Interdisciplinary Integrated Statistics Practice | 跨学科综合统计题型训练

📚 Interdisciplinary Integrated Statistics Practice | 跨学科综合统计题型训练

Interdisciplinary problems in SQA Statistics require applying statistical techniques to real-world contexts—from environmental science to economics. This article provides a comprehensive set of practice exercises covering descriptive statistics, probability, scatter graphs, and more, each set in a cross-curricular scenario. Mastering these will sharpen your analytical skills and prepare you for exam success.

SQA 统计中的跨学科问题要求将统计技术应用于环境科学、经济学等现实情境。本文提供一套全面的练习，涵盖描述性统计、概率、散点图等内容，每道题都设置在跨学科场景中。熟练掌握这些将提升你的分析技能并助力考试成功。

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1. Environmental Science: Descriptive Statistics of Water Quality | 环境科学：水质的描述性统计

A limnologist collected water samples from five sites along a Highland stream. The pH readings are: 6.5, 7.0, 6.8, 7.2, 5.9.

一位湖沼学家从一条高地溪流的五个地点采集了水样。pH 读数为：6.5、7.0、6.8、7.2、5.9。

Site	A	B	C	D	E
pH	6.5	7.0	6.8	7.2	5.9

Mean = Σxᵢ / n = (6.5+7.0+6.8+7.2+5.9) ÷ 5 = 33.4 ÷ 5 = 6.68

The mean pH is 6.68, indicating slightly acidic water on average.

平均 pH 为 6.68，表明水体平均呈弱酸性。

To find the median, order the values: 5.9, 6.5, 6.8, 7.0, 7.2. The middle value is 6.8.

求中位数，将数值排序：5.9、6.5、6.8、7.0、7.2。中间值为 6.8。

Range = Maximum - Minimum = 7.2 - 5.9 = 1.3

A range of 1.3 highlights notable variation in pH across the sites; the lowest reading (5.9) may indicate a local pollution input.

极差为 1.3，凸显各地点 pH 存在显著差异；最低读数 5.9 可能表明局部有污染输入。

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2. Biology: Probability and Genetic Traits | 生物：概率与遗传性状

In Mendelian pea plants, the probability of a round-seed phenotype (dominant) is 3/4, and that of a wrinkled seed (recessive) is 1/4. Two offspring are produced independently.

在孟德尔豌豆植株中，圆粒表型（显性）的概率为 3/4，皱粒表型（隐性）为 1/4。两个后代独立产生。

P(both round) = (3/4) × (3/4) = 9/16

The probability that both offspring have round seeds is 9/16.

两个后代都是圆粒的概率为 9/16。

P(at least one wrinkled) = 1 - P(both round) = 1 - 9/16 = 7/16

Alternatively, you could add the probabilities of exactly one wrinkled and both wrinkled. The result is 7/16.

或者，你可以将恰好一个皱粒和两个都是皱粒的概率相加。结果为 7/16。

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3. Business Studies: Scatter Graphs and Sales Trends | 商业研究：散点图与销售趋势

A small coffee shop records weekly advertising expenditure (£100s) and weekly sales (£100s) over six weeks:

一家小咖啡店记录了六周的每周广告支出（百英镑）和每周销售额（百英镑）：

Advertising (£100s)	1.0	1.5	2.0	2.5	3.0	3.5
Sales (£100s)	5.2	6.1	7.3	8.0	9.4	10.1

Plot the data on a scatter graph with advertising on the x-axis. The points show a strong positive correlation: as advertising rises, sales tend to increase linearly.

在散点图上绘制数据，以广告支出为 x 轴。各点显示出强正相关：随着广告支出增加，销售额呈线性增长。

Draw a line of best fit that passes through the middle of the points. Use your line to estimate sales when advertising is £250 (2.5 on the scale). A reasonable reading gives approximately £8200 (8.2).

画一条穿过点群中间的最佳拟合线。利用该线估计当广告支出为 250 英镑（尺度上 2.5）时的销售额。合理的读值大约为 8200 英镑（8.2）。

If the trend continues, a further increase in advertising should yield higher sales, but always consider external factors outside the model.

如果趋势持续，进一步增加广告支出应当带来更高销售额，但始终要考虑到模型之外的外部因素。

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广告 / Sponsored

4. Physical Education: Box Plots for Athlete Performance | 体育：运动表现的箱线图比较

A basketball coach analysed points scored per game by two players across nine matches.

一位篮球教练分析了两位球员在九场比赛中的每场得分。

Player A	68	72	75	78	80	82	85	88	90
Player B	65	70	74	77	79	81	84	86	95

For each player, find the five-figure summary: minimum, Q1, median, Q3, maximum.

为每位球员计算五数总结：最小值、Q1、中位数、Q3、最大值。

Player A ordered: 68,72,75,78,80,82,85,88,90. Min=68, Q1=75, Median=80, Q3=85, Max=90.

球员 A 排序后：68,72,75,78,80,82,85,88,90。最小值=68，Q1=75，中位数=80，Q3=85，最大值=90。

Player B: 65,70,74,77,79,81,84,86,95. Min=65, Q1=74, Median=79, Q3=84, Max=95.

球员 B：65,70,74,77,79,81,84,86,95。最小值=65，Q1=74，中位数=79，Q3=84，最大值=95。

IQR(A) = 85 - 75 = 10, IQR(B) = 84 - 74 = 10

Both players have the same interquartile range, suggesting similar consistency in the middle half of their performances. However, Player B's maximum score is a high outlier (95) which inflates the range. Box plots would reveal that Player A's distribution is more symmetric, while Player B's shows a slight positive skew.

两位球员的四分位距相同，意味着他们中间一半表现的一致性相似。不过球员 B 的最高分是一个高异常值（95），扩大了全距。箱线图将显示球员 A 的分布更对称，而球员 B 的分布略显正偏。

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5. Geography: Standard Deviation of Temperature Data | 地理：温度数据的标准差

A weather station records daily maximum temperatures (°C) for a week in July: 22, 24, 19, 26, 23, 21, 20.

某气象站记录了一周七月的每日最高气温（°C）：22、24、19、26、23、21、20。

Mean x̄ = (22+24+19+26+23+21+20) ÷ 7 = 155 ÷ 7 ≈ 22.14°C

Compute the deviations and squared deviations:

计算偏差与平方偏差：

4)=3.86 → 14.90, (23-22.14)=0.86 → 0.74, (21-22.14)=-1.14 → 1.30, (20-22.14)=-2.14 → 4.58.

(26-22.14)=3.86→14.90, (23-22.14)=0.86→0.74, (21-22.14)=-1.14→1.30, (20-22.14)=-2.14→4.58。

s = √[ Σ(xᵢ - x̄)² / (n-1) ] = √[ (0.02+3.46+9.86+14.90+0.74+1.30+4.58) / 6 ] = √34.86/6 = √5.81 ≈ 2.41°C

The standard deviation of about 2.41°C tells us that on most days the temperature deviates from the mean by about 2.4 degrees. The relatively small spread is characteristic of a stable summer weather pattern.

标准差约为 2.41°C，说明多数日子的气温偏离平均值约 2.4 度。这种相对较小的波动是稳定夏季天气模式的特征。

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6. Sociology: Sampling and Mean Estimation from Surveys | 社会学：调查中的抽样与均值估计

A school of 1200 students conducts a survey on daily social media time. A simple random sample of 100 students yields a sample mean of 2.5 hours and a sample standard deviation of 0.8 hours.

一所拥有 1200 名学生的学校开展了一项关于每日社交媒体

Published by TutorHao | Year 11 统计 Revision Series | aleveler.com

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