📚 Year 8 SQA Statistics: Cross-curricular Integrated Problem Solving | 八年级 SQA 统计：跨学科综合题型训练
Statistics is not an isolated subject – it appears in geography when comparing city sizes, in science when plotting temperature changes, and in everyday decisions like choosing the safest route to school. This article provides a carefully designed set of cross-curricular practice questions, blending data handling skills with real-world contexts across multiple subjects. Each section models how to interpret charts, calculate averages, assess probability, and critique data presentation, helping you build confidence for SQA assessments.
统计并不是一门孤立的学科——它出现在地理课的城市人口比较中，出现在科学课的温度变化折线图里，也出现在选择最安全上学路线的日常决策中。本文提供了一套精心设计的跨学科练习题，将数据处理技能与多个学科的真实情境相结合。每个小节都示范如何解读图表、计算平均数、评估概率以及批判性地审视数据呈现方式，帮助你为 SQA 评估建立信心。
1. Interpreting Bar Charts with Geography Data | 用地理数据解读柱状图
In a geography project, students recorded the number of visitors to four Scottish landmarks over one weekend. The bar chart shows the results.
在地理课题中，学生们记录了一个周末内四处苏格兰地标的游客数量。柱状图显示了结果。
Edinburgh Castle: 85, Stirling Castle: 65, Loch Ness Visitor Centre: 110, Glenfinnan Viaduct: 40.
爱丁堡城堡：85，斯特灵城堡：65，尼斯湖游客中心：110，格伦芬南高架桥：40。
Question: How many more visitors went to Loch Ness than to Stirling Castle? What is the total number of visitors?
问题：到尼斯湖的游客比到斯特灵城堡的多多少人？游客总数是多少？
Solution: Loch Ness had 110 visitors and Stirling Castle had 65. The difference is 110 − 65 = 45 visitors. The total is 85 + 65 + 110 + 40 = 300 visitors.
解答：尼斯湖有 110 名游客，斯特灵城堡有 65 名。差值为 110 − 65 = 45 名游客。总数为 85 + 65 + 110 + 40 = 300 名游客。
2. Line Graphs in Science: Temperature Change Experiment | 科学中的折线图：温度变化实验
A science class heated water and recorded the temperature every minute. The line graph shows an initial temperature of 20°C, rising to 50°C at 3 minutes, 70°C at 5 minutes, and staying at 100°C from 8 minutes onward.
科学课上，学生加热水并每分钟记录温度。折线图显示初始温度 20°C，第 3 分钟升至 50°C，第 5 分钟 70°C，从第 8 分钟起保持在 100°C。
Question: During which 2-minute interval did the temperature increase the most? Describe what happened after 8 minutes in terms of state change.
问题：在哪一个 2 分钟区间内温度上升最快？描述第 8 分钟后从物态变化的角度发生了什么。
Solution: Between minutes 1 and 3, the rise was 30°C (20 to 50). Between 3 and 5, it was 20°C; between 5 and 7, it was 15°C, and between 7 and 8, it was 15°C. The steepest segment was from minute 1 to 3, with a gradient of 15°C per minute. After 8 minutes the temperature plateaued at 100°C because water reached its boiling point and started changing into steam.
解答：第 1 到第 3 分钟上升了 30°C，第 3 到第 5 分钟上升 20°C，第 5 到第 7 分钟上升 15°C，第 7 到第 8 分钟上升 15°C。最陡的线段是第 1 到第 3 分钟，斜率为每分钟 15°C。第 8 分钟后温度稳定在 100°C，因为水达到了沸点并开始变为水蒸气。
3. Pie Charts in School Life: Favourite Sports Survey | 学校生活中的饼图：最喜爱运动调查
A school surveyed 120 pupils about their favourite sport. The pie chart shows: Football 45 pupils, Swimming 30, Basketball 25, Tennis 20.
学校对 120 名学生进行了最喜爱运动调查。饼图显示：足球 45 人，游泳 30 人，篮球 25 人，网球 20 人。
Question: What fraction of pupils chose swimming? What angle represents tennis on the pie chart?
问题：选择游泳的学生占几分之几？在饼图中，代表网球的扇形角度是多少？
Solution: Swimming fraction = 30/120 = 1/4. Tennis fraction = 20/120 = 1/6. The angle for tennis = 1/6 × 360° = 60°.
解答：游泳所占比例为 30/120 = 1/4。网球所占比例为 20/120 = 1/6。网球的扇形角度 = 1/6 × 360° = 60°。
4. Calculating Mean, Median and Mode in PE: Basketball Scores | 体育中的平均数、中位数和众数：篮球得分
A basketball player scored the following points in eight games: 12, 15, 12, 18, 21, 12, 10, 16.
一名篮球运动员在八场比赛中的得分如下：12, 15, 12, 18, 21, 12, 10, 16。
Question: Find the mean, median and mode of these scores. Which average best represents the player’s typical performance? Why?
问题：求这些得分的平均数、中位数和众数。哪一个平均数最能代表这位球员的典型表现？为什么？
Solution: Mean = (12+15+12+18+21+12+10+16) ÷ 8 = 116 ÷ 8 = 14.5 points. To find the median, order the data: 10, 12, 12, 12, 15, 16, 18, 21. Median = (12+15)/2 = 13.5 points. Mode = 12 (most frequent). The mode of 12 shows the most common score, but the mean of 14.5 is slightly higher due to one high game. The median is less affected by the 21 and may be the best typical value.
解答：平均数 = (12+15+12+18+21+12+10+16) ÷ 8 = 116 ÷ 8 = 14.5 分。求中位数时排序：10, 12, 12, 12, 15, 16, 18, 21。中位数 = (12+15)/2 = 13.5 分。众数 = 12（出现最频繁）。众数 12 显示了最常见的得分，而平均数 14.5 因为一场高分而偏高。中位数受 21 的影响较小，可能是最典型的代表值。
5. Range and Data Interpretation in Geography: Daily Rainfall | 地理中的极差与数据解读：逐日降雨量
A weather station recorded rainfall (in mm) over one week in Glasgow: Mon 4, Tue 7, Wed 3, Thu 8, Fri 12, Sat 9, Sun 5.
气象站记录了格拉斯哥一周的降雨量（毫米）：周一 4，周二 7，周三 3，周四 8，周五 12，周六 9，周日 5。
Question: Calculate the range. On how many days was the rainfall above the weekly mean? Why might Friday’s reading be an outlier in a typical week?
问题：计算极差。有多少天的降雨量高于周平均值？为什么周五的数据在典型的一周中可能是一个异常值？
Solution: Range = 12 − 3 = 9 mm. Total = 4+7+3+8+12+9+5 = 48 mm, mean = 48 ÷ 7 ≈ 6.86 mm. Days above the mean: Tue (7), Thu (8), Fri (12), Sat (9) – that is 4 days. Friday’s 12 mm could be an outlier if a sudden storm occurred, whereas other days show more typical Glasgow drizzle. In statistics, an outlier can skew the mean upwards.
解答：极差 = 12 − 3 = 9 毫米。总降雨量 = 4+7+3+8+12+9+5 = 48 毫米，平均数 = 48 ÷ 7 ≈ 6.86 毫米。高于平均数的日子：周二（7）、周四（8）、周五（12）、周六（9）——共 4 天。周五的 12 毫米可能是异常值，如果当天发生突发暴雨，而其他日子显示的是格拉斯哥常见的毛毛雨。在统计中，异常值会拉高平均数。
6. Probability in Games: Spinner Experiment | 游戏中的概率：转盘实验
A games stall has a spinner divided into four equal sectors: red, blue, green, yellow. Players win a prize if the spinner lands on yellow.
一个游戏摊位有一个转盘，分成四个相等的扇形：红色、蓝色、绿色、黄色。转到黄色即获奖。
Question: What is the theoretical probability of winning in one spin? If the spinner is spun 60 times, about how many yellows would you expect? Explain why actual results may differ.
问题：单次转动转盘的理论获奖概率是多少？如果转动转盘 60 次，你预计大约出现多少次黄色？解释为什么实际结果可能不同。
Solution: Theoretical probability = 1/4 = 0.25 or 25%. Expected frequency = 1/4 × 60 = 15 yellows. In an actual experiment, the number might be 12 or 17 because of random variation. Probability describes long-term behaviour, not exact short-term outcomes.
解答：理论概率 = 1/4 = 0.25 或 25%。预期频数 = 1/4 × 60 = 15 次黄色。实际实验中，这个数字可能是 12 或 17，因为存在随机波动。概率描述的是长期行为，而非短期的精确结果。
7. Scatter Plots and Correlation in Health: Height vs Arm Span | 健康课中的散点图与相关性：身高与臂展
A biology class measured height (cm) and arm span (cm) for ten pupils. The data points generally go upwards together. One pupil, however, has a noticeably longer arm span for their height.
生物课上，学生测量了十名同学的身高（厘米）和臂展（厘米）。数据点总体上呈一同上升的趋势。但有一位学生的臂展相对于身高明显偏长。
Question: Describe the correlation shown by the scatter plot. What might cause the one point to deviate from the pattern?
问题：描述散点图显示的相关性。什么原因可能导致那个点偏离整体模式？
Solution: The scatter plot shows positive correlation – as height increases, arm span tends to increase. This is because limbs are generally proportional to height. The outlying point could be due to measurement error, an individual with a different build, or a pupil still growing. Identifying outliers helps in understanding data quality.
解答：散点图显示正相关——身高增加时，臂展往往也随之增加。这是因为四肢通常与身高成比例。那个偏离的点可能是由于测量误差、不同的体型，或者该学生仍处于生长期。识别异常值有助于理解数据质量。
8. Two-way Tables in Social Studies: Transport to School | 社会课中的双向表：上学交通方式
A survey asked 100 pupils in Year 8 about their usual transport method. The two-way table shows: Walking – 40 (25 boys, 15 girls); Bus – 35 (10 boys, 25 girls); Cycling – 25 (15 boys, 10 girls).
一项调查询问了 100 名八年级学生通常的上学交通方式。双向表显示：步行——40 人（25 名男生，15 名女生）；公交车——35 人（10 名男生，25 名女生）；骑车——25 人（15 名男生，10 名女生）。
Question: What percentage of girls take the bus? Which transport method is most popular among boys? Can you suggest a reason for the gender differences shown?
问题：乘坐公交车的女生占女生总数的百分比是多少？哪一种交通方式在男生中最受欢迎？你能否对其中显示的性别差异提出一个理由？
Solution: Total girls = 15+25+10 = 50. Girls taking bus = 25, so 25/50 = 1/2 = 50%. Boys most often walk (25 out of 50). The difference could be due to personal preference, distance, or after-school activity patterns – but without more evidence we cannot assume a definite cause.
解答：女生总数 = 15+25+10 = 50。乘公交车的女生 = 25，因此 25/50 = 1/2 = 50%。男生最常步行（50 人中有 25 人）。差异可能来自个人偏好、距离或课后活动模式——但缺少更多证据时，我们不能假定一个确定的原因。
9. Stem-and-Leaf Diagrams in Health and Wellbeing: Pupils’ Heights | 健康课中的茎叶图：学生的身高
A class recorded heights in centimetres: 142, 146, 147, 151, 153, 155, 158, 161, 164, 169. They displayed these in a stem-and-leaf plot with stems 14, 15, 16.
一个班级记录了身高（厘米）：142, 146, 147, 151, 153, 155, 158, 161, 164, 169。他们用茎叶图展示，茎为 14、15、16。
Question: From the stem-and-leaf, find the median height. What is the modal stem? What can you say about the distribution shape?
问题：从茎叶图中找出身高的中位数。众数茎是多少？关于分布形状你能得出什么结论？
Solution: Data ordered: 142,146,147 | 151,153,155,158 | 161,164,169. Total 10 values, median = (5th+6th)/2 = (153+155)/2 = 154 cm. Modal stem is 15 as it has four leaves. The distribution is roughly symmetric and slightly clustered around 150-160 cm.
解答：数据排序：142,146,147 | 151,153,155,158 | 161,164,169。共 10 个值，中位数 = (第5+第6)/2 = (153+155)/2 = 154 厘米。众数茎是 15，因为有四片叶。分布大致对称，略微聚集在 150–160 厘米之间。
10. Critiquing Misleading Graphs in Media: A Marketing Chart | 媒体中的误导性图表辨析：一张营销图表
A phone company advertises “battery life has doubled!” using a bar chart where the first bar starts at 8 hours and ends at 10, while the second bar starts at 0 and zooms to 10, creating a visual impression that the increase is far more dramatic than the actual difference of 2 hours.
一家手机公司用柱状图宣传“电池续航翻倍！”，图中第一条柱从 8 小时开始到 10 小时结束，而第二条柱从 0 开始突增至 10 小时，造成视觉上提升极其巨大的假象，而实际不过增加了 2 小时。
Question: Identify two ways this graph misleads the viewer. How could you redraw the graph to show the data fairly?
问题：指出这张图误导观众的两个方面。如何重绘这张图才能公正地展示数据？
Solution: The y-axis does not start at zero, exaggerating the height difference. The scale is inconsistent between bars, making the first increase look tiny and the second huge. To fix it, draw both bars from zero with a consistent scale, so the heights are 8 and 10, showing only a 25% improvement.
解答：纵轴不是从零开始，夸大了高度差。两个柱形图的尺度不一致，使得第一个增加看起来极小，而第二个极大。修正时可让两根柱子均从零开始并使用一致的尺度，这样高度分别为 8 和 10，仅显示出 25% 的提升。
11. Comparing Data Sets Using Averages and Range: Two Classes’ Test Scores | 利用平均数和极差比较数据集：两个班级的测验成绩
Class A scored: 55, 60, 65, 70, 75, 80, 85. Class B scored: 50, 55, 65, 70, 85, 90, 95.
A 班成绩：55, 60, 65, 70, 75, 80, 85。B 班成绩：50, 55, 65, 70, 85, 90, 95。
Question: Calculate the mean, median and range for each class. Which class performed more consistently? Justify your answer.
问题：计算每个班的平均数、中位数和极差。哪个班级的表现更稳定？请说明理由。
Solution: Class A mean = (55+60+65+70+75+80+85) ÷ 7 = 490 ÷ 7 = 70. Median = 70, range = 85−55 = 30. Class B mean = (50+55+65+70+85+90+95) ÷ 7 = 510 ÷ 7 ≈ 72.9. Median = 70, range = 95−50 = 45. Class A has a smaller range, indicating more consistent scores, while Class B has a slightly higher mean but much wider spread.
解答：A 班平均数 = (55+60+65+70+75+80+85) ÷ 7 = 490 ÷ 7 = 70。中位数 = 70，极差 = 85 − 55 = 30。B 班平均数 = (50+55+65+70+85+90+95) ÷ 7 = 510 ÷ 7 ≈ 72.9。中位数 = 70，极差 = 95 − 50 = 45。A 班的极差更小，表明成绩更稳定，而 B 班平均数略高但数据分布更广。
12. Drawing Conclusions from Statistical Investigations: A Full Cycle | 从统计调查中得出结论：一个完整的研究循环
Pupils designed a question: “Does the amount of sleep affect reaction time?” They collected data from 20 peers, recording hours of sleep and time to catch a ruler (cm). A scatter plot showed a weak negative trend – more sleep linked to slightly faster reactions, but with many exceptions.
学生们设计了一个问题：“睡眠时间会影响反应速度吗？”他们收集了 20 位同学的数据，记录睡眠小时数和接住尺子的距离（厘米）。散点图显示了一个弱的负相关趋势——睡眠更多略关联于更快的反应，但有许多例外。
Question: What conclusion can they draw? Why might the evidence not be strong enough? Suggest a way to improve the investigation.
问题：他们可以得出什么结论？为什么证据可能不够有力？给出改进调查的建议。
Solution: They can state there seems to be a slight negative association, but they cannot claim causation because other factors like diet, exercise or stress could affect reaction time. The sample of 20 is also small. To improve, increase sample size, control other variables (same time of day, no caffeine), and use more precise timing equipment.
解答：他们可以声称似乎存在轻微的负相关，但不能断言因果关系，因为饮食、运动或压力等其他因素都可能影响反应时间。20 人的样本也偏小。改进方法包括扩大样本量、控制其他变量（同一时段、无咖啡因），并使用更精确的计时设备。
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