Year 10 CCEA Statistics: Quick Reference Formula & Theorem Handbook | CCEA Year 10 统计公式定理速查手册

📚 Year 10 CCEA Statistics: Quick Reference Formula & Theorem Handbook | CCEA Year 10 统计公式定理速查手册

This handbook brings together every essential formula and theorem you need to master CCEA Year 10 Statistics. Whether you are calculating averages, measuring spread, or working with probability rules, this page gives you clear paired explanations and ready‑to‑use notation. Keep it handy while revising, doing homework, or preparing for end‑of‑topic tests.

本手册汇集了掌握 CCEA Year 10 统计学所需的所有核心公式和定理。无论你在计算平均数、衡量离散程度，还是应用概率规则，本页都提供了清晰配对的解释和即用的符号。复习、写作业或准备单元测验时，请随时参考。

1. Mean of Raw Data | 原始数据的平均数

The arithmetic mean (often just called the average) is found by adding all the data values and dividing by the number of values. It is the most common measure of central tendency and works well when the data has no extreme outliers.

算术平均数（通常简称为平均值）是将所有数据值相加后除以数据个数得到的。它是最常用的集中趋势度量，在数据没有极端离群值时效果很好。

x̄ = Σx / n

Here Σx is the sum of all observations and n is the total number of observations. For example, the mean of 3, 7, 8, 11 is (3+7+8+11) ÷ 4 = 7.25.

其中 Σx 是所有观测值的总和，n 是观测值的总数。例如，3、7、8、11 的平均值为 (3+7+8+11) ÷ 4 = 7.25。

2. Median and Mode | 中位数与众数

The median is the middle value when the data is arranged in order. If there is an odd number of observations, the median is the centre value. If n is even, the median is the mean of the two centre values. The mode is simply the value that occurs most frequently.

中位数是将数据排序后位于中间的值。若观测值个数为奇数，中位数就是正中间的那个值；若 n 为偶数，中位数则是中间两个值的平均数。众数就是出现次数最多的值。

Median position = (n + 1) / 2

For the data set 2, 3, 5, 5, 7, the median is the 3rd value = 5, and the mode is 5. For 1, 4, 6, 9, the median is (4+6)/2 = 5, and there is no mode.

对于数据集 2、3、5、5、7，中位数为第 3 个值 5，众数为 5。对于 1、4、6、9，中位数为 (4+6)/2 = 5，且没有众数。

3. Range and Interquartile Range (IQR) | 极差与四分位距

The range is the simplest measure of spread: it is the difference between the largest and smallest values. The interquartile range gives a more robust view of spread by focusing on the middle 50% of the data.

极差是最简单的离散度量：它是最大值与最小值的差。四分位距通过关注中间 50% 的数据，提供了更稳健的离散程度视角。

Range = Xₘₐₓ − Xₘᵢₙ

IQR = Q₃ − Q₁

Q₁ (lower quartile) is the median of the lower half of the data, and Q₃ (upper quartile) is the median of the upper half. For the ordered list 2, 4, 5, 8, 11, 13, 14, the range is 14−2=12, Q₁=4.5, Q₃=12, so IQR=7.5.

下四分位数 Q₁ 是数据下半部分的中位数，上四分位数 Q₃ 是数据上半部分的中位数。对于排序后的列表 2、4、5、8、11、13、14，极差为 14−2=12，Q₁=4.5，Q₃=12，因此 IQR=7.5。

4. Variance and Standard Deviation (Population) | 总体方差与标准差

Variance and standard deviation measure how spread out the data are around the mean. A small standard deviation means the data points are close to the mean; a large one indicates they are more spread out.

方差和标准差衡量数据围绕平均值的分散程度。标准差小表示数据点接近平均值；标准差大则表示数据分布更分散。

σ² = Σ(x − x̄)² / n

σ = √[Σ(x − x̄)² / n]

For a population of size n, calculate each deviation (x−x̄), square it, sum the squares, divide by n, then take the square root for standard deviation. In CCEA Year 10, you may also meet the sample standard deviation where division is by (n−1). Always check whether the question refers to a population or a sample.

对于大小为 n 的总体，计算每个离差 (x−x̄)，将其平方后求和，除以 n，然后开平方即得标准差。在 CCEA Year 10 中，你也可能遇到样本标准差，其分母为 (n−1)。请务必确认题目指的是总体还是样本。

5. Mean from a Frequency Table | 由频数表求平均值

When data is grouped into a frequency table, we multiply each value by its frequency, sum these products, and divide by the total frequency. For grouped continuous data, use the midpoint of each class interval as x.

当数据被整理成频数表时，我们将每个值乘以其频数，求和后除以总频数。对于分组连续数据，用每个组区间的中点值作为 x。

x̄ = Σ(f × x) / Σf

Score (x)	Frequency (f)	f × x
2	3	6
5	4	20
8	2	16

Here Σf = 9, Σ(f × x) = 6+20+16 = 42, so the mean is 42 ÷ 9 ≈ 4.67.

这里 Σf = 9，Σ(f × x) = 6+20+16 = 42，因此平均值为 42 ÷ 9 ≈ 4.67。

6. Basic Probability | 基础概率

Probability measures how likely an event is to happen. It is always a number between 0 (impossible) and 1 (certain). For equally likely outcomes, the probability of an event A is the number of favourable outcomes divided by the total number of possible outcomes.

概率衡量事件发生的可能性，它是介于 0（不可能）和 1（必然）之间的一个数。对于等可能结果，事件 A 的概率等于有利结果的数目除以所有可能结果的总数。

P(A) = Number of favourable outcomes / Total number of possible outcomes

If a fair six‑sided die is rolled, P(rolling a 4) = 1/6. The sum of probabilities of all mutually exclusive outcomes in a sample space is 1.

如果掷一枚公平的六面骰子，P(掷出 4 点) = 1/6。样本空间中所有互斥结果的概率之和为 1。

7. Addition Rule for Probability | 概率的加法法则

For mutually exclusive events (events that cannot happen at the same time), the probability that either A or B occurs is simply the sum of their individual probabilities. When events are not mutually exclusive, we must subtract the intersection to avoid double counting.

对于互斥事件（不能同时发生的事件），A 或 B 发生的概率就是它们各自概率之和。当事件不是互斥时，必须减去交集部分以避免重复计算。

Mutually exclusive: P(A ∪ B) = P(A) + P(B)

Not mutually exclusive: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Drawing a king or a queen from a deck of cards are mutually exclusive, so P(king ∪ queen) = 4/52 + 4/52 = 8/52. Drawing a king or a heart are not mutually exclusive, so P(king ∪ heart) = 4/52 + 13/52 − 1/52 = 16/52.

从一副扑克牌中抽到 K 或 Q 是互斥事件，因此 P(K ∪ Q) = 4/52 + 4/52 = 8/52。抽到 K 或红桃则不是互斥的，因此 P(K ∪ 红桃) = 4/52 + 13/52 − 1/52 = 16/52。

8. Multiplication Rule and Independent Events | 乘法法则与独立事件

Two events are independent if the occurrence of one does not affect the probability of the other. For independent events A and B, the probability that both occur is the product of their individual probabilities.

如果一个事件的发生不影响另一个事件发生的概率，则这两个事件是独立的。对于独立事件 A 和 B，两者同时发生的概率是各自概率的乘积。

P(A ∩ B) = P(A) × P(B)

Tossing a coin and rolling a die are independent. The probability of getting heads and a 6 is P(heads) × P(6) = ½ × ⅙ = 1/12. Always confirm independence before multiplying.

抛硬币和掷骰子是独立事件。得到正面且掷出 6 点的概率为 P(正面) × P(6) = ½ × ⅙ = 1/12。在相乘之前务必确认事件独立性。

9. Conditional Probability | 条件概率

Conditional probability is the probability of event A given that event B has already happened. It allows us to update our probability estimate based on new information.

条件概率是指在事件 B 已经发生的前提下事件 A 发生的概率。它让我们能够基于新信息更新概率估计。

P(A | B) = P(A ∩ B) / P(B)

For example, in a class of 30 students, 12 study French and 8 study both French and Spanish. If a student studies French, the probability they also study Spanish is 8/12 = 2/3. This formula is the foundation of tree‑diagram calculations.

例如，某班有 30 名学生，其中 12 人学习法语，8 人同时学习法语和西班牙语。如果一名学生学习法语，那么他也学习西班牙语的概率为 8/12 = 2/3。该公式是树形图计算的基础。

10. Probability Tree Diagrams | 概率树形图

A tree diagram shows all possible outcomes of two or more events and their probabilities. The probabilities along each branch must sum to 1. To find the probability of a combined event, multiply the probabilities along its branch. To find the probability of an event that can occur in several ways, add the final probabilities of the relevant branches.

树形图展示两个或多个事件的所有可能结果及其概率。每条分支上的概率之和必须为 1。要找到复合事件的概率，沿着其分支将概率相乘。若某一事件可通过多种方式发生，则将相关分支的最终概率相加。

A bag contains 3 red and 2 blue counters. Two counters are drawn without replacement. The tree diagram will have first‑stage branches with P(R)=3/5 and P(B)=2/5. The second‑stage probabilities change depending on the first outcome. For example, P(B | B) = 1/4. Thus P(B then B) = 2/5 × 1/4 = 2/20 = 1/10.

一个袋子里有 3 个红色和 2 个蓝色计数器，无放回地抽取两次。树形图第一级分支为 P(红)=3/5 和 P(蓝)=2/5。第二级概率取决于第一次的结果，例如 P(第二次蓝 | 第一次蓝) = 1/4。因此 P(蓝 then 蓝) = 2/5 × 1/4 = 2/20 = 1/10。

11. Counting Principles: Permutations | 计数原理：排列

A permutation is an arrangement of objects in a specific order. The number of ways to arrange n distinct objects in a line is n factorial. When only r objects are chosen from n and the order matters, we use the permutation formula.

排列是指对象按特定顺序的排列方式。将 n 个不同对象排成一列的方式数为 n 的阶乘。当从 n 个对象中选取 r 个且顺序重要时，我们使用排列公式。

n! = n × (n−1) × … × 2 × 1

ⁿPᵣ = n! / (n−r)!

How many ways can a gold, silver and bronze medal be awarded to 8 athletes? Here order matters, so ⁸P₃ = 8! / 5! = 8×7×6 = 336. CCEA Year 10 may also introduce the use of nPr on a calculator.

将金、银、铜牌授予 8 名运动员有多少种方式？这里顺序重要，因此 ⁸P₃ = 8! / 5! = 8×7×6 = 336。CCEA Year 10 也可能介绍在计算器上使用 nPr 功能。

12. Counting Principles: Combinations | 计数原理：组合

A combination is a selection of objects where the order does not matter. The number of ways to choose r objects from n distinct objects is given by the binomial coefficient, often read as ‘n choose r’.

组合是指顺序不重要的对象选取方式。从 n 个不同对象中选取 r 个的方式数由二项式系数给出，通常读作“n 选 r”。

ⁿCᵣ = n! / [r! × (n−r)!]

Choosing a committee of 4 people from 10 candidates does not require an order, so the number of ways is ¹⁰C₄ = 10! / (4! × 6!) = (10×9×8×7) / (4×3×2×1) = 210. Combinations are often used in probability questions where you count the number of favourable selections.

从 10 名候选人中选出 4 人组成委员会不需要考虑顺序，因此方式数为 ¹⁰C₄ = 10! / (4! × 6!) = (10×9×8×7) / (4×3×2×1) = 210。组合常用于需要计算有利选取数目的概率问题中。

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Year 10 CCEA Statistics: Quick Reference Formula & Theorem Handbook | CCEA Year 10 统计公式定理速查手册