器或只允许使用基础科学计算器时，这个递推关系特别有用。它可以避免单独计算较大的幂和阶乘。

6. Cumulative Probabilities | 累积概率

Cumulative probabilities give the probability that X is less than or equal to a given value. For example, P(X ≤ k) means the probability of observing k or fewer events. It is the sum of all probabilities from x = 0 to x = k.

累积概率给出 X 小于或等于某个给定值的概率。例如，P(X ≤ k) 表示观察到 k 个或更少事件的概率。它是 x = 0 到 x = k 的所有概率之和。

P(X ≤ k) = Σᵣ₌₀ᵏ e⁻λ λʳ / r!

In Edexcel exams, you may use a calculator or statistical tables to find cumulative Poisson probabilities. However, you should still show the expression you are evaluating, especially in written questions.

在 Edexcel 考试中，你可以使用计算器或统计表来计算泊松累积概率。但在书面题中，你仍应写出所计算的表达式，尤其是在需要展示过程的题目里。

To find P(X > k), use the complement rule: P(X > k) = 1 − P(X ≤ k). Similarly, for an interval such as P(a ≤ X ≤ b), calculate P(X ≤ b) − P(X ≤ a − 1).

要求 P(X > k)，可使用补集规则：P(X > k) = 1 − P(X ≤ k)。类似地，对于区间 P(a ≤ X ≤ b)，可计算 P(X ≤ b) − P(X ≤ a − 1)。

7. Poisson Approximation to the Binomial Distribution | 泊松分布对二项分布的近似

When a binomial random variable has a large number of trials n and a small probability of success p, the binomial distribution can be approximated by a Poisson distribution with parameter λ = np. This approximation reduces the calculation burden and is especially useful when n is large and p is small.

当二项随机变量的试验次数 n 很大且成功概率 p 很小时，可以用参数 λ = np 的泊松分布来近似二项分布。这种近似可以减轻计算负担，在 n 大且 p 小时特别有用。

If X ~ B(n, p), n is large, p is small, then X ≈ Po(np)

A common rule of thumb is that the approximation works well when n is greater than about 50 and p is less than about 0.1, with np moderate. The larger n is and the smaller p is, the better the approximation.

一个常用的经验法则是，当 n 大于约 50 且 p 小于约 0.1，并且 np 适中时，这种近似效果较好。n 越大、p 越小，近似效果越好。

For example, if X ~ B(200, 0.01), then λ = 200 × 0.01 = 2, so X ≈ Po(2). This makes it much easier to calculate probabilities such as P(X = 3).

例如，若 X ~ B(200, 0.01)，则 λ = 200 × 0.01 = 2，因此 X ≈ Po(2)。这使得计算 P(X = 3) 等概率变得容易得多。

8. Modelling and Changing the Interval | 建模与改变时间或空间区间

In many Edexcel questions, the rate λ is given for one interval, but the question asks about a different interval. If events occur at a constant average rate, the parameter λ must be scaled in proportion to the length of the new interval.

在许多 Edexcel 题目中，给出的率 λ 对应的是一个区间，但题目问的是另一个区间。如果事件以恒定的平均率发生，则必须按新区间的长度成比例地调整参数 λ。

For instance, if calls arrive at an average rate of 5 per hour, then over a two-hour period the Poisson parameter is λ = 5 × 2 = 10. If the question asks about a 30-minute period, λ = 5 × 0.5 = 2.5. Always check that the units match the interval in the question.

例如，如果电话以平均每小时 5 个的速率到达，那么在两个小时的时间段内，泊松参数为 λ = 5 × 2 = 10。如果题目问的是 30 分钟的时间段，则 λ = 5 × 0.5 = 2.5。务必检查单位是否与题目中的区间一致。

9. Worked Example 1: Direct Probability Calculation | 例题 1：直接概率计算

Problem: A call centre receives calls at an average rate of 4 calls per hour. Let X be the number of calls received in one hour. Find the probability that exactly 6 calls are received in a randomly selected hour.

问题：某呼叫中心平均每小时接到 4 个电话。设 X 为一小时内接到的电话数。求在随机选择的一小时内恰好接到 6 个电话的概率。

Solution: First define X ~ Po(4). We require P(X = 6). Using the probability mass function, substitute λ = 4 and x = 6.

解答：首先定义 X ~ Po(4)。我们需要求 P(X = 6)。使用概率质量函数，代入 λ = 4 和 x = 6。

P(X = 6) = e⁻⁴ × 4⁶ / 6! = 0.1042 (4 d.p.)

Therefore, the probability of receiving exactly 6 calls in one hour is approximately 0.1042. In Edexcel answers, you should state the distribution and the formula before giving the numerical result.

因此，一小时内恰好接到 6 个电话的概率约为 0.1042。在 Edexcel 的答案中，你应先说明分布和公式，再给出数值结果。

10. Worked Example 2: Cumulative Probability | 例题 2：累积概率计算

Problem: Flaws in a roll of fabric occur at an average rate of 2 flaws per metre. Find the probability that a randomly selected 3-metre length contains at most 2 flaws.

问题：一卷布料上的瑕疵平均每米出现 2 个。求随机选择的 3 米布料中最多有 2 个瑕疵的概率。

Solution: First scale the rate for 3 metres: λ = 2 × 3 = 6. Let X ~ Po(6). We require P(X ≤ 2), which is P(X = 0) + P(X = 1) + P(X = 2).

解答：首先将率换算为 3 米：λ = 2 × 3 = 6。设 X ~ Po(6)。我们需要求 P(X ≤ 2)，即 P(X = 0) + P(X = 1) + P(X = 2)。

P(X ≤ 2) = e⁻⁶ (1 + 6 + 6²/2) = 25e⁻⁶ = 0.0620 (4 d.p.)

So the probability that a 3-metre length has at most 2 flaws is about 0.0620. This is a typical cumulative probability question and shows why it is important to scale λ before calculating.

因此，3 米布料中最多有 2 个瑕疵的概率约为 0.0620。这是一道典型的累积概率题，也说明了为何在计算前先调整 λ 非常重要。

11. Worked Example 3: Binomial Approximation | 例题 3：二项分布近似

Problem: A machine produces electronic components, and each component has a 1% chance of being defective. A sample of 200 components is selected. Use a Poisson approximation to find the probability that exactly 3 components are defective.

问题：一台机器生产电子元件，每个元件有 1% 的概率存在缺陷。随机抽取 200 个元件。使用泊松近似求恰好有 3 个元件有缺陷的概率。

Solution: Let X be the number of defective components. Then X ~ B(200, 0.01). The binomial has n = 200 and p =

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The Poisson Distribution | 泊松分布