📚 Interdisciplinary Statistics Problems for Year 8 | 八年级跨学科统计综合题型训练
Statistics is not a subject that exists in isolation; it is a powerful toolkit for understanding and solving problems across all areas of learning. In the WJEC Year 8 curriculum, students are increasingly expected to apply statistical skills — such as calculating averages, interpreting charts, and using probability — to real-world contexts drawn from science, geography, sport, and everyday life. This article provides a comprehensive set of interdisciplinary practice questions designed to strengthen your ability to transfer statistical thinking across subjects. Each section introduces a different context, presents a scenario, asks targeted questions, and discusses key methods. By working through these exercises, you will not only improve your statistical fluency but also see how numbers help us make sense of the world around us.
统计学不是一门孤立的学科；它是一个强大的工具箱，能帮助我们理解并解决各个学习领域中的问题。在 WJEC 八年级课程中，学生需要越来越多地将统计技能——比如计算平均数、解读图表、使用概率——应用到来自科学、地理、体育和日常生活的真实情境中。本文提供了一整套跨学科综合练习题，旨在强化你将统计思维迁移到不同学科的能力。每一节都引入一个不同的背景，呈现一个场景，提出针对性的问题，并讨论关键方法。通过这些练习，你不仅会提高统计的熟练程度，还会看到数字如何帮助我们理解身边的世界。
1. Statistics in Science: Analysing Experiment Data | 科学中的统计：分析实验数据
A biology class measured the growth of cress seedlings over 10 days. The heights (in mm) recorded for one group were: 12, 15, 14, 16, 18, 13, 19, 15, 17, 20. Compute the mean, median, mode, and range. Explain which average best represents the typical growth and why. Then, draw a line graph to show the daily growth if the initial height was 0 mm on day 0, assuming linear growth between the average daily increase.
生物课上，一组学生测量了水芹幼苗在 10 天内的生长情况。记录的高度（单位：毫米）为：12, 15, 14, 16, 18, 13, 19, 15, 17, 20。请计算平均数、中位数、众数和极差。解释哪一个平均量最能代表典型的生长情况，并说明理由。然后，假设第 0 天初始高度为 0 毫米，且每天按平均日增长量线性生长，绘制一张折线图来展示每日的生长情况。
First, sort the data: 12, 13, 14, 15, 15, 16, 17, 18, 19, 20. Mean = (sum) ÷ 10 = 159 ÷ 10 = 15.9 mm. Median = (15 + 16) ÷ 2 = 15.5 mm. Mode = 15 mm (appears twice). Range = 20 − 12 = 8 mm. The mean (15.9) is slightly pulled up by the higher values, while the median (15.5) is less affected by extremes. In this small dataset, the mean is appropriate because there are no outliers, but the median is also reliable. For the line graph, if the total growth over 10 days is 159 mm, the average daily increase is 15.9 mm/day. Plot points at (0,0), (1,15.9), (2,31.8), … (10,159) and connect them.
首先，将数据排序：12, 13, 14, 15, 15, 16, 17, 18, 19, 20。平均数 = 总和 ÷ 10 = 159 ÷ 10 = 15.9 毫米。中位数 = (15 + 16) ÷ 2 = 15.5 毫米。众数 = 15 毫米（出现两次）。极差 = 20 − 12 = 8 毫米。平均数（15.9）被较高的数值稍微拉高，而中位数（15.5）受极端值影响较小。在这个小数据集中，平均数合适因为没有异常值，但中位数同样可靠。对于折线图，如果 10 天的总生长量为 159 毫米，则平均日增长量为 15.9 毫米/天。绘制点 (0,0), (1,15.9), (2,31.8), … (10,159) 并连线。
2. Statistics in Geography: Population Pyramids | 地理中的统计：人口金字塔
A village has the following population by age group: 0–14: 120, 15–29: 200, 30–44: 180, 45–59: 150, 60–74: 100, 75+: 50. Construct a population pyramid (back-to-back bar chart) splitting each group into males (48%) and females (52%). Calculate the dependency ratio using the formula: ((population aged 0–14 + population aged 60+) ÷ population aged 15–59) × 100. Interpret what this ratio tells us about the village.
一个村庄各年龄组的人口如下：0–14 岁：120 人，15–29 岁：200 人，30–44 岁：180 人，45–59 岁：150 人，60–74 岁：100 人，75 岁以上：50 人。请构建一个人口金字塔（背对背条形图），将每个组分为男性（48%）和女性（52%）。用公式计算抚养比：((0–14 岁人口 + 60 岁以上人口) ÷ 15–59 岁人口) × 100。解释这个比值说明了该村庄的什么情况。
Total 0–14 = 120; 60+ = 100 + 50 = 150; working-age (15–59) = 200 + 180 + 150 = 530. Dependency ratio = ((120 + 150) ÷ 530) × 100 ≈ (270 ÷ 530) × 100 ≈ 50.9%. This means for every 100 working-age people, there are about 51 dependents (young and old). For the pyramid, calculate males and females per group e.g., 0–14: male = 120×0.48 ≈ 58, female = 120×0.52 ≈ 62. Plot males on the left (negative direction) and females on the right. The shape will be wider in the middle (working ages) indicating a developing or growing population structure.
0–14 岁总计 = 120；60 岁以上 = 100 + 50 = 150；劳动年龄（15–59 岁）= 200 + 180 + 150 = 530。抚养比 = ((120 + 150) ÷ 530) × 100 ≈ (270 ÷ 530) × 100 ≈ 50.9%。这意味着每 100 名劳动年龄人口对应约 51 名受抚养人口（年幼和年长者）。对于金字塔图表，计算每个组的男女人数，例如 0–14 岁：男性 = 120×0.48 ≈ 58，女性 = 120×0.52 ≈ 62。将男性绘制在左侧（负数方向），女性在右侧。图形会在中间（劳动年龄）较宽，表明一种发展中或正在增长的人口结构。
3. Statistics in Physical Education: Tracking Athletic Performance | 体育中的统计：追踪运动表现
A Year 8 student recorded her 100-metre sprint times (in seconds) over 8 weeks: 15.2, 15.0, 14.8, 14.9, 14.6, 14.7, 14.5, 14.4. Calculate the mean and range for the first 4 weeks and the last 4 weeks. What is the percentage improvement in the mean time from the first half to the second half? Draw a time-series graph and describe the trend. Is the improvement consistent?
一名八年级学生记录了她 8 周内的 100 米短跑时间（单位：秒）：15.2, 15.0, 14.8, 14.9, 14.6, 14.7, 14.5, 14.4。分别计算前 4 周和后 4 周的平均数和极差。从前半段到后半段，平均时间的提高百分比是多少？绘制时间序列图并描述趋势。进步是否稳定？
First 4 weeks: 15.2, 15.0, 14.8, 14.9. Mean = (15.2+15.0+14.8+14.9) ÷ 4 = 59.9 ÷ 4 = 14.975 s. Range = 15.2 − 14.8 = 0.4 s. Last 4 weeks: 14.6, 14.7, 14.5, 14.4. Mean = (14.6+14.7+14.5+14.4) ÷ 4 = 58.2 ÷ 4 = 14.55 s. Range = 14.7 − 14.4 = 0.3 s. Percentage improvement = ((14.975 − 14.55) ÷ 14.975) × 100 ≈ (0.425 ÷ 14.975) × 100 ≈ 2.84%. The time-series graph shows a clear downward trend, with times decreasing nearly every week. The improvement is fairly consistent, though there is a slight plateau between weeks 3 and 4. The decreasing range indicates more consistent performance.
前 4 周：15.2, 15.0, 14.8, 14.9。平均数 = (15.2+15.0+14.8+14.9) ÷ 4 = 59.9 ÷ 4 = 14.975 秒。极差 = 15.2 − 14.8 = 0.4 秒。后 4 周：14.6, 14.7, 14.5, 14.4。平均数 = (14.6+14.7+14.5+14.4) ÷ 4 = 58.2 ÷ 4 = 14.55 秒。极差 = 14.7 − 14.4 = 0.3 秒。提高百分比 = ((14.975 − 14.55) ÷ 14.975) × 100 ≈ (0.425 ÷ 14.975) × 100 ≈ 2.84%。时间序列图显示出明显的下降趋势，几乎每周时间都在减少。进步相当稳定，不过第 3 周和第 4 周之间略有停滞。极差的减小表明表现越来越稳定。
4. Statistics in Business: Cost, Revenue and Break-Even | 商业中的统计：成本、收入与盈亏平衡
The school enterprise club sells handmade bracelets. Material cost per bracelet is £1.20, and the stall fee is £15 per day. Selling price is £3.50 per bracelet. Construct a table showing total cost, total revenue, and profit for 0, 5, 10, 15, 20, 25 sales. Find the break-even point (number of bracelets to sell to make zero loss). Plot both cost and revenue on the same axes and identify the break-even point graphically. What is the probability of making a profit if past data shows that on any day sales of 0–5, 6–10, 11–15, 16–20, 21–25 bracelets occur with relative frequencies 0.1, 0.2, 0.35, 0.25, 0.1?
学校创业社团出售手工编织手链。每条手链的材料成本为 1.20 英镑，摊位费每天 15 英镑。售价为每条 3.50 英镑。构建一个表格，显示销售量为 0、5、10、15、20、25 时的总成本、总收入和利润。找出盈亏平衡点（即零亏损所需销售的手链数量）。在同一坐标系中绘制成本和收入图形，并在图上标出盈亏平衡点。如果过去的数据显示，在任何一天，销售 0–5 条、6–10 条、11–15 条、16–20 条、21–25 条手链的相对频率分别为 0.1、0.2、0.35、0.25、0.1，那么获得盈利的概率是多少？
Total cost = 15 + 1.20 × n. Revenue = 3.50 × n. Profit = Revenue − Cost = 2.30n − 15. Set profit = 0 → n = 15 ÷ 2.30 ≈ 6.52, so the break-even point is 7 bracelets (since you can’t sell a fraction). For the table: n=0: cost=15, rev=0, profit=−15; n=5: cost=21, rev=17.5, profit=−3.5; n=10: cost=27, rev=35, profit=8; n=15: cost=33, rev=52.5, profit=19.5; n=20: cost=39, rev=70, profit=31; n=25: cost=45, rev=87.5, profit=42.5. Graphically, the cost line starts at (£0, £15) and cost at n=25 is £45; revenue line from (0,0) to (25, 87.5). They intersect at n≈6.5. Profit occurs when n≥7. From the frequency distribution, profit-making ranges are 11–15, 16–20, 21–25 with probabilities 0.35, 0.25, 0.10; also 6–10 range partly, but safe to consider only those above break-even. Probabilities for n≥11: 0.35+0.25+0.10 = 0.70. So P(profit) = 0.70 or 70%.
总成本 = 15 + 1.20 × n。收入 = 3.50 × n。利润 = 收入 − 成本 = 2.30n − 15。设利润为零 → n = 15 ÷ 2.30 ≈ 6.52，所以盈亏平衡点是 7 条手链（因为不能卖零头）。表格：n=0：成本=15，收入=0，利润=−15；n=5：成本=21，收入=17.5，利润=−3.5；n=10：成本=27，收入=35，利润=8；n=15：成本=33，收入=52.5，利润=19.5；n=20：成本=39，收入=70，利润=31；n=25：成本=45，收入=87.5，利润=42.5。在图形中，成本线从 (0 条, £15) 开始，到 n=25 时成本为 £45；收入线从 (0,0) 到 (25, 87.5)。两条线在 n≈6.5 处相交。当 n≥7 时盈利。根据频率分布，盈利的区间为 11–15、16–20、21–25，概率分别为 0.35、0.25、0.10；此外 6–10 区间部分盈利，但为保险起见只考虑超过盈亏平衡点的部分。n≥11 的概率：0.35+0.25+0.10 = 0.70。所以 P(盈利) = 0.70 或 70%。
5. Statistics in Health: Body Mass Index (BMI) Analysis | 健康中的统计：体质指数分析
The school nurse recorded the weights and heights of five Year 8 pupils:
| Pupil | Weight (kg) | Height (m) |
|---|---|---|
| A | 45 | 1.55 |
| B | 52 | 1.60 |
| C | 38 | 1.40 |
| D | 60 | 1.65 |
| E | 48 | 1.58 |
BMI = weight (kg) ÷ (height (m))². Calculate the BMI for each pupil. Classify each as underweight (<18.5), normal (18.5–24.9), overweight (25–29.9) or obese (≥30) using adult thresholds (for practice). Then find the mean BMI and comment on the group’s health. Would the mean be affected if a rugby player with weight 80 kg and height 1.75 m joined?
BMI = 体重 (kg) ÷ (身高 (m))²。计算每位学生的 BMI。使用成人阈值进行分类（作为练习）：偏瘦 (<18.5)、正常 (18.5–24.9)、超重 (25–29.9) 或肥胖 (≥30)。然后计算平均 BMI，并对该组的健康状况进行评价。如果一名体重 80 公斤、身高 1.75 米的橄榄球运动员加入，平均值会受到影响吗？
Pupil A: BMI = 45 ÷ (1.55²) = 45 ÷ 2.4025 ≈ 18.73 (normal). B: 52 ÷ (1.60²) = 52 ÷ 2.56 = 20.31 (normal). C: 38 ÷ (1.40²) = 38 ÷ 1.96 ≈ 19.39 (normal). D: 60 ÷ (1.65²) = 60 ÷ 2.7225 ≈ 22.04 (normal). E: 48 ÷ (1.58²) = 48 ÷ 2.4964 ≈ 19.23 (normal). Mean BMI = (18.73+20.31+19.39+22.04+19.23) ÷ 5 = 99.7 ÷ 5 ≈ 19.94. The group has a healthy average. Adding a rugby player with BMI = 80 ÷ (1.75²) = 80 ÷ 3.0625 ≈ 26.12 (overweight) would raise the mean to (99.7+26.12)÷6 ≈ 20.97, still normal, but the mean hides individual variation. A box plot would reveal the outlier.
学生 A：BMI = 45 ÷ (1.55²) = 45 ÷ 2.4025 ≈ 18.73（正常）。B：52 ÷ (1.60²) = 52 ÷ 2.56 = 20.31（正常）。C：38 ÷ (1.40²) = 38 ÷ 1.96 ≈ 19.39（正常）。D：60 ÷ (1.65²) = 60 ÷ 2.7225 ≈ 22.04（正常）。E：48 ÷ (1.58²) = 48 ÷ 2.4964 ≈ 19.23（正常）。平均 BMI = (18.73+20.31+19.39+22.04+19.23) ÷ 5 = 99.7 ÷ 5 ≈ 19.94。该组的平均健康水平良好。加入橄榄球运动员后，BMI = 80 ÷ (1.75²) = 80 ÷ 3.0625 ≈ 26.12（超重），平均 BMI 会上升到 (99.7+26.12)÷6 ≈ 20.97，仍属正常，但平均值掩盖了个体差异。箱线图可以显示出异常值。
6. Statistics in Environmental Studies: Temperature Change | 环境研究中的统计：气温变化
Over one week, the daily maximum temperatures (°C) in a town were: Mon 12, Tue 14, Wed 11, Thu 13, Fri 15, Sat 16, Sun 17. The same week last year had: Mon 10, Tue 11, Wed 10, Thu 12, Fri 13, Sat 14, Sun 14. Use a back-to-back stem-and-leaf diagram to compare the two weeks. Calculate the mean temperature for each week and the percentage increase in the overall mean. What conclusion can you draw? If the data were grouped into intervals of 10–12, 13–15, 16–18, construct a histogram for this year’s temperatures and explain why area is proportional to frequency.
在某一周内，一个城镇的日最高气温（°C）为：周一 12，周二 14，周三 11，周四 13，周五 15，周六 16，周日 17。去年同一周为：周一 10，周二 11，周三 10，周四 12，周五 13，周六 14，周日 14。使用背对背的茎叶图来对比这两周。计算每周的平均气温以及总体平均值的增长百分比。你能得出什么结论？如果将数据分组为 10–12、13–15、16–18 的区间，为今年的气温构建直方图，并解释为什么面积与频数成比例。
This year: data sorted 11,12,13,14,15,16,17; mean = (11+12+13+14+15+16+17)÷7 = 98÷7 = 14°C. Last year: 10,10,11,12,13,14,14; mean = (10+10+11+12+13+14+14)÷7 = 84÷7 = 12°C. Percentage increase = ((14−12)÷12)×100 ≈ 16.7%. Stem-and-leaf: stem = tens (1), leaves = units. This year leaves: 1|1 2 3 4 5 6 7 ; last year on left: 10,10,11,12,13,14,14 → 1|0 0 1 2 3 4 4. Back-to-back. The increase suggests consistently warmer days this year. Grouped histogram: intervals: 10≤t<13, 13≤t<16, 16≤t<19. Frequencies: 10–12: values 11,12 (2 days); 13–15: 13,14,15 (3 days); 16–18: 16,17 (2 days). In a histogram with equal class widths, area ∝ frequency because area = class width × frequency density, and class width constant, so area directly reflects frequency.
今年：数据排序 11,12,13,14,15,16,17；平均值 = (11+12+13+14+15+16+17)÷7 = 98÷7 = 14°C。去年：10,10,11,12,13,14,14；平均值 = (10+10+11+12+13+14+14)÷7 = 84÷7 = 12°C。增长百分比 = ((14−12)÷12)×100 ≈ 16.7%。茎叶图：茎为十位数 (1)，叶为个位数。今年：1|1 2 3 4 5 6 7；去年在左侧：10,10,11,12,13,14,14 → 1|0 0 1 2 3 4 4。背对背展示。增长表明今年的天气持续更温暖。分组直方图：区间：10≤t<13, 13≤t<16, 16≤t<19。频数：10–12：数值 11,12（2 天）；13–15：13,14,15（3 天）；16–18：16,17（2 天）。在组距相等的直方图中，面积与频数成比例，因为面积 = 组距 × 频率密度，而组距恒定，因此面积直接反映频数。
7. Statistics in Social Studies: Survey Design and Bias | 社会科学中的统计：调查设计与偏差
A student wants to find out the most popular social media platform among Year 8 pupils. She asks the first 30 students who enter the school library on a Monday morning. Identify the sampling method and explain why it might be biased. Suggest a better sampling method and describe how to implement it. If the results from the biased sample are 60% TikTok, 25% Instagram, 10% Snapchat, 5% YouTube, calculate the angle for each sector in a pie chart. How might the true percentages differ?
一名学生想了解八年级学生中最受欢迎的社交媒体平台。她询问了周一早上最早进入学校图书馆的 30 名学生。请指出这种抽样方法并解释为什么可能存在偏差。提出一种更好的抽样方法，并描述如何实施。如果该偏差样本的结果为 TikTok 60%、Instagram 25%、Snapchat 10%、YouTube 5%，请计算饼图中每个扇区的角度。真实百分比可能会有什么不同？
This is convenience sampling. It may be biased because students who visit the library on Monday morning might not represent the entire year group (e.g., they may be more studious and less likely to use certain apps). A better method is stratified random sampling: divide Year 8 into tutor groups or genders in proportion to the population, then randomly select the required number from each stratum. For the pie chart: TikTok angle = 0.60 × 360° = 216°; Instagram = 0.25 × 360° = 90°; Snapchat = 0.10 × 360° = 36°; YouTube = 0.05 × 360° = 18°. The true percentages might show a lower share for TikTok if the library users are less typical, possibly more YouTube and less TikTok. Random sampling variation would reduce such bias.
这是便利抽样。可能存在偏差，因为周一早上进入图书馆的学生可能无法代表整个年级（例如，他们可能更爱学习，不太可能使用某些应用程序）。更好的方法是分层随机抽样：按比例将八年级分成注册小组或性别层，然后从每一层中随机选择所需数量的学生。饼图角度：TikTok = 0.60 × 360° = 216°；Instagram = 0.25 × 360° = 90°；Snapchat = 0.10 × 360° = 36°；YouTube = 0.05 × 360° = 18°。如果图书馆使用者不太具有代表性，真实百分比可能会显示 TikTok 份额较低，YouTube 份额可能更高，TikTok 则更低。随机抽样可减少此类偏差。
8. Statistics in Food Technology: Recipe Scaling and Ratios | 食品技术中的统计：食谱缩放与比例
A basic biscuit recipe uses 250 g flour, 125 g butter, 75 g sugar, 1 egg (≈50 g). This makes 20 biscuits. A student wants to make 50 biscuits for a bake sale. Write the scaling factor and calculate the new ingredient quantities. If 20% of biscuits might break during transport, how many extra should be baked to ensure 50 intact ones are available? What is the ratio of flour to butter in its simplest form? Represent the ingredient proportions in a component bar chart for the original and scaled recipes.
一个基本的饼干食谱使用 250 克面粉、125 克黄油、75 克糖、1 个鸡蛋（约 50 克）。可制作 20 块饼干。一名学生想在烘焙义卖中制作 50 块饼干。请写出缩放系数，并计算新的配料用量。如果运输过程中 20% 的饼干可能破碎，为了确保有 50 块完好的饼干，应额外烘烤多少块？面粉与黄油的最简比例是多少？用分量条形图表示原始食谱和缩放后食谱的配料比例。
Scaling factor = 50 ÷ 20 = 2.5. New flour = 250 × 2.5 = 625 g; butter = 125 × 2.5 = 312.5 g; sugar = 75 × 2.5 = 187.5 g; eggs = 1 × 2.5 = 2.5 eggs (use 2 large eggs ≈ 100 g or beat 2.5 small). To ensure 50 intact biscuits, let x be number baked; 0.8x ≥ 50 → x ≥ 62.5, so bake 63 biscuits. Flour:butter ratio = 250:125 = 2:1. Component bar chart: original total mass = 250+125+75+50 = 500 g; scaled total = 500×2.5 = 1250 g. Draw bars with segments proportional to each ingredient’s mass.
缩放系数 = 50 ÷ 20 = 2.5。新面粉用量 = 250 × 2.5 = 625 克；黄油 = 125 × 2.5 = 312.5 克；糖 = 75 × 2.5 = 187.5 克；鸡蛋 = 1 × 2.5 = 2.5 个（可用 2 个大鸡蛋约 100 克，或搅打 2.5 个小鸡蛋）。为确保有 50 块完好的饼干，设烘烤数量为 x；0.8x ≥ 50 → x ≥ 62.5，因此应烘烤 63 块饼干。面粉与黄油的比例 = 250:125 = 2:1。分量条形图：原始总质量 = 250+125+75+50 = 500 克；缩放后总质量 = 500×2.5 = 1250 克。绘制条形，各段与每种配料的质量成比例。
9. Statistics in Art and Design: Colour Preference and Pie Charts | 艺术与设计中的统计：颜色偏好与饼图
An art teacher surveyed 40 Year 8 students about their favourite primary colour. Results: Red 12, Blue 18, Yellow 10. Create a pie chart and a bar chart for this data. Which representation makes it easier to see that Blue is the most popular? Calculate the relative frequency of each colour and use them to estimate how many students in a year group of 200 would prefer Blue. What sampling issues might arise if only one art class was surveyed?
一位美术老师调查了 40 名八年级学生最喜欢的三原色。结果：红色 12 人，蓝色 18 人，黄色 10 人。为这组数据创建一个饼图和一个条形图。哪种表示方式更容易看出蓝色最受欢迎？计算每种颜色的相对频率，并用它们估计在一个 200 人的年级组中会有多少人偏好蓝色。如果只调查了一个美术班，可能会出现哪些抽样问题？
Pie chart angles: Red = (12/40)×360° = 108°; Blue = (18/40)×360° = 162°; Yellow = (10/40)×360° = 90°. Bar chart: vertical bars with heights 12, 18, 10. The bar chart makes it instantly clear that Blue is the tallest bar. Relative frequencies: Red 0.30, Blue 0.45, Yellow 0.25. Estimated Blue lovers in 200: 0.45 × 200 = 90 students. Surveying only one art class is a convenience sample; it may not represent the whole year group because art students might have different colour preferences than, say, sports enthusiasts. This would lead to bias.
饼图角度：红色 = (12/40)×360° = 108°；蓝色 = (18/40)×360° = 162°；黄色 = (10/40)×360° = 90°。条形图：垂直条形，高度分别为 12、18、10。条形图可以一目了然地看出蓝色是最高的条形。相对频率：红色 0.30，蓝色 0.45，黄色 0.25。估计 200 人年级中喜欢蓝色的人数：0.45 × 200 = 90 人。只调查一个美术班属于便利样本；它可能无法代表整个年级，因为美术学生对颜色的偏好可能与体育爱好者等不同。这将导致偏差。
10. Interdisciplinary Problem-Solving: Combining Statistics with Logic | 跨学科问题解决：统计与逻辑的结合
A theme park wants to reduce queue times. They record the number of visitors per hour: 200, 250, 300, 280, 220, 180. Each ride can process 50 people per ride and takes 5 minutes per ride. How many rides are needed to keep the average queue time under 10 minutes? Use the mean number of visitors per hour. If the park decides to open a new ride every time the hourly visitor count exceeds 250, on which hours would new rides open? Calculate the percentage of operating hours that would have the extra ride. This problem requires you to use mean, threshold logic, and percentages together.
一家主题公园希望减少排队时间。他们记录了每小时的游客数量：200、250、300、280、220、180。每个游乐设施每次可容纳 50 人，每次运行需要 5 分钟。为使平均排队时间保持在 10 分钟以内，需要多少个游乐设施？请使用每小时游客的平均数。如果公园决定每当每小时游客数超过 250 时就开设一个新的游乐设施，那么在哪些时段会开设新设施？计算需要额外设施运营时间的百分比。这道题要求你同时运用平均数、阈值逻辑和百分比。
First, mean visitors per hour = (200+250+300+280+220+180) ÷ 6 = 1430 ÷ 6 ≈ 238.33. Each ride does 12 rides per hour (60 ÷ 5), so capacity per ride per hour = 50 × 12 = 600 people. To keep queue under 10 minutes, the throughput should match visitor arrival rate so that no large buildup occurs. With mean arrival 238, one ride is enough theoretically. However, to handle peaks, consider maximum hour: 300. Using one ride, capacity 600 > 300, so one ride suffices. Hence number of rides needed =
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