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A-Level Further Mathematics: Difference of Means of Two Independent Normal Distributions | A-Level 进阶数学:两独立正态分布均值之差

📚 A-Level Further Mathematics: Difference of Means of Two Independent Normal Distributions | A-Level 进阶数学:两独立正态分布均值之差

In many real-world experiments, we compare two populations by looking at the difference between their means. For example, do students using teaching method A score higher on average than students using method B? If both populations are normal and the samples are independent, we can model the difference between the sample means as a single normal variable. This article explains the key results, derivations, and exam-style applications for Edexcel A-Level Further Mathematics.

在许多现实实验中,我们需要通过比较两个总体的均值来得出结论。例如,使用方法A的学生是否在平均分上高于使用方法B的学生?如果两个总体都服从正态分布,且样本相互独立,我们就可以把样本均值之差看作一个单一的正态变量。本文将讲解 Edexcel 进阶数学中与此相关的核心结论、推导过程以及考试型应用。


1. Why the Difference is Normal | 为什么差值服从正态分布

The most important fact is that a linear combination of independent normal random variables is itself normally distributed. This is a special property of the normal distribution and is one of the reasons it is so widely used in statistical modelling.

最重要的事实是:独立正态随机变量的线性组合仍然服从正态分布。这是正态分布的一个特殊性质,也是它在统计建模中被广泛使用的原因之一。

Suppose X ~ N(μ_X, σ_X²) and Y ~ N(μ_Y, σ_Y²), with X and Y independent. Then the difference X − Y obeys:

设 X ~ N(μ_X, σ_X²) 与 Y ~ N(μ_Y, σ_Y²) 相互独立,则差值 X − Y 服从:

X − Y ~ N(μ_X − μ_Y, σ_X² + σ_Y²)

Notice that the variance of the difference is the sum of the two variances, not the difference. This is a common trap: subtracting one variable from another does not reduce uncertainty; it increases it.

注意:差值的方差是两个方差之和,而不是两个方差之差。这是一个常见陷阱:将一个变量减去另一个变量并不会减少不确定性,反而会增加不确定性。


2. The Distribution of the Difference Between Sample Means | 样本均值之差的分布

In exam questions, we are usually comparing two sample means rather than two individual observations. Let X₁, X₂, …, Xₙ be a random sample from N(μ_X, σ_X²), and let Y₁, Y₂, …, Yₘ be an independent random sample from N(μ_Y, σ_Y²). The sample means are:

在考试题中,我们通常比较两个样本均值,而不是两个单个观测值。设 X₁, X₂, …, Xₙ 是来自 N(μ_X, σ_X²) 的随机样本,Y₁, Y₂, …, Yₘ 是来自 N(μ_Y, σ_Y²) 的独立随机样本,则样本均值为:

X̄ = (X₁ + X₂ + … + Xₙ) / n, Ȳ = (Y₁ + Y₂ + … + Yₘ) / m

The sampling distributions of the sample means are:

这两个样本均值的抽样分布为:

Quantity Mean Variance
μ_X σ_X² / n
Ȳ μ_Y σ_Y² / m
X̄ − Ȳ μ_X − μ_Y σ_X² / n + σ_Y² / m

Therefore, the sampling distribution of X̄ − Ȳ is:

因此,X̄ − Ȳ 的抽样分布为:

X̄ − Ȳ ~ N(μ_X − μ_Y, σ_X²/n + σ_Y²/m)

The negative sign in front of Ȳ does not change the variance formula, because squaring the coefficient −1 gives +1.

Ȳ 前面的负号不会改变方差公式,因为系数 −1 的平方等于 +1。


3. Deriving the Mean | 均值的推导

The mean of the difference follows directly from the linearity of expectation. For any random variables A and B, E[A − B] = E[A] − E[B]. Applying this to the sample means gives:

差值均值可以直接由期望的线性性质得到。对任意随机变量 A 和 B,有 E[A − B] = E[A] − E[B]。将其应用于样本均值,得到:

E[X̄ − Ȳ] = E[X̄] − E[Ȳ] = μ_X − μ_Y

This makes intuitive sense: the average difference between the sample means should be centred at the difference between the population means.

这很符合直觉:样本均值之差的平均值应以总体均值之差为中心。


4. Deriving the Variance | 方差的推导

For independent random variables A and B, Var(A − B) = Var(A) + Var(B). More generally, for constants a and b:

对独立随机变量 A 和 B,有 Var(A − B) = Var(A) + Var(B)。更一般地,对常数 a 和 b:

Var(aA + bB) = a²Var(A) + b²Var(B)

If A and B are not independent, an extra covariance term appears. Independence is therefore essential. For the difference of sample means, taking a = 1 and b = −1 gives:

如果 A 与 B 不独立,则会出现额外的协方差项,因此独立性至关重要。对于样本均值之差,取 a = 1,b = −1,得到:

Var(X̄ − Ȳ) = Var(X̄) + (−1)²Var(Ȳ) = σ_X²/n + σ_Y²/m

This is why the two variances are added even though we are subtracting the means.

这就是为什么虽然我们在做均值相减,但两个方差仍然要相加。


5. Standardising the Difference | 对差值进行标准化

Once we know that X̄ − Ȳ is normal, we can calculate probabilities by standardising. If we know the population variances, the standardised statistic is:

一旦知道 X̄ − Ȳ 服从正态分布,我们就可以通过标准化来计算概率。如果我们已知总体方差,标准化统计量为:

Z = [(X̄ − Ȳ) − (μ_X − μ_Y)] / √(σ_X²/n + σ_Y²/m)

This Z value follows a standard normal distribution, N(0, 1). In exam solutions, always show the denominator clearly before using the normal distribution table.

该 Z 值服从标准正态分布 N(0, 1)。在考试解答中,务必清楚写出分母,然后再使用正态分布表。


6. Worked Example: Probability | 例题:求概率

The independent random variables X ~ N(80, 12²) and Y ~ N(75, 10²). Let D = X − Y. Find P(D > 0).

设独立随机变量 X ~ N(80, 12²),Y ~ N(75, 10²)。令 D = X − Y,求 P(D > 0)。

Step 1: Write down the distribution of D.

第一步:写出 D 的分布。

D ~ N(80 − 75, 12² + 10²) = N(5, 244)

Step 2: Standardise the value 0.

第二步:将数值 0 标准化。

P(D > 0) = P(Z > (0 − 5) / √244) = P(Z > −0.320)

Step 3: Use the normal distribution table. Since P(Z > −0.320) = P(Z < 0.320), the required probability is approximately 0.6255.

第三步:查标准正态分布表。因为 P(Z > −0.320) = P(Z < 0.320),所以所求概率约为 0.6255。


7. Linear Combinations of Independent Normal Variables | 独立正态变量的线性组合

The result for the difference of two variables is a special case of a more general theorem. If X₁, X₂, …, Xₙ are independent normal variables and a₁, a₂, …, aₙ are constants, then:

两个变量之差的分布是一个更一般定理的特例。如果 X₁, X₂, …, Xₙ 是独立正态变量,a₁, a₂, …, aₙ 是常数,则:

Σ aᵢXᵢ ~ N(Σ aᵢμᵢ, Σ aᵢ²σᵢ²)

For a single sample mean, we take aᵢ = 1/n for every observation. If Xᵢ ~ N(μ, σ²), then:

对于单个样本均值,我们取每个观测值的系数 aᵢ = 1/n。若 Xᵢ ~ N(μ, σ²),则:

X̄ = (1/n) Σ Xᵢ ~ N(μ, σ²/n)

This general formula is powerful because it also applies to sums and differences of more than two independent sample means.

这个通用公式非常强大,因为它也适用于两个以上独立样本均值之和与差。


8. Comparing Two Sample Means | 比较两个样本均值

Suppose machine A produces components with diameter X ~ N(10.0, 0.20²), and machine B produces components with diameter Y ~ N(9.9, 0.30²). Independent samples of sizes n = 25 and m = 36 are taken. Find P(X̄ > Ȳ).

假设机器A生产的零件直径 X ~ N(10.0, 0.20²),机器B生产的零件直径 Y ~ N(9.9, 0.30²)。分别取独立样本 n = 25,m = 36。求 P(X̄ > Ȳ)。

The difference D = X̄ − Ȳ has mean 10.0 − 9.9 = 0.1 and variance:

差值 D = X̄ − Ȳ 的均值为 10.0 − 9.9 = 0.1,方差为:

0.20²/25 + 0.30²/36 = 0.0016 + 0.0025 = 0.0041

P(D > 0) = P(Z > (0 − 0.1) / √0.0041) = P(Z > −1.561)

From the standard normal table, this probability is approximately 0.9408.

查标准正态分布表,这个概率约为 0.9408。


9. Hypothesis Testing for μ_X − μ_Y | 关于 μ_X − μ_Y 的假设检验

The same standardised statistic is used in hypothesis testing. For example, to test whether two population means are equal, we set:

同样的标准化统计量也用于假设检验。例如,要检验两个总体均值是否相等,我们设定:

H₀: μ_X − μ_Y = 0

H₁: μ_X − μ_Y > 0Published by TutorHao | A-Level 进阶数学 Revision Series | aleveler.com

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