Investigation 3 - The mean and standard deviation of a binomial distribution | 探究3：二项分布的均值与标准差

📚 Investigation 3 - The mean and standard deviation of a binomial distribution | 探究3：二项分布的均值与标准差

d count the total number of successes. While the scenario may be simple, two key summaries — the mean and the standard deviation — give us deep insight into what we should expect 'on average' and build an intuitive understanding of why they take the forms μ = np and σ = √(np(1-p)).

二项分布源自于将独立的伯努利试验重复固定次数并统计成功总次数。虽然情景简单，但两个关键汇总量——均值和标准差——能让我们深刻理解“平均而言”应期待什么，以及典型的变异程度。在本次探究中，我们将推导这些度量，利用模拟数据进行探索，并建立对为何μ = np、σ = √(np(1-p))的直观理解。

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1. Recap of the Binomial Setting | 二项分布条件回顾

A random variable X follows a binomial distribution, denoted X ~ B(n, p), if it counts the number of successes in n independent trials, each having the same probability of success p. The four essential conditions are: a fixed number of trials n, only two possible outcomes per trial (success or failure), a constant probability p of success, and independent trials.

若随机变量X表示在n次独立试验中成功的次数，且每次试验成功的概率p相同，则X服从二项分布，记作X ~ B(n, p)。四个基本条件是：试验次数n固定、每次试验仅有两种可能结果（成功或失败）、成功概率p恒定、以及各次试验相互独立。

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2. Defining Mean and Standard Deviation | 均值与标准差的定义

dard deviation σ = √Var(X) measures typical deviation from the mean in the original units.

离散随机变量X的均值（或期望值）为E(X) = Σ x·P(X = x)，表示长期平均结果。方差Var(X) = E[(X − μ)²]量化离散程度，而标准差σ = √Var(X)以原单位度量偏离均值的典型幅度。

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3. Intuitive Idea of the Binomial Mean | 二项分布均值的直观理解

s naturally n × p. This linear scaling reflects the additive nature of independent trials.

若抛掷一枚均匀硬币100次，直觉告诉自己大约期待50次正面。每次试验成功概率为p时，n次试验中预期的成功次数自然为n × p。这种线性倍增反映了独立试验的可加性。

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4. Derivation of the Mean using Indicator Variables | 借助指示变量推导均值

(X) = E(I₁) + E(I₂) + ... + E(Iₙ) = np. No independence assumption is needed for the mean.

为第i次试验定义指示变量Iᵢ：若试验成功则Iᵢ = 1，否则为0。于是E(Iᵢ) = 1·p + 0·(1-p) = p。因为X = I₁ + I₂ + ... + Iₙ，由期望的线性性得E(X) = E(I₁) + E(I₂) + ... + E(Iₙ) = np。推导均值时不需要独立性假设。

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5. Example Calculation of the Mean | 均值的计算示例

a large number of 30-shot series, the long-term average number of hits will be close to 6.

假设一名弓箭手命中靶心的概率p = 0.2，射击30次。预期命中次数为μ = 30 × 0.2 = 6。在进行大量30箭为一系列的实验时，长期平均命中数将接近6。

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6. Intuition behind the Binomial Variance | 二项分布方差的直观基础

p(1-p). Note that the variance is largest when p = 0.5, where outcomes are most uncertain.

二项计数的变异性源自每次试验的不可预测性。对单次试验，方差为p(1-p)。对多次独立试验，总方差是各次方差之和，由此得到σ² = np(1-p)。注意当p = 0.5时方差最大，此时结果最不确定。

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7. Derivation of the Variance | 方差的推导

les, so Var(X) = Σ Var(Iᵢ) = n·p(1-p). Therefore the standard deviation is σ = √(np(1-p)).

对于指示变量Iᵢ，Var(Iᵢ) = E(Iᵢ²) − [E(Iᵢ)]² = p − p² = p(1-p)。由于各试验相互独立，指示变量也相互独立，故Var(X) = Σ Var(Iᵢ) = n·p(1-p)。因此标准差为σ = √(np(1-p))。

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8. Worked Example: Mean and Standard Deviation | 计算实例：均值和标准差

A die is rolled 60 times; success is rolling a '6' The variance is 60 × (1/6) × (5/6) = 60 × 5/36 = 300/36 ≈ 8.33. The standard deviation is √8.33 e between 7 and 13 successes most of the time (within one standard deviation of the mean).

掷一粒骰子60次，设掷出“6”为成功（p = 1/6）。n = 60，故μ = 60 × (1/6) = 10。方差为60 × (1/6) × (5/6) = 60 × 5/36 = 300/36 ≈ 8.33。标准差约为√8.33 ≈ 2.89。通常多数时候会观察到介于7到13次成功（均值加减一个标准差范围内）。

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9. Investigating with Simulated Data | 利用模拟数据探究

eans will cluster around 30 × 0.3 = 9, and the sample standard deviations will approximate √30 × 0.3 × 0.7 rious n and p confirms the formulas and helps visualize the effect of changing parameters.

使用GDC或软件，从B(30, 0.3)中生成200个样本，并计算样本均值与标准差。这些样本均值将聚集在30 × 0.3 = 9附近，样本标准差将接近√30 × 0.3 × 0.7 ≈ 2.51。对不同的n和p重复实验可印证公式，并帮助直观感受参数变化的影响。

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10. Effect of n and p on the Distribution Shape | n和p对分布形态的影响

a fixed p, the mean np increases linearly, and the standard deviation √(np(1-p)) grows as √n. The distribution becomes more symmetric and bell-shaped when np and n(1-p) are both sufficiently large (typically >5), linking to the normal approximation.

固定p时，随着n增大，均值np线性增大，标准差√(np(1-p))以√n的速度增长。当np和n(1-p)都足够大（通常大于5）时，分布形态会变得更对称并呈钟形，这与正态近似相关联。

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11. Connection to Sampling Proportions | 与样本比例的关系

If we consider the sample proportion of successes p̂ = X/n, then E(p̂) = p and Var(p̂) = p(1-p)/n, giving standard deviation σ_p̂ proportion is fundamental in confidence intervals and hypothesis testing for proportions.

若考虑样本成功比例p̂ = X/n，则E(p̂) = p，Var(p̂) = p(1-p)/n，标准差为σ_p̂ = √(p(1-p)/n)。这个比例的标准误差在比例的置信区间和假设检验中至关重要。

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12. Real-world Application and Conclusion | 现实应用与总结

np and σ = √(np(1-p)) equips us with powerful tools for decision-making under uncertainty.

二项分布的均值和标准差使制造商能够预测次品率，让民调机构估计误差幅度，并让健康科研人员对响应治疗的患者人数建模。理解了μ = np和σ = √(np(1-p))，我们便掌握了在不确定性下进行决策的有力工具。

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