OxfordAQA FM03 Statistics Final Mark Scheme v1.0 Key Concepts | OxfordAQA FM03 统计评分标准重要知识点精讲

📚 OxfordAQA FM03 Statistics Final Mark Scheme v1.0 Key Concepts | OxfordAQA FM03 统计评分标准重要知识点精讲

The OxfordAQA Further Mathematics Unit 03 (FM03) paper focuses on Statistics, and the final mark scheme for June 2023 (v1.0) provides clear insights into how examiners assess key concepts. This article breaks down the core topics and markscheme requirements, helping students understand exactly what is needed to score high marks.

OxfordAQA 高等数学第三单元（FM03）聚焦统计学，2023年6月最终评分标准（v1.0）清晰展示了考官如何评价关键知识点。本文将剖析核心主题与评分要求，帮助学生准确掌握高分要领。

1. Poisson Distribution: Definition and Conditions | 泊松分布：定义与条件

The Poisson distribution models the number of random events occurring in a fixed interval of time or space. It is characterised by the parameter λ, which represents the mean rate of occurrence.

泊松分布用于模拟固定时间或空间间隔内随机事件发生的次数，由参数 λ 表征，λ 表示事件发生的平均速率。

The probability mass function is: P(X = x) = (e⁻λ λˣ) / x! for x = 0,1,2,… For example, if λ = 3, then P(X = 2) = e⁻³ × 3² / 2! ≈ 0.2240.

概率质量函数为：P(X = x) = (e⁻λ λˣ) / x!，其中 x = 0,1,2,…。例如，当 λ = 3 时，P(X = 2) = e⁻³ × 3² / 2! ≈ 0.2240。

Key conditions are: events occur singly in an interval, they are independent, and the rate λ is constant. In the FM03 markscheme, always state these conditions when justifying a Poisson model to secure the method marks.

关键条件是：事件在区间内单个发生、彼此独立，且速率 λ 恒定。在 FM03 评分标准中，在证明泊松模型适用性时务必陈述这些条件以获取方法分。

2. Poisson Approximation to the Binomial | 泊松近似二项分布

When n is large and p is small, a binomial distribution B(n, p) can be approximated by a Poisson distribution with λ = np. The approximation is valid when n > 50 and np < 5, or more liberally n > 100 and np < 10.

当 n 很大且 p 很小时，二项分布 B(n, p) 可用参数 λ = np 的泊松分布近似。近似有效条件通常为 n > 50 且 np < 5，或更宽松地 n > 100 且 np < 10。

Define λ = np explicitly. For instance, if X ~ B(200, 0.02) then λ = 4 and X ≈ Po(4). The mark scheme rewards this clear declaration.

需明确写出 λ = np。例如，若 X ~ B(200, 0.02)，则 λ = 4 且 X ≈ Po(4)。评分标准奖励这种清晰声明。

3. Continuous Random Variables: PDF and CDF | 连续随机变量：概率密度函数与累积分布函数

A continuous random variable X has a probability density function f(x) where the total area under the curve equals 1. The cumulative distribution function is F(x) = P(X ≤ x) = ∫₋∞ˣ f(t) dt.

连续随机变量 X 具有概率密度函数 f(x)，曲线下总面积为 1。累积分布函数为 F(x) = P(X ≤ x) = ∫₋∞ˣ f(t) dt。

To find probabilities, integrate f(x) or use F(x). For example, if f(x) = 2x for 0 < x < 1, then F(x) = x² for 0 ≤ x ≤ 1. P(X < 0.5) = F(0.5) = 0.25. The median m solves F(m) = 0.5, so m² = 0.5, m = √0.5 ≈ 0.707.

若要计算概率，可积分 f(x) 或使用 F(x)。例如，若 f(x) = 2x，0 < x < 1，则 F(x) = x²，0 ≤ x ≤ 1。P(X < 0.5) = F(0.5) = 0.25。中位数 m 满足 F(m) = 0.5，即 m² = 0.5，m = √0.5 ≈ 0.707。

Mark schemes penalise forgetting piecewise domains and incorrect integration limits. Always check that F(x) increases to 1.

评分标准中对忽略分段定义域和积分限错误会扣分。一定要校验 F(x) 是否递增至 1。

4. Expectation and Variance of Continuous Random Variables | 连续随机变量的期望与方差

The expected value E(X) of a continuous random variable is given by ∫ x f(x) dx over the domain. Variance is Var(X) = ∫ (x – μ)² f(x) dx = E(X²) – [E(X)]².

连续随机变量的期望 E(X) 为 ∫ x f(x) dx 在整个定义域积分。方差为 Var(X) = ∫ (x – μ)² f(x) dx = E(X²) – [E(X)]²。

For the previous f(x) = 2x on (0,1), E(X) = ∫₀¹ 2x² dx = 2/3, E(X²) = ∫₀¹ 2x³ dx = 1/2, so Var(X) = 1/2 – (2/3)² = 1/18.

对于之前 f(x) = 2x 在 (0,1) 上的例子，E(X) = ∫₀¹ 2x

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