📚 Year 10 SQA Statistics: Case Study Practice and Applications | Year 10 SQA 统计：案例分析实战演练
Welcome to this focused revision resource for SQA Year 10 Statistics. The aim is to build your confidence through realistic case studies that mirror the style of exam questions. Each case study presents a scenario, requires you to select and apply appropriate statistical techniques, and helps you learn to interpret results in context. By working through these examples, you will sharpen your skills in handling data, calculating summary statistics, drawing diagrams, and understanding probability — all essential for success.
欢迎来到这篇专为 SQA Year 10 统计设计的复习文章。我们的目标是通过贴近真实考题风格的案例分析，帮助你建立信心。每一个案例都呈现一个具体情境，要求你选择合适的统计方法，并在情境中解释结果。通过演练这些例子，你将提升处理数据、计算汇总统计量、绘制图表和理解概率的能力——这些都是考试成功的关键。
1. Introduction to Case Study Approach | 案例研究方法导论
In SQA Statistics, you are often given a real‑life problem and asked to investigate it using data. A case study approach means you need to identify what question is being asked, decide which data are relevant, and then choose the most suitable statistical tools. Typical tools include measures of central tendency (mean, median, mode), measures of spread (range, interquartile range, standard deviation), graphical representations (bar charts, pie charts, scatter graphs), and probability calculations. Always keep the original context in mind when interpreting results — a mean height of 160 cm means something very different in a class of 11‑year‑olds than in a group of 17‑year‑olds.
在 SQA 统计考试中，你经常会遇到一个实际问题，要求用数据来探究。案例分析的方法意味着你需要明确被问及的问题，判断哪些数据是相关的，然后选择最合适的统计工具。常用的工具包括集中趋势的度量（平均数、中位数、众数）、离散程度的度量（全距、四分位距、标准差）、统计图示（条形图、饼图、散点图）以及概率计算。在解释结果时，务必始终结合原始背景——在11岁班级中得到的平均身高160厘米，与在17岁群体中得到的含义截然不同。
2. Case Study 1: Analysing Pupil Heights | 案例一：学生身高分析
A teacher measured the heights (in cm) of a sample of 14 Year 10 pupils. The data are: 155, 162, 158, 163, 149, 170, 165, 158, 161, 160, 157, 159, 163, 148. Your task is to describe the distribution of heights using appropriate statistics and to comment on any unusual values.
一位老师测量了14名Year 10学生的身高（单位：厘米）。数据如下：155, 162, 158, 163, 149, 170, 165, 158, 161, 160, 157, 159, 163, 148。你的任务是使用恰当的统计量来描述身高的分布，并对任何异常值进行评论。
4 = 2238 ÷ 14 = 159.9 cm. The median is the average of the 7th and 8th values: (159 + 160) ÷ 2 = 159.5 cm. The mode is 158 cm and 163 cm, both appearing twice. The range is 170 − 148 = 22 cm. Because the mean and median are very close, we can say the distribution is roughly symmetric. The value 170 looks slightly high, but with a range of 22 cm it is not an extreme outlier. You might also mention that the interquartile range would provide a better measure of spread, but the range gives a quick indication.
首先，将数据按顺序排列：148, 149, 155, 157, 158, 158, 159, 160, 161, 162, 163, 163, 165, 170。平均数计算为（所有身高之和）÷ 14 = 2238 ÷ 14 = 159.9 cm。中位数是第7和第8个值的平均数：（159 + 160）÷ 2 = 159.5 cm。众数是158 cm和163 cm，均出现两次。全距为 170 − 148 = 22 cm。由于平均数和中位数非常接近，我们可以说分布大致对称。170 这个值看起来略高，但考虑到全距为 22 cm，它并不是极端异常值。你还可以指出，四分位距能提供更好的离散度量，但全距给出了一个快速的把握。
When answering an exam question, always show your working clearly. For the mean, write the formula explicitly:
在考试作答时，务必清晰地展示计算过程。对于平均数，明确写出公式：
Mean = (x₁ + x₂ + … + xₙ) / n = 2238 / 14 ≈ 159.9 cm
3. Case Study 2: Examining Quiz Scores with a Frequency Table | 案例二：用频数表分析测验成绩
A quiz out of 10 marks was taken by 50 pupils. The results are grouped as follows: 0-2 marks (5 pupils), 3-5 (12), 6-8 (24), 9-10 (9). You need to estimate the mean score and find the modal class and the median class. Discuss which average is most representative.
一次满分为10分的测验有50名学生参加。成绩分组如下：0-2分（5人），3-5分（12人），6-8分（24人），9-10分（9人）。你需要估算平均分，并找出众数所在组和中位数所在组。讨论哪一个平均指标最具代表性。
d mean = 306.5 ÷ 50 = 6.13 marks. The modal class is 6-8 (the highest frequency). To find the median class, calculate cumulative frequencies: 5, 17, 41, 50. The 25th and 26th values fall in the 6-8 class, so the median class is also 6-8. Because the distribution is slightly skewed (more high scores), the mean may be pulled a little downwards, so the median class could be more representative. In context, most pupils scored between 6 and 8, suggesting the quiz was fairly easy.
对于分组数据，使用组中值：1, 4, 7, 9.5。将每个组中值乘以频数：1×5=5, 4×12=48, 7×24=168, 9.5×9=85.5。总和为 306.5。估计平均分 = 306.5 ÷ 50 = 6.13 分。众数所在组是 6-8（最高频数）。为了找到中位数所在组，计算累积频数：5, 17, 41, 50。第25和26个值落在 6-8 组，因此中位数所在组也是 6-8。由于分布略不平滑（高分较多），平均数可能略微拉低，所以中位数所在组可能更具代表性。在这个情境中，大多数学生的成绩在6到8分之间，表明这次测验比较简单。
You could extend this by drawing a cumulative frequency diagram to estimate the median and quartiles precisely.
你可以进一步绘制累积频数图，来精确估算中位数和四分位数。
4. Case Study 3: Probability in a Dice Game | 案例三：掷骰子游戏中的概率
Two fair six‑sided dice are rolled. A player wins if the sum is 7 or 11, loses if the sum is 2, 3 or 12, and otherwise rerolls. Calculate the probability of winning on the first roll. Is this game fair? Explain.
同时掷两个公平的六面骰子。如果点数和为7或11，玩家获胜；如果点数和为2、3或12，玩家输掉；其他情况则重新掷。计算第一次掷就获胜的概率。这个游戏是否公平？请解释。
omes = 8. P(win) = 8/36 = 2/9. Losing outcomes: sum 2 (1 way), 3 (2 ways), 12 (1 way) → total 4. P(lose) = 4/36 = 1/9. The remaining outcomes (reroll) = 36 − 8 − 4 = 24, probability 24/36 = 2/3. The game is not fair because the chance of winning (≈ 22.2%) is not equal to the chance of losing (≈ 11.1% probabilities are unequal. A fair game would have P(win) = P(lose) on each decisive roll.
共有36种等可能的结果。点数和为7的组合有6种（1+6, 2+5, 3+4, 4+3, 5+2, 6+1）。和为11的组合有2种（5+6, 6+5）。所以获胜结果共8种。获胜概率 = 8/36 = 2/9。输的结果：和为2（1种），3（2种），12（1种）→ 共4种。输的概率 = 4/36 = 1/9。剩余的结果（重掷） = 36 − 8 − 4 = 24，概率为 24/36 = 2/3。这个游戏不公平，因为获胜的概率（约22.2%）不等于失败的概率（约11.1%）。长期来看，游戏设计者可能占有优势或劣势，但就玩家而言，每次决定性掷骰的概率并不相等。一个公平的游戏应当在每次产生结果时，P(获胜) = P(失败)。
You can also calculate the expected number of rolls to finish the game using geometric distribution concepts, though this goes slightly beyond Year 10 core material.
你还可以利用几何分布的概念，计算完成这场游戏所需要的期望掷骰次数，不过这略微超出了Year 10的核心内容。
5. Case Study 4: Scatter Graphs and Correlation | 案例四：散点图与相关性
A class collected data on the number of hours spent revising and the test score (out of 50) for eight pupils. The data pairs (hours, score): (2, 18), (4, 24), (5, 30), (3, 22), (8, 45), (7, 40), (6, 35), (9, 48). Plot a scatter graph, describe the correlation, and draw a line of best fit by eye. Use your line to estimate the score for a pupil who revised for 5.5 hours.
一个班级收集了8名学生复习小时数和测试成绩（满分50）。数据对（小时，成绩）：(2, 18), (4, 24), (5, 30), (3, 22), (8, 45), (7, 40), (6, 35), (9, 48)。画散点图，描述相关性，并凭目测画出最佳拟合线。用你的线估计一位复习了5.5小时的学生的成绩。
ave roughly the same number of points above and below the line. One possible line passes near (2,18) and (9,48). The gradient is approximately (48−18) / (9−2) = 30/7 ≈ 4.3. The equation could be roughly score = 4.3 × hours + 9.4 (by reading off the intercept). Using the line, for 5.5 hours the estimated score is about 4.3×5.5 + 9.4 ≈ 33. Make sure you label axes, use a sensible scale, and mark the units. In exams, you are usually given graph paper — draw carefully.
首先，描点。点从图形左下方向右上方移动，表明存在强正相关。画最佳拟合线时，力保上下两侧的点数大致相等。一条可能的拟合线通过 (2,18) 和 (9,48) 附近。斜率约为 (48−18) / (9−2) = 30/7 ≈ 4.3。方程大致为 成绩 = 4.3 × 小时 + 9.4（通过读取截距得到）。利用这条线，复习5.5小时的预计成绩约为 4.3×5.5 + 9.4 ≈ 33 分。确保标注坐标轴，使用合理的刻度并标明单位。在考试中通常会提供坐标纸，请细致绘图。
Correlation does not imply causation — just because two variables move together does not mean one causes the other. A clear comment on this will boost your exam marks.
相关性并不意味着因果关系——两个变量一起变动，并不代表一个引起了另一个。清晰地对此进行评论会为你的考试加分。
6. Case Study 5: Sampling Methods in Action | 案例五：实际操作中的抽样方法
A school wants to estimate the proportion of pupils who eat breakfast before school. There are 1200 pupils: 400 in S1–S2, 500 in S3–S4, 300 in S5–S6. The survey team decides to interview 100 pupils. Explain how a stratified sample could be taken and why it is better than a simple random sample here.
一所学校希望估算学生中上学前吃早餐的比例。全校有1200名学生：S1-S2年级400人，S3-S4年级500人，S5-S6年级300人。调查小组决定采访100名学生。请解释如何抽取分层样本，并说明为何此处分层抽样优于简单随机抽样。
oups) and a random sample is taken from each stratum in proportion to its size. For the S1–S2 stratum, 400/1200 = 1/3 of the sample → about 33 pupils. S3–S4: 500/1200 ≈ 42 pupils. S5– sented. A simple random sample might, by chance, include too many senior pupils or not enough junior ones, leading to biased results. Stratified sampling improves precision when the characteristic (eating breakfast) may differ between age groups.
在分层抽样中，先将总体分为不同的层（这里为年级组），然后从每一层中按比例随机抽取样本。S1-S2层：400/1200 = 1/3 的样本，约33名学生。S3-S4层：500/1200 ≈ 42名。S5-S6层：300/1200 = 25名。在每个组内，使用随机方法（如随机数生成器）抽取。这种方法确保了各年级组都被公平地代表。而简单随机样本可能由于偶然地包含过多高年级学生或覆盖不足的低年级学生，从而导致偏倚的结果。当特征（如吃早餐）可能在不同年龄组间存在差异时，分层抽样能提高估计的精确性。
Another common method in SQA questions is systematic sampling, where you pick every nth person from a list. Always be ready to compare sampling methods, discussing bias and practicality.
SQA考题中另一种常见的方法是系统抽样，即从名单中每隔n个人抽取一位。随时做好准备比较不同的抽样方法，讨论其偏差和可行性。
7. Case Study 6: Interpreting Statistical Diagrams | 案例六：解读统计图
You are shown a comparative bar chart of favourite sports for boys and girls in Year 10. In another question, a pie chart displays the percentage of pupils cycling, walking, taking the bus or being driven to school. Explain what conclusions can be drawn and what limitations these diagrams have.
你面前有一幅对比条形图，显示了Year 10男女生最喜爱的运动。另一道题中，一幅饼图展示了学生骑车、步行、乘公交或家长开车送其上学的人数百分比。请解释可以得出哪些结论，以及这些图表有哪些局限。
From the comparative bar chart, you might notice that football is more popular with boys while netball is favoured by girls. However, the chart only shows counts or relative frequencies; it does not tell you why these preferences exist. The pie chart shows proportions of travel modes, but if the total number of pupils is not given, it is impossible to know the actual frequencies. Pie charts also become hard to read when there are many small slices. Always look for missing labels, non‑standard colour use, or a lack of key. When interpreting, phrase your conclusions tentatively: 'The pie chart suggests that most pupils walk to school, but the data may hide seasonal differences.'
从对比条形图中，你可能会发现足球在男生中更受欢迎，而无挡板篮球在女生中受青睐。然而，图表只展示了计数或相对频率；它并没有告诉你这些偏好为何存在。饼图展示了出行方式的比例，但如果未提供学生总数，便无法知道实际的频数。当出现许多细小扇形时，饼图也难以阅读。始终留意是否有缺失的标签、不标准的颜色用法或是缺少图例。解读时，请以留有余地的方式表达结论：'饼图表明大多数学生步行上学，但数据可能隐藏了季节性差异。'
Be prepared to sketch a simple bar chart or pie chart from a given frequency table. Neatness and clear labels earn marks.
做好准备，根据给定的频数表绘制简单的条形图或饼图。整洁的绘图和清晰的标注能赢得分数。
8. Case Study 7: Mean, Median and Mode — Which to Use? | 案例七：平均数、中位数与众数——选哪一个？
A small company has nine employees with salaries (in £1000s): 18, 20, 22, 23, 24, 25, 25, 26, 95. The £95k figure is the director's salary. Calculate the mean, median and mode. Which measure best describes the typical salary? Why?
一家小型公司有九名员工，薪水（单位：千英镑）如下：18, 20, 22, 23, 24, 25, 25, 26, 95。其中95千英镑是总监的薪水。计算平均数、中位数和众数。这三个测量值中，哪一个最能代表典型的薪金水平？为什么？
Sum = 18+20+22+23+24+25+25+26+95 = 278. Mean = 278 ÷ 9 ≈ 30.89 (≈ £0,890). Median (5th value) = 24. Mode = 25. The mean is heavily influenced by the director's high salary and is not typical of most employees. The median (£24,000) and mode (£ tains outliers, the median is more robust. The mode is useful for identifying the most common value, but may not reflect central tendency if data is varied.
总和 = 18+20+22+23+24+25+25+26+95 = 278。平均数 = 278 ÷ 9 ≈ 30.89（约 £30,890）。中位数（第5个值）= 24。众数 = 25。平均数受到总监高薪水的严重影响，不能代表大多数员工的典型水平。中位数（£24,000）和众数（£25,000）与大多数据值更为接近。因此，中位数或众数能更准确地描绘典型薪水。一般而言，当数据含有异常值时，中位数更具稳健性。众数有助于识别最常见的值，但如果数据差异较大，它可能无法反映集中趋势。
Always explain your choice clearly, linking back to the context. This shows examiners you are thinking statistically rather than just calculating.
务必清晰地解释你的选择，并联系到实际情境。这能让考官看出你是在进行统计思考，而非仅仅是计算。
9. Case Study 8: Comparing Data Sets Using Box Plots | 案例八：使用箱线图比较数据集
Two Year 10 classes took the same statistics test. Class A's marks (out of 40): 12, 15, 18, 20, 22, 25, 28, 30, 32, 35. Class B's marks: 8, 12, 14, 19, 21, 24, 27, 31, 33, 37. Construct box plots and compare the performances.
两个Year 10班级参加了相同的统计测验。A班成绩（满分40）：12, 15, 18, 20, 22, 25, 28, 30, 32, 35。B班成绩：8, 12, 14, 19, 21, 24, 27, 31, 33, 37。构建箱线图并比较两班的成绩表现。
For Class A, the five‑number summary: min=12, Q1=17 (average of 15 and 18? Wait, with 10 values, Q1 is the 2.5th value → between 15 and 18, so mid = 16.5? Let' pper half: 25,28,30,32,35 → Q3 = 30. So Q1=18, median=23.5, Q3=30, min=12, max=35. For B: 8,12,14,19,21,24,27,31,33,37. Median = (21+24)/2=22.5. Q1: lower half 8,12,14,19,21 → Q1=14. Q3: upper half 24,27,31,33,37 → Q3=31. Min=8, max=37.
对于A班，五数概括：最小值=12，Q1=18，中位数=23.5，Q3=30，最大值=35。对于B班：最小值=8，Q1=14，中位数=22.5，Q3=31，最大值=37。
Drawing box plots (on the same scale) shows that Class B has a slightly lower median but a wider spread. The interquartile range for A is 12, for B it is 17, so B's marks are more varied. The box plots also reveal that the top 25% in B extends higher than A, despite a lower minimum. A good comparison cites both central tendency and spread, and comments on overlap. This approach is common in SQA exams — you might be asked to write and compare box plots from raw data.
在相同标尺上绘制箱线图，可以发现B班的中位数略低，但分布更广。A班的四分位距为12，B班为17，因此B班成绩的离散程度更大。箱线图还显示，B班的前25%成绩高于A班，尽管其最小值更低。良好的比较既要提到集中趋势，也要提及离散程度，并对重叠部分加以评论。在SQA考试中，这很常见——你可能需要从原始数据出发，画出并比较箱线图。
10. Case Study 9: Introduction to Probability Distributions | 案例九：概率分布初步
A fair coin is tossed 4 times. Let X be the number of heads obtained. Write down the probability distribution of X, calculate the expected number of heads, and find the probability of getting at least 2 heads. This introduces the binomial distribution.
一枚公平的硬币抛掷4次。设X为得到正面的次数。写出X的概率分布，计算期望正面数，并求出至少得到2个正面的概率。这引入了二项分布。
cient C(4,k). The probability P(X=k) = C(4,k) × (0.5)⁴. Values: P(X=0)=1/16, P(X=1)=4/16, P(X=2)=6/16, P(X=3)=4/16, P(X=4)=1/16. The expected value E(X) = np = 4 × 0.5 = 2 heads. P(X ≥ 2) = P(2) + P(3) + P(4) = (6+4+1)/16 = 11/16. This distribution is symmetric because p=0.5.
每次抛掷相互独立，P(正面)=0.5。可能的结果总数为 2⁴ = 16。得到k个正面的方法数由二项式系数 C(4,k) 给出。概率 P(X=k) = C(4,k) × (0.5)⁴。具体值：P(X=0)=1/16, P(X=1)=4/16, P(X=2)=6/16, P(X=3)=4/16, P(X=4)=1/16。期望值 E(X) = np = 4 × 0.5 = 2个正面。P(X ≥ 2) = P(2) + P(3) + P(4) = (6+4+1)/16 = 11/16。该分布因 p=0.5 而对称。
In an exam, you could be asked to use a tree diagram to verify these probabilities. Practise drawing clear, labelled trees. The ability to recognise a binomial situation and apply the formula without listing all outcomes is a key skill for higher marks.
考试中可能要求你用树状图来印证这些概率。请练习绘制清晰、标注明确的树状图。识别二项分布情境，并能直接套用公式而无需罗列所有结果，是获取高分的关键能力。
11. Case Study 10: Using Data to Make Predictions | 案例十：利用数据进行预测
his equation to predict the score for a pupil who revises for 0 hours and for 10 hours. Discuss the reliability of these predictions.
利用案例四中复习小时数与测试成绩的数据，一名学生计算出回归线方程为 成绩 = 4.7 × 小时 + 10.2。用这个方程预测一位不复习（0小时）和复习10小时的学生的成绩。讨论这些预测的可靠性。
reality a pupil with zero revision should still score around 18 based on the original data — score = 4.7×10 + 10.2 = 57.2, but the test is out of 50, so this prediction is impossible! data are unreliable. The regression line fits the observed points well within the range 2– eads to nonsensical results. Always warn that extrapolation can produce invalid estimates.
复习0小时：成绩 = 10.2，这或许可以理解为基础分数，但实际上根据原始数据，不复习的学生成绩也在18分左右——模型在极端处可能不准确。复习10小时：成绩 = 4.7×10 + 10.
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