📚 CCEA Year 10 Statistics: Case Study Practice | CCEA Year 10 统计案例分析实战演练
Welcome to this hands-on case study designed for CCEA Year 10 Statistics. In this article, you will follow the journey of a small cafe, Sunshine Cafe, and use real data to practise every major topic in your syllabus – from frequency tables and box plots to scatter graphs and probability. By working through the exercises, you will learn how to think like a statistician, turning raw numbers into meaningful conclusions.
欢迎来到这个为 CCEA Year 10 统计学设计的实战案例分析。本文将带你走进一家名为阳光咖啡馆的小店，利用真实数据练习考纲中的每一个重要主题——从频数表和箱线图到散点图和概率。通过一步步的练习，你将学会像统计学家一样思考，把原始数字转化为有意义的结论。
1. Introducing the Sunshine Café Dataset | 介绍阳光咖啡馆数据集
Sunshine Cafe is a small coffee shop that has recorded its daily customer numbers over a two-week period. The owner wants to understand customer flow to make better staffing decisions. The raw data for 14 consecutive days is: 45, 52, 38, 61, 55, 48, 59, 63, 40, 50, 57, 45, 49, 54. Each value represents the total number of customers served in a day.
阳光咖啡馆是一家小型咖啡店，记录了连续两周的每日顾客数量。店主希望了解客流情况以便做出更合理的排班决策。连续14天的原始数据为：45, 52, 38, 61, 55, 48, 59, 63, 40, 50, 57, 45, 49, 54。每个数值代表一天接待的顾客总数。
2. Organising Data into Frequency Tables | 将数据组织为频数表
We first group the data into class intervals of width 10, starting with 30–39. Tallying the numbers gives the following frequency distribution: 30–39: 1 day; 40–49: 5 days; 50–59: 6 days; 60–69: 2 days. A grouped frequency table makes it much easier to see the spread at a glance.
我们首先将数据按组距 10 分组，从 30–39 开始。清点数据后得到如下频数分布：30–39: 1 天；40–49: 5 天；50–59: 6 天；60–69: 2 天。分组频数表让我们能够一目了然地看出数据的分布情况。
| Class Interval | Frequency |
|---|---|
| 30–39 | 1 |
| 40–49 | 5 |
| 50–59 | 6 |
| 60–69 | 2 |
When constructing a grouped frequency table, ensure that class boundaries do not overlap and that every data point belongs to exactly one class. The class width here is uniform, which is essential for drawing a correct histogram later.
在构建分组频数表时，要确保组与组之间没有重叠，且每个数据点恰好属于一个组。这里的组距是一致的，这对后续绘制正确的直方图至关重要。
3. Visualising Data: Bar Charts and Histograms | 数据可视化：条形图与直方图
For discrete count data like daily customers, a bar chart is appropriate because the horizontal axis shows individual categories. However, when data is grouped into continuous intervals, a histogram is used – the area of each bar is proportional to frequency. Here, because all class widths are equal, the histogram’s bar heights directly represent the frequencies.
对于像每日顾客数这样的离散计数数据，条形图是合适的，因为横轴展示的是单独的类别。然而，当数据被分成连续区间时，就要使用直方图——每个条形的面积与频数成正比。在这里，由于所有组距都相等，直方图的条形高度直接代表频数。
Imagine drawing a histogram with the class intervals on the x-axis. The bar for 40–49 reaches a height of 5, while the modal class 50–59 reaches 6. A frequency polygon can be added by joining the midpoints of the tops of the bars: 34.5 (height 1), 44.5 (5), 54.5 (6), 64.5 (2). This helps to visualise the shape of the distribution, which appears roughly symmetric with a slight right skew.
设想画一个直方图，横轴为组区间。40–49 的条形高度为 5，而众数所在组 50–59 的高度为 6。还可以通过连接各条形顶部中点来添加频数多边形，中点分别为 34.5 (高度1), 44.5 (5), 54.5 (6), 64.5 (2)。这有助于直观地看出分布的形状，大致对称，略带右偏。
4. Measures of Central Tendency | 集中趋势的度量
5+52+38+61+55+48+59+63+40+50+57+45+49+54 = 716, so the mean is 716 ÷ 14 ≈ 51.14 customers.
集中趋势让我们能用一个典型值来概括数据。首先是平均数：将所有 14 个数值相加再除以 14。总和为 45+52+38+61+55+48+59+63+40+50+57+45+49+54 = 716，因此平均数约为 716 ÷ 14 ≈ 51.14 名顾客。
an of the 7th and 8th values: (50 + 52) ÷ 2 = 51 customers. The mode, the most frequent value, is 45, which occurs twice.
要找出中位数，我们将数据排序：38, 40, 45, 45, 48, 49, 50, 52, 54, 55, 57, 59, 61, 63。因为有 14 个数值（偶数个），中位数是第 7 与第 8 个数值的平均数：(50 + 52) ÷ 2 = 51 名顾客。众数，即出现最频繁的值，是 45，出现了两次。
Mean = Σxᵢ / n = 716/14 ≈ 51.14
5. Measures of Spread: Range, Quartiles and IQR | 离散程度的度量：范围、四分位数及四分位距
tells us how scattered the data is. The range is simply the maximum minus the minimum: 63 – (Q₂) is 51. The lower quartile Q₁ is the median of the first half: 38, 40, 45, 45, 48, 49 – the middle two are 45 and 45, so Q₁ = 45.
离散程度告诉我们数据有多分散。范围就是最大值减去最小值：63 – 38 = 25。四分位数是将排序后的数据分成四等份。中位数 (Q₂) 为 51。下四分位数 Q₁ 是前半部分数据的中位数：38, 40, 45, 45, 48, 49 —— 中间两个数是 45 和 45，因此 Q₁ = 45。
The upper quartile Q₃ is the median of the second half: 52, 54, 55, 57, 59, 61, 63 – the middle two are 55 and 57, so Q₃ = 56. The interquartile range (IQR) = Q₃ – Q₁ = 56 – 45 = 11. The IQR measures the spread of the middle 50% of customers, which is more robust than the range.
上四分位数 Q₃ 是后半部分数据的中位数：52, 54, 55, 57, 59, 61, 63 —— 中间两个数是 55 和 57，因此 Q₃ = 56。四分位距 (IQR) = Q₃ – Q₁ = 56 – 45 = 11。四分位距衡量的是中间 50% 顾客数的分布情况，它比范围更不容易受极端值影响。
| Min | Q₁ | Median | Q₃ | Max |
|---|---|---|---|---|
| 38 | 45 | 51 | 56 | 63 |
6. Box Plots and Outliers | 箱线图和异常值
end to the minimum (38) and maximum (63) because there are no outliers. To check for outliers, we compute the lower fence: Q₁ – 1.5 × IQR = 45 – ence: Q₃ + 1.5 × IQR = 56 + 16.5 = 72.5. All data points lie well within these boundaries.
箱线图利用上一节的五数概括来绘制。盒子从 Q₁ (45) 画到 Q₃ (56)，并在中位数 (51) 处画一条线。由于没有异常值，须线一直延伸到最小值 (38) 和最大值 (63)。为了检验异常值，我们计算下界：Q₁ – 1.5 × IQR = 45 – 16.5 = 28.5，上界：Q₃ + 1.5 × IQR = 56 + 16.5 = 72.5。所有数据点都稳稳地落在这个范围内。
Drawing a box plot helps the cafe owner see that although the median is 51, the daily count could be as low as 38 or as high as 63, giving a realistic expectation of customer variability.
画出箱线图能帮助咖啡店老板直观地看到，尽管中位数是 51，但每日客流量可能低至 38 或高达 63，从而对客流波动有一个现实的预期。
7. Introduction to Probability Using Data | 利用数据介绍概率
at is the probability that on a randomly chosen day the cafe serves more than 55 customers? Counting from our list: 61, 59, 63, 57 are greater than 55 – that is 4 days. So, P(customers > 55) = 4/14 ≈ 0.286 or 28.6%.
概率可以通过相对频数从历史数据中估算出来。例如，随机选取一天，咖啡馆服务超过 55 名顾客的概率是多少？从列表中数出：61, 59, 63, 57 共 4
Published by TutorHao | Year 10 统计 Revision Series | aleveler.com
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