Binomial Distribution Mean & Variance Formulas | 二项分布的均值与方差公式

📚 Binomial Distribution Mean & Variance Formulas | 二项分布的均值与方差公式

The binomial distribution is one of the most important discrete probability distributions in A-Level Further Mathematics. Understanding its mean and variance formulas is essential for solving exam questions efficiently and interpreting real-world data.

二项分布是 A-Level 进阶数学中最重要的离散概率分布之一。理解其均值与方差公式，对于高效解答考试题目以及解释现实数据都至关重要。

1. What is a Binomial Distribution? | 什么是二项分布？

A binomial distribution models the number of successes in a fixed number of independent trials, where each trial has the same probability of success. We write X ~ B(n, p), where n is the number of trials and p is the probability of success on each trial.

二项分布用于描述在固定次数的独立试验中“成功”的次数，每次试验的成功概率相同。我们记为 X ~ B(n, p)，其中 n 是试验次数，p 是每次试验成功的概率。

The probability of exactly k successes is given by the formula:

P(X = k) = ⁿCₖ × pᵏ × (1 − p)ⁿ⁻ᵏ

确切发生 k 次成功的概率由以下公式给出：

P(X = k) = ⁿCₖ × pᵏ × (1 − p)ⁿ⁻ᵏ

Here, ⁿCₖ is the binomial coefficient, also written as C(n, k) or “n choose k”.

这里的 ⁿCₖ 是二项式系数，也可写作 C(n, k) 或“n 选 k”。

2. The Mean of a Binomial Distribution | 二项分布的均值

The mean, or expected value, of a binomial distribution is remarkably simple. If X ~ B(n, p), then the expected value of X is given by:

二项分布的均值（或期望值）非常简单。如果 X ~ B(n, p)，那么 X 的期望值为：

E(X) = np

This makes intuitive sense: if you roll a fair die 60 times, the expected number of sixes is 60 × (1/6) = 10.

这在直观上非常合理：如果你掷一颗公平的骰子 60 次，出现 6 点的期望次数就是 60 × (1/6) = 10。

The derivation uses the linearity of expectation. Let X = I₁ + I₂ + … + Iₙ, where each Iᵢ is an indicator variable that equals 1 if trial i is a success and 0 otherwise. Since E(Iᵢ) = p for each trial, we have E(X) = np.

该公式的推导利用了期望的线性性质。令 X = I₁ + I₂ + … + Iₙ，其中每个 Iᵢ 是指示变量：如果第 i 次试验成功则取 1，否则取 0。由于每次试验的 E(Iᵢ) = p，因此 E(X) = np。

3. The Variance of a Binomial Distribution | 二项分布的方差

The variance measures the spread of the distribution around the mean. For X ~ B(n, p), the variance is:

方差衡量分布围绕均值的离散程度。对于 X ~ B(n, p)，方差为：

Var(X) = np(1 − p)

It is often written as Var(X) = npq, where q = 1 − p is the probability of failure.

它常写作 Var(X) = npq，其中 q = 1 − p 是失败的概率。

The standard deviation is the square root of the variance:

标准差是方差的平方根：

SD(X) = √[np(1 − p)]

Notice that the variance is maximised when p = 0.5, and it becomes zero when p = 0 or p = 1. This makes sense: if an event is certain or impossible, there is no spread.

注意，当 p = 0.5 时方差最大；当 p = 0 或 p = 1 时方差为零。这很合理：如果事件必然发生或必然不发生，就不存在离散度。

4. Deriving the Variance Formula | 方差的推导

To derive Var(X) = np(1 − p), we first compute E(X²). Using the identity Var(X) = E(X²) − [E(X)]², we need to find E(X²).

为了推导 Var(X) = np(1 − p)，我们首先计算 E(X²)。利用恒等式 Var(X) = E(X²) − [E(X)]²，我们需要求出 E(X²)。

Using the indicator variable approach, X = ΣIᵢ, we have:

使用指示变量的方法，X = ΣIᵢ，我们有：

E(X²) = E[(ΣIᵢ)²] = E[ΣIᵢ² + 2ΣΣᵢ<ⱼ IᵢIⱼ]

Since Iᵢ² = Iᵢ (because 0² = 0 and 1² = 1), we get E(Iᵢ²) = p. Also, for i ≠ j, Iᵢ and Iⱼ are independent, so E(IᵢIⱼ) = p².

因为 Iᵢ² = Iᵢ（0² = 0，1² = 1），所以 E(Iᵢ²) = p。此外，当 i ≠ j 时，Iᵢ 与 Iⱼ 相互独立，因此 E(IᵢIⱼ) = p²。

There are n terms of Iᵢ² and n(n − 1) terms of IᵢIⱼ. Therefore:

其中 Iᵢ² 有 n 项，IᵢIⱼ 有 n(n − 1) 项。因此：

E(X²) = np + n(n − 1)p²

Subtracting [E(X)]² = (np)² gives:

减去 [E(X)]² = (np)²，得到：

Var(X) = np + n(n − 1)p² − n²p² = np − np² = np(1 − p)

5. Worked Example: Calculating Mean and Variance | 实例：计算均值与方差

Suppose a multiple-choice test has 20 questions, each with 4 options. A student guesses every answer. Let X be the number of correct answers. Here X ~ B(20, 0.25).

假设一份选择题试卷有 20 道题，每题有 4 个选项。一名学生完全随机猜测。设 X 为答对的题数。这里 X ~ B(20, 0.25)。

Mean = np = 20 × 0.25 = 5

均值 = np = 20 × 0.25 = 5

Variance = np(1 − p) = 20 × 0.25 × 0.75 = 3.75

方差 = np(1 − p) = 20 × 0.25 × 0.75 = 3.75

Standard deviation = √3.75 ≈ 1.936

标准差 = √3.75 ≈ 1.936

So the student would expect about 5 correct answers, with a standard deviation of about 1.94.

因此，该学生预计答对约 5 题，标准差约为 1.94。

6. Relation Between Binomial and Bernoulli Distributions | 二项分布与伯努利分布的关系

A Bernoulli distribution is a binomial distribution with n = 1. If Y ~ Bernoulli(p), then E(Y) = p and Var(Y) = p(1 − p).

伯努利分布是 n = 1 时的二项分布。如果 Y ~ Bernoulli(p)，则 E(Y) = p，Var(Y) = p(1 − p)。

A binomial random variable is the sum of n independent and identically distributed Bernoulli random variables. This explains why the mean is np and the variance is np(1 − p): both are n times the corresponding Bernoulli values.

二项随机变量是 n 个独立同分布的伯努利随机变量之和。这就解释了为什么均值是 np、方差是 np(1 − p)：二者都是相应伯努利值的 n 倍。

This relationship is useful in proofs and in understanding why the formulas take the form they do. It also connects to the Central Limit Theorem for large n.

这种关系在证明中和理解公式形式时非常有用。它还与 n 较大时的中心极限定理相联系。

7. Special Cases and Boundary Behaviour | 特殊情况与边界行为

When p = 0, every trial is a failure, so X is always 0. Then E(X) = 0 and Var(X) = 0, which matches the formulas.

当 p = 0 时，每次试验都失败，因此 X 恒为 0。此时 E(X) = 0，Var(X) = 0，与公式一致。

When p = 1, every trial is a success, so X is always n. Then E(X) = n and Var(X) = 0.

当 p = 1 时，每次试验都成功，因此 X 恒为 n。此时 E(X) = n，Var(X) = 0。

When p = 0.5, the distribution is symmetric about the mean, and the variance attains its maximum value n/4.

当 p = 0.5 时，分布关于均值对称，方差达到最大值 n/4。

These boundary cases are quick checks for the validity of the formulas in exam situations.

这些边界情况是考试中检验公式正确性的快速方法。

8. Using Properties of Expected Value and Variance | 利用期望与方差的性质

If X ~ B(n, p), then a linear transformation of X, such as Y = aX + b, has:

如果 X ~ B(n, p)，那么 X 的线性变换，例如 Y = aX + b，具有以下性质：

E(Y) = aE(X) + b = anp + b

Var(Y) = a²Var(X) = a²np(1 − p)

These properties are frequently tested in Edexcel exam papers. For example, if X ~ B(50, 0.2) and Y = 3X + 2, then E(Y) = 3 × 10 + 2 = 32 and Var(Y) = 9 × 8 = 72.

这些性质在 Edexcel 考卷中经常考查。例如，若 X ~ B(50, 0.2) 且 Y = 3X + 2，则 E(Y) = 3 × 10 + 2 = 32，Var(Y) = 9 × 8 = 72。

Notice that adding a constant does not change the variance, but multiplying by a constant scales the variance by the square of that constant.

注意，加常数不改变方差，但乘以常数会使方差按该常数的平方缩放。

9. Common Exam Pitfalls | 常见考试陷阱

One common mistake is using the formula for the mean of a binomial distribution on a problem that is not binomial. Always check the conditions: fixed n, independent trials, two outcomes, and constant p.

一个常见错误是将二项分布的均值公式用于并非二项分布的问题。务必检查条件：固定的 n、独立试验、两种结果、p 恒定。

Another mistake is confusing p with 1 − p in the variance formula. Remember that the variance is np(1 − p), not np² or n(1 − p).

另一个错误是在方差公式中将 p 与 1 − p 混淆。记住方差是 np(1 − p)，而不是 np² 或 n(1 − p)。

Students also sometimes forget to take the square root when asked for the standard deviation. Read the question carefully: mean, variance, or standard deviation?

学生有时在需要求标准差时忘记取平方根。仔细审题：要求的是均值、方差还是标准差？

Quantity	Formula	Example (n=10, p=0.3)
Mean	np	3
Variance	np(1−p)	2.1
Standard deviation	√[np(1−p)]	√2.1 ≈ 1.449

10. Proving the Mean Using the Probability Mass Function | 用概率质量函数证明均值

For a more rigorous proof, we can start directly from the definition of expectation:

为了更严格的证明，我们可以直接从期望的定义出发：

E(X) = Σₖ₌₀ⁿ k × ⁿCₖ pᵏ(1 − p)ⁿ⁻ᵏ

Using the identity k × ⁿCₖ = n × ⁿ⁻¹Cₖ₋₁, we obtain:

利用恒等式 k × ⁿCₖ = n × ⁿ⁻¹Cₖ₋₁，我们得到：

E(X) = np × Σₖ₌₁ⁿ ⁿ⁻¹Cₖ₋₁ pᵏ⁻¹(1 − p)ⁿ⁻ᵏ

Let j = k − 1. Then the sum becomes Σⱼ₌₀ⁿ⁻¹ ⁿ⁻¹Cⱼ pʲ(1 − p)ⁿ⁻¹⁻ʲ = 1, since it is the total probability of a binomial distribution B(n − 1, p). Hence E(X) = np.

令 j = k − 1，则求和变为 Σⱼ₌₀ⁿ⁻¹ ⁿ⁻¹Cⱼ pʲ(1 − p)ⁿ⁻¹⁻ʲ = 1，因为这是二项分布 B(n − 1, p) 的总概率。因此 E(X) = np。

This proof demonstrates the elegance of the binomial coefficient identities and is a good exercise for further mathematics students.

这个证明展示了二项式系数恒等式的优雅之处，是进阶数学学生很好的练习。

11. Generalising to the Negative Binomial and Poisson Distributions | 推广到负二项分布与泊松分布

For a negative binomial distribution, which counts the number of trials needed to achieve r successes, the mean is r/p and the variance is r(1 − p)/p².

对于负二项分布（统计达到 r 次成功所需的试验次数），均值为 r/p，方差为 r(1 − p)/p²。

For a Poisson distribution, which approximates the binomial when n is large and p is small with λ = np, the mean and variance are both equal to λ.

对于泊松分布（当 n 较大且 p 较小时近似二项分布，λ = np），其均值与方差都等于 λ。

The fact that the Poisson distribution has equal mean and variance is a key property. In contrast, the binomial distribution has variance smaller than the mean when p ≠ 0.5, because np(1 − p) < np for p > 0.

泊松分布均值等于方差是一个关键性质。相比之下，当 p ≠ 0.5 时，二项分布的方差小于均值，因为对于 p > 0，np(1 − p) < np。

These comparisons help students build a broader framework of discrete distributions in Edexcel A-Level Further Mathematics.

这些比较帮助学生建立 Edexcel A-Level 进阶数学中离散分布更广阔的框架。

12. Summary and Quick Revision | 总结与快速复习

For X ~ B(n, p), the two essential formulas are: E(X) = np and Var(X) = np(1 − p). These are listed in the Edexcel formula booklet, but understanding their derivation helps avoid misapplication.

对于 X ~ B(n, p)，两个基本公式是：E(X) = np 和 Var(X) = np(1 − p)。它们列在 Edexcel 公式手册中，但理解其推导有助于避免误用。

Mean = np — the expected number of successes. | 均值 = np — 成功的期望次数。
Variance = np(1 − p) — measures spread. | 方差 = np(1 − p) — 衡量离散程度。
Standard deviation = √[np(1 − p)]. | 标准差 = √[np(1 − p)]。
Always verify binomial conditions before applying formulas. | 在应用公式前务必验证二项分布条件。

In an exam, you may be asked to find the mean and variance directly, or to use them in a normal approximation with continuity correction. Mastery of these formulas is a stepping stone to more advanced topics such as hypothesis testing and confidence intervals.

在考试中，你可能被要求直接求均值与方差，或者用它们进行带连续性修正的正态近似。掌握这些公式是迈向假设检验和置信区间等更高级主题的基石。

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Binomial Distribution Mean & Variance Formulas | 二项分布的均值与方差公式