A-Level Further Mathematics: Mean and Variance of Distributions | A-Level 进阶数学：分布的均值与方差

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📚 A-Level Further Mathematics: Mean and Variance of Distributions | A-Level 进阶数学：分布的均值与方差

In A-Level Further Mathematics, the mean and variance of a distribution are essential tools for summarising random behaviour. They allow us to describe the centre and spread of a probability distribution in a precise, mathematical way, forming the foundation for inference, hypothesis testing, and statistical modelling.

在进阶数学中，分布的均值与方差是概括随机行为的重要工具。它们使我们能够以精确的数学方式描述概率分布的中心位置与离散程度，为统计推断、假设检验与统计建模奠定基础。

1. Expected Value of a Discrete Random Variable | 离散型随机变量的期望值

For a discrete random variable X with probability mass function P(X = x_i) = p_i, the mean, also called the expected value, is defined as the sum of each outcome multiplied by its probability.

对于概率质量函数为 P(X = x_i) = p_i 的离散型随机变量 X，其均值（也称期望值）定义为每个结果与其概率乘积之和。

E(X) = Σ xᵢpᵢ = Σ xᵢP(X = xᵢ)

This is a weighted average, where outcomes with higher probability contribute more to the mean. The expected value is often denoted by μ.

这是一个加权平均，概率越高的结果对均值的贡献越大。期望值通常用 μ 表示。

probabilities 0.2, 0.3, 0.5, then E(X) = 1(0.2) + 2(0.3) + 3(0.5) = 0.2 + 0.6 + 1.5 = 2.3.
若 X 取值为 1、2、3，对应概率为 0.2、0.3、0.5，则 E(X) = 1(0.2) + 2(0.3) + 3(0.5) = 0.2 + 0.6 + 1.5 = 2.3。
The expected value does not have to be a value that X can actually take; it is a long-run average.
期望值不一定是 X 实际能取到的值，它是长期平均意义上的值。

2. Variance of a Discrete Random Variable | 离散型随机变量的方差

Variance measures the spread of a distribution around the mean. It is defined as the expected value of the squared deviation from the mean.

方差度量分布围绕均值的离散程度，其定义为离均差平方的期望值。

Var(X) = E[(X − μ)²] = Σ (xᵢ − μ)² pᵢ

A more convenient computational formula is widely used in examinations:

考试中更常用的计算公式为：

Var(X) = E(X²) − [E(X)]²

ared units, the standard deviation is simply the square root of variance: SD(X) = √Var(X).

由于方差涉及平方单位，标准差就是方差的平方根：SD(X) = √Var(X)。

3. Key Properties of Expectation and Variance | 期望与方差的重要性质

Linear transformations of random variables follow specific rules for mean and variance. For constants a and b:

随机变量的线性变换遵循特定的均值与方差规律。对于常数 a 与 b：

E(aX + b) = aE(X) + b
E(aX + b) = aE(X) + b
Var(aX + b) = a²Var(X)
Var(aX + b) = a²Var(X)

The constant b shifts the distribution but does not affect variance; the constant a scales the spread by a factor of a² in the variance.

常数 b 平移分布但不影响方差；常数 a 对方差产生 a² 倍的比例缩放。

E(aX + bY) = aE(X) + bE(Y)

For independent random variables X and Y, expectation is always additive, and variance is also additive:

对于独立随机变量 X 与 Y，期望始终可加，方差同样可加：

Var(X ± Y) = Var(X) + Var(Y) (for independent X, Y)

4. Mean and Variance of a Binomial Distribution | 二项分布的均值与方差

If X ~ B(n, p), where n is the number of independent trials and p is the probability of success on each trial, then the mean and variance are:

若 X ~ B(n, p)，其中 n 为独立试验次数，p 为每次试验的成功概率，则均值与方差为：

E(X) = np, Var(X) = np(1 − p) = npq

For example, if a fair coin is tossed 50 times and X is the number of heads, then X ~ B(50, 0.5), so E(X) = 50 × 0.5 = 25 and Var(X) = 50 × 0.5 × 0.5 = 12.5.

例如，将一枚均匀硬币抛掷 50 次，设 X 为正面出现的次数，则 X ~ B(50, 0.5)，于是 E(X) = 50 × 0.5 = 25，Var(X) = 50 × 0.5 × 0.5 = 12.5。

The derivation uses the fact that a binomial random variable is the sum of n independent Bernoulli random variables, each with mean p and variance p(1 − p).

该推导利用了二项分布随机变量是 n 个独立伯努利随机变量之和这一事实，每个伯努利变量的均值为 p，方差为 p(1 − p)。

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

If X ~ Po(λ), where λ is the average rate of occurrence in a fixed interval, then:

若 X ~ Po(λ)，其中 λ 为固定区间内的平均发生率，则：

E(X) = λ, Var(X) = λ

A distinctive feature of the Poisson distribution is that the mean equals the variance. This property is known as equidispersion.

泊松分布的一个显著特征是均值等于方差，这一性质称为等离散性。

ency calls arriving at a hospital in one hour follows Po(8). Then E(X) = 8 and Var(X) = 8.
示例：某医院一小时内收到的紧急呼叫次数服从 Po(8)，则 E(X) = 8，Var(X) = 8。
This result is derived from the Poisson probability mass function P(X = x) = e⁻λ λˣ / x!.
该结果可由泊松分布的概率质量函数 P(X = x) = e⁻λ λˣ / x! 推导得出。

6. Mean and Variance of a Geometric Distribution | 几何分布的均值与方差

If X ~ Geo(p), where X is the number of trials needed to obtain the first success, then:

若 X ~ Geo(p)，即 X 为取得首次成功所需的试验次数，则：

E(X) = 1/p, Var(X) = (1 − p) / p²

Here p is the probability of success on each independent trial. The geometric distribution models waiting time for the first success.

其中 p 为每次独立试验的成功概率。几何分布用于刻画等待首次成功所需的次数。

Distribution \| 分布	E(X)	Var(X)
B(n, p)	np	np(1 − p)
Po(λ)	λ	λ
Geo(p)	1/p	(1 − p)/p²

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7. Mean and Variance of a Continuous Uniform Distribution | 连续均匀分布的均值与方差

For a continuous random variable X uniformly distributed on the interval [a, b], the probability density function is constant over the interval.

对于在区间 [a, b] 上均匀分布的连续随机变量 X，其概率密度函数在区间内为常数。

f(x) = 1/(b − a) for a ≤ x ≤ b

The mean and variance are given by:

其均值与方差为：

E(X) = (a + b)/2, Var(X) = (b − a)² / 12

The mean is simply the midpoint of the interval. The variance grows with the square of the interval length.

均值就是区间的中点。方差随区间长度的平方增长。

8. Mean and Variance of a Normal Distribution | 正态分布的均值与方差

The Normal distribution is the most important continuous distribution in statistics. If X ~ N(μ, σ²), then:

正态分布是统计学中最重要的连续分布。若 X ~ N(μ, σ²)，则：

E(X) = μ, Var(X) = σ²

The parameters μ and σ² directly define the mean and variance of the distribution. The standard Normal distribution is N(0, 1).

参数 μ 与 σ² 直接定义了分布的均值与方差。标准正态分布为 N(0, 1)。

If X ~ N(10, 4), then E(X) = 10 and SD(X) = √4 = 2.
若 X ~ N(10, 4)，则 E(X) = 10，SD(X) = √4 = 2。
The Empirical Rule states that approximately 68% of data lies within μ ± σ, 95% within μ ± 2σ, and 99.7% within μ ± 3σ.
经验法则指出，约 68% 的数据落在 μ ± σ 以内，95% 落在 μ ± 2σ 以内，99.7% 落在 μ ± 3σ 以内。

9. Mean and Variance of a Linear Combination | 线性组合的均值与方差

When dealing with linear combinations of independent random variables, the rules for expectation and variance are essential for solving examination problems.

处理独立随机变量的线性组合时，期望与方差的运算规则对于解题至关重要。

E(aX + bY) = aE(X) + bE(Y)

Var(aX + bY) = a²Var(X) + b²Var(Y) (X, Y independent)

Consider X ~ N(5, 9) and Y ~ , independent. Let W = 2X − 3Y. Then E(W) = 2(5) − 3(3) = 1 and Var(W) = 4(9) + 9(4) = 72.

设 X ~ N(5, 9)，Y ~ N(3, 4)，且二者独立。令 W = 2X − 3Y，则 E(W) = 2(5) − 3(3) = 1，Var(W) = 4(9) + 9(4) = 72。

ependent, the covariance term must be included: Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y).

若 X 与 Y 不独立，则必须包含协方差项：Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)。

10. Using Probability Generating Functions to Find Mean and Variance | 利用概率生成函数求均值与方差

ability generating function (PGF) of a discrete random variable X is defined as G(t) = E(tˣ) = Σ P(X = x)tˣ. It provides an elegant way to obtain moments.

离散型随机变量 X 的概率生成函数定义为 G(t) = E(tˣ) = Σ P(X = x)tˣ。它提供了求矩的优雅方法。

E(X) = G'(1), E(X(X − 1)) = G''(1)

From these we can derive the variance using:

由此可推导方差：

Var(X) = G''(1) + G'(1) − [G'(1)]²

For example, for X ~ B(n, p), G(t) = (q + pt)ⁿ where q = 1 − p. Then G'(t) = np(q + pt)ⁿ⁻¹ and G'(1) = np; G''(t) = n(n−1)p²(q + pt)ⁿ⁻², so G''(1) = n(n−1)p². Hence Var(X) = n(n−1)p² + np − (np)² = np(1−p).

例如，对于 X ~ B(n, p)，G(t) = (q + pt)ⁿ，其中 q = 1 − p。则 G'(t) = np(q + pt)ⁿ⁻¹，G'(1) = np；G''(t) = n(n−1)p²(q + pt)ⁿ⁻²，因此 G''(1) = n(n−1)p²。于是 Var(X) = n(n−1)p² + np − (np)² = np(1−p)。

11. Using Moment Generating Functions | 使用矩生成函数

(MGF) is another powerful tool. For a random variable X, the MGF is defined as M(t) = E(eᵗˣ).

矩生成函数是另一个强大工具。对于随机变量 X，其矩生成函数定义为 M(t) = E(eᵗˣ)。

E(X) = M'(0), E(X²) = M''(0)

Var(X) = M''(0) − [M'(0)]²

For X ~ N(μ, σ²), it is known that M(t) = exp(μt + σ²t²/2). Then M'(t) = (μ + σ²t)M(t), so M'(0) = μ. Differentiating again gives M''(0) = σ² + μ², confirming Var(X) = σ².

对于 X ~ N(μ, σ²)，已知 M(t) = exp(μt + σ²t²/2)。则 M'(t) = (μ + σ²t)M(t)，因此 M'(0) = μ。再次求导得 M''(0) = σ² + μ²，验证了 Var(X) = σ²。

12. Worked Examination-Style Problems | 典型考试题型精讲

Let us apply these formulas to problems commonly encountered in Edexcel A-Level Further Mathematics examinations.

下面将上述公式应用于 Edexcel 考试中常见题型。

Problem 1: ion: P(X = 0) = 0.1, P(X = 1) = 0.4, P(X = 2) = 0.3, P(X = 3) = 0.2. Find E(X) and Var(X).

题目1：离散型随机变量 X 的分布为：P(X = 0) = 0.1，P(X = 1) = 0.4，P(X = 2) = 0.3，P(X = 3) = 0.2。求 E(X) 与 Var(X)。

E(X) = 0(0.1) + 1(0.4) + 2(0.3) + 3(0.2) = 0 + 0.4 + 0.6 + 0.6 = 1.6.

E(X²) = 0²(0.1) + 1²(0.4) + 2²(0.3) + 3²(0.2) = 0 + 0.4 + 1.2 + 1.8 = 3.4.

Var(X) = 3.4 − (1.6)² = 3.4 − 2.56 = 0.84.
Problem 2: X ~ Po(6) and Y ~ N(10, 4) are independent. Find E(3X − 2Y) and Var(3X − 2Y).

题目2：已知 X ~ Po(6) 与 Y ~ N(10, 4) 相互独立。求 E(3X − 2Y) 和 Var(3X − 2Y)。

E(3X − 2Y) = 3(6) − 2(10) = 18 − 20 = −2.

Var(3X − 2Y) = 9Var(X) + 4Var(Y) = 9(6) + 4(4) = 54 + 16 = 70.
Problem 3: The number of defects in a length of fabric follows a Poisson distribution with mean 0.5 per metre. Find the mean and standard deviation of the number of defects in a 10-metre length.

题目3：某织物上的缺陷数服从泊松分布，每米平均 0.5 个。求 10 米长度上缺陷数的均值与标准差。

For 10 metres, λ = 0.5 × 10 = 5, so E(X) = 5 and Var(X) = 5, thus SD(X) = √5 ≈ 2.236.

对于 10 米，λ = 0.5 × 10 = 5，因此 E(X) = 5，Var(X) = 5，故 SD(X) = √5 ≈ 2.236。

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