📚 Year 10 SQA Statistics: International Competition Preparation Guide | Year 10 SQA 统计：国际竞赛备战攻略
Statistics and probability questions appear in almost every international mathematics competition, from UKMT to AMC and beyond. Building a strong foundation with the Year 10 SQA Statistics curriculum gives you the perfect launchpad to tackle these challenges with confidence. This guide walks you through the essential statistical concepts and shows how to apply them to competition-style problems, helping you sharpen your analytical skills and avoid common traps.
统计与概率类题目几乎出现在所有国际数学竞赛中，无论是 UKMT 还是 AMC 都不例外。Year 10 SQA 统计课程为你打下扎实的基础，让你能够自信地应对这些挑战。本攻略将带你梳理核心统计概念，并展示如何将其应用到竞赛题目中，帮你提升分析能力、避开常见陷阱。
1. Understanding Probability Fundamentals | 理解概率基础
tions, decimals and percentages, and you learn the rule that for equally likely outcomes, probability = favourable outcomes / total outcomes.
概率衡量事件发生的可能性，用一个介于 0（不可能）到 1（必然）之间的数表示。在 SQA 统计中，你使用分数、小数和百分比，并学习对于等可能结果，概率 = 有利结果数 / 总结果数。
Example: A jar contains 5 green, 3 yellow and 2 blue sweets. Find the probability of taking a yellow sweet at random.
例题：一个罐子里有 5 颗绿色糖、3 颗黄色糖和 2 颗蓝色糖。随机取出一颗糖，求拿到黄色糖的概率。
Solution: Total sweets = 5 + 3 + 2 = 10. P(yellow) = 3/10 = 0.3.
解答：总糖数 = 5 + 3 + 2 = 10。P(黄色) = 3/10 = 0.3。
In competitions, always check whether outcomes are equally likely. Misinterpreting 'at random' is a classic error.
在竞赛中，务必核实结果是否等可能。误判“随机”一词是典型的错误。
2. Counting Principles and Combinatorics | 计数原理与组合
occur in m ways and another in n ways, the number of ways both events can occur is m × n.
计数问题是许多竞赛概率题的骨架。基本计数原理指出，如果一个事件有 m 种发生方式，另一个有 n 种，则两个事件连续发生的方式数为 m × n。
ons. The number of ways to choose r items from n is written as C(n, r) or nCr and equals n! / (r! × (n-r)!).
在顺序不重要时，使用组合。从 n 个物品中选 r 个的方法数记作 C(n, r) 或 nCr，等于 n! / (r! × (n-r)!)。
Example: How many different teams of 3 can be formed from 8 players?
例题：从 8 名球员中选出 3 人组成一队，共有多少种不同选法？
Solution: C(8,3) = 8! / (3! × 5!) = (8×7×6) / (3×2×1) = 56.
解答：C(8,3) = 8! / (3! × 5!) = (8×7×6) / (3×2×1) = 56。
竞赛中常需要你把计数原理与概率结合，先算出总组合数再求概率。
In contests you will often combine counting with probability – find the total number of possibilities first, then calculate the target probability.
3. Conditional Probability and Independence | 条件概率与独立性
Conditional probability P(A| already occurred. It is calculated as P(A and B) / P(B). Two events are independent if P(A|B) = P(A), meaning the occurrence of B does not affect the probability of A.
条件概率 P(A|B) 是在事件 B 已经发生的条件下，事件 A 发生的概率，计算公式为 P(A 且 B) / P(B)。如果 P(A|B) = P(A)，则两事件独立，说明 B 的发生不影响 A 的概率。
Example: In a class, 60% study Art, 40% study Biology, and 25% study both. Find the probability that a student studies Art given they study Biology.
例题：某班 60% 的学生学艺术，40% 学生物，25% 两门都学。已知一个学生学生物，求该生学艺术的概率。
Solution: P(Art|Biology) = P(Art and Biology) / P(Biology) = 0.25 / 0.40 = 5/8 = 0.625.
解答：P(艺术|生物) = P(艺术且生物) / P(生物) = 0.25 / 0.40 = 5/8 = 0.625。
Recognising independence from a description or a table can save you from doing unnecessary calculations in a timed contest.
在限时竞赛中，能够从文字或表格中识别独立性，可以省去不必要的计算。
4. Tree Diagrams and Probability Tables | 树形图与概率表
Tree diagrams help visualise multi-stage experiments. You multiply probabilities along branches and add the probabilities of the relevant end results. SQA questions often involve probability trees with and without replacement.
树形图有助于直观展现多步试验。你沿着分支乘概率，然后将相关末端结果的概率相加。SQA 题目常常涉及放回与不放回的树形图。
Example: A bag contains 3 red and 5 blue discs. Two discs are drawn without replacement. Draw the tree and find P(both red).
例题：袋中有 3 个红色和 5 个蓝色碟片，不放回地抽取两次。画出树形图并求 P(两个都是红色)。
Solution: First draw: P(R) = 3/8. Second: P(R|R) = 2/7. Multiply: 3/8 × 2/7 = 6/56 = 3/28.
解答：第一次：P(红) = 3/8。第二次：P(红|第一次红) = 2/7。相乘：3/8 × 2/7 = 6/56 = 3/28。
竞赛中经常出现不放回的多步试验，此时条件概率就会改变。
Multi-stage experiments without replacement often appear in contests, where conditional probabilities shift at each step.
5. Descriptive Statistics: Mean, Median, Mode, Range | 描述性统计：均值、中位数、众数、极差
he middle value when data is ordered, and mode is the most frequent value. Range = maximum – minimum. SQA emphasises interpreting these measures, not just computing them.
均值是算术平均数，中位数是将数据排序后中间的值，众数是出现次数最多的值。极差 = 最大值 - 最小值。SQA 强调解读这些度量，而不仅仅是计算。
When a competition asks 'which average best describes this data?', consider outliers. The median is often a better summary when data is skewed.
若竞赛题问“哪一个平均数最能描述这组数据？”，考虑离群值。数据偏斜时，中位数通常是更好的概括。
Example: The weights of five packages are 2 kg, 3 kg, 4 kg, 4 kg, 100 kg. Which measure is misleading?
例题：五个包裹的重量为 2 kg、3 kg、4 kg、4 kg、100 kg。哪个度量具误导性？
n: Mean = 22.6 kg, heavily influenced by the outlier. Median = 4 kg gives a truer picture.
解答：均值 = 22.6 kg，受离群值严重影响。中位数 = 4 kg 更能反映实际情况。
6. Frequency Distributions and Histograms | 频数分布与直方图
Data can be organised into frequency tables. Histograms display the distribution, with bar area proportional to frequency. In SQA, you learn to estimate the mean and median from a grouped frequency table.
数据可以整理成频数表。直方图展示分布情况，柱子面积与频数成正比。在 SQA 课程中，你学习从分组频数表估算均值和中位数。
To estimate the mean from a grouped table: multiply each midpoint by its frequency, sum them, then divide by total frequency.
要从分组表格估算均值：将每组中点值乘以其频数，求和，再除以总频数。
Example: Estimate the mean from the table. (0-10: 5 items, 10-20: 8 items, 20-30: 7 items).
例题：根据表格估算均值。(0-10：5 个，10-20：8 个，20-30：7 个)。
n: Midpoints 5, 15, 25. Sum = 5×5 + 8×15 + 7×25 = 25 + 120 + 175 = 320. n = 20. Mean ≈ 16.
解答：中点 5、15、25。总和 = 5×5 + 8×15 + 7×25 = 320。n = 20。均值 ≈ 16。
Competition histograms may have unequal class widths – always check the frequency density.
竞赛中的直方图可能有不等的组距，务必检查频数密度。
7. Interquartile Range and Box Plots | 四分位距与箱线图
The interquartile range (IQR) = Q3 − Q1, measuring the spread of the middle 50% summary: minimum, Q1, median, Q3, maximum. They are excellent for comparing distributions.
四分位距 (IQR) = Q3 − Q1，衡量中间 50% 数据的分散程度。箱线图提供五数概括：最小值、Q1、中位数、Q3、最大值，非常适合比较分布。
mmetry, skewness and potential outliers (values more than 1.5 × IQR beyond the quartiles).
若竞赛题出现箱线图，留意对称性、偏斜度以及可能的离群值（超出四分位数 1.5 倍 IQR 的值）。
Example: Data has Q1=10, Q3=18. Is 32 an outlier? Check 1.5×IQR = 12, upper fence = 18+12=30. 32 > 30, so yes.
例题：数据 Q1=10，Q3=18。32 是离群值吗？检验 1.5×IQR=12，上须 = 18+12=30。32 > 30，所以是离群值。
8. Expected Value: Making Decisions | 期望值：决策依据
he long-run average outcome of a random process. SQA introduces this with spinners, dice and games of chance. It is central to many competition probability problems.
期望值 E(X) = Σ x·P(x) 表示随机过程的长期平均结果。SQA 通过转盘、骰子和机会游戏引入此概念。它也是许多竞赛概率问题的核心。
Example: A game costs £2 to play. You roll a fair die. If you roll a 6 you win £10; otherwise you win nothing. Is it worth playing?
例题：游戏费 2 英镑。掷一个公平骰子，掷出 6 赢 10 英镑，否则不赢钱。值不值得玩？
gain=10-2=8. P(lose)=5/6, net gain=−2. E(X) = (1/6)×8 + (5/6)×(−2) = 8/6 − 10/6 = −2/6 ≈ −£0.33. On average you lose 33p, so not favourable.
解答：P(赢)=1/6，净收益=8；P(输)=5/6，净收益=−2。E(X) = (1/6)×8 + (5/6)×(−2) = −2/6 ≈ −0.33 英镑。平均损失 33 便士，不划算。
竞赛常要求比较两个游戏哪个更优，正确计算期望值就能快速判断。
Contest items often ask you to compare two games – correct expected value computation gives a quick verdict.
9. Variance and Standard Deviation | 方差与标准差
ula σ² = Σ(x − μ)² / n for a population. These quantities describe how spread out data is.
方差衡量各数据与均值之差的平方的平均值。标准差 σ = √方差。SQA 中你使用总体公式 σ² = Σ(x − μ)² / n。这些量描述数据的离散程度。
. Squared deviations: (2−4)²=4, (4−4)²=0, (6−4)²=4. Variance = (4+0+4)/3 = 8/3 ≈ 2.67. σ ≈ √2.67 ≈ 1.63.
例题：数据：2, 4, 6。均值=4。方差 = (4+0+4)/3 = 8/3 ≈ 2.67，σ ≈ 1.63。
In competitions, knowing the relationship between range and standard deviation can help check for sensible answers.
在竞赛中，了解极差与标准差之间的关系有助于检查答案是否合理。
10. Introduction to the Binomial Distribution | 二项分布入门
is P(X=k) = C(n,k) pᵏ (1-p)ⁿ⁻ᵏ. While this may extend beyond Year 10, many competition problems assume basic binomial reasoning.
一项二项试验有固定的独立试验次数 n，每次成功概率 p 相同。恰好 k 次成功的概率为 P(X=k) = C(n,k) pᵏ (1-p)ⁿ⁻ᵏ。虽然这可能超出 Year 10 范围，但许多竞赛题假设你具备基本的二项推理能力。
Example: A fair coin is tossed 4 times. Find the probability of exactly 2 heads.
例题：抛一枚公平硬币 4 次，求恰好出现 2 次正面的概率。
Solution: n=4, k=2, p=0.5. C(4,2)=6. P = 6 × (0.5)² × (0.5)² = 6 × 0.25 × 0.25 = 0.375.
解答：n=4, k=2, p=0.5, C(4,2)=6。P = 6 × 0.5² × 0.5² = 0.375。
You can often short-cut binomial calculations by recognising symmetry or using known percentages.
通过识别对称性或使用已知百分比，你常常可以快速简化二项计算。
11. Interpreting Graphs and Statistical Charts | 统计图表的解读
Competition papers frequently present data as bar charts, pie charts, cumulative frequency graphs or scatter diagrams. Your SQA training in reading scales, identifying trends and extracting data points is invaluable.
竞赛试卷常以条形图、饼图、累积频数图或散点图呈现数据。你在 SQA 中学到的读取刻度、识别趋势及提取数据点的技能非常宝贵。
When a graph shows misleading scales or a broken axis, question the impression it creates – contest designers love testing data literacy.
当图表展示有误导性的刻度或断裂坐标轴时，要质疑其造成的印象——竞赛出题人喜欢测试数据素养。
Example: A bar chart shows sales of 200, 210, 220 over three months but the y-axis starts at 190, exaggerating the increase. The true percentage rise is only 10%.
例题：一个条形图展示三个月销售额为 200、210、220，但 y 轴从 190 开始，夸大了增幅。实际增长百分比仅为 10%。
12. Practice Strategies and Common Pitfalls | 练习策略与常见陷阱
To excel in competition statistics, mix timed past-paper drills with untimed conceptual deep dives. Always re-draw tree diagrams if you get confused, and double-check whether events are independent or conditional.
想在竞赛统计中出类拔萃，把限时的真题演练与不限时的概念深究结合起来。一旦思路混乱，马上重画树形图，并且反复核实事件是独立的还是条件相关的。
Common pitfalls include: confusing 'at least one' with 'exactly one', forgetting to square probabilities for combined events, and using the wrong formula for combinations when order matters.
常见陷阱包括：混淆“至少一个”和“恰好一个”，忘记对组合事件平方概率，以及顺序重要时误用组合公式。
Build a mistake log of every error you make in practice. Patterns will emerge, and you will learn to read questions more carefully.
建立一个错题本，记录练习中犯的每一个错误。规律会浮现出来，你会学会更加仔细地读题。
Published by TutorHao | Statistics Revision Series | aleveler.com
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