AS CAIE Statistics: Case Study Walkthrough and Practice | AS CAIE 统计：案例分析实战演练

📚 AS CAIE Statistics: Case Study Walkthrough and Practice | AS CAIE 统计：案例分析实战演练

Case studies are the secret weapon for mastering AS Statistics. By stepping through a real dataset, you can connect isolated techniques – from frequency tables to normal approximations – into a single, flowing analysis. This walkthrough will build your confidence for CAIE exam questions, where contextual application is everything.

案例分析是攻克 AS 统计的秘密武器。通过一个真实的数据集，你可以把原本孤立的技巧（从频数表到正态近似）串联成一次完整的分析。这次实战演练将帮助你建立应对 CAIE 考试情境题的信心，因为归根结底，理论要在应用中闪光。

1. Understanding the Case | 理解案例

A coffee shop manager recorded the waiting times (in seconds) of 30 randomly selected customers during a busy morning rush. The raw data are: 38, 42, 45, 50, 55, 57, 60, 62, 65, 66, 68, 70, 72, 73, 75, 78, 80, 82, 85, 88, 90, 92, 95, 97, 100, 105, 108, 110, 115, 120. The goal is to describe the distribution of waiting times and use probability models to predict service levels.

一家咖啡店的经理在繁忙的早高峰记录了 30 位随机顾客的等待时间（秒）。原始数据为：38, 42, 45, 50, 55, 57, 60, 62, 65, 66, 68, 70, 72, 73, 75, 78, 80, 82, 85, 88, 90, 92, 95, 97, 100, 105, 108, 110, 115, 120。目标是描述等待时间的分布，并利用概率模型预测服务水平。

2. Organising Data into a Frequency Table | 将数据整理为频数表

We group the continuous data into equal-width intervals of 20 seconds: 30–49, 50–69, 70–89, 90–109, 110–129. The boundaries are 29.5, 49.5, 69.5, 89.5, 109.5, 129.5. The frequency table captures the counts at a glance.

我们将连续数据分成组距为 20 秒的等宽区间：30–49、50–69、70–89、90–109、110–129。组界为 29.5、49.5、69.5、89.5、109.5、129.5。频数表能够一目了然地展示计数。

Here we used continuous class boundaries to ensure that every possible waiting time belongs to exactly one interval. Always check that the total frequency sums to 30.

这里我们使用连续组界，确保每一个可能的等待时间恰好属于一个区间。务必检查总频数是否为 30。

3. Visualising with a Histogram | 使用直方图可视化

Since all intervals have the same width, frequency density equals frequency. The histogram simply plots frequency on the vertical axis against waiting time on the horizontal axis. Bars of equal width represent each interval, with no gaps between them because data are continuous.

由于所有区间的宽度相等，频数密度就等于频数。直方图只需在纵轴上标出频数，横轴上标出等待时间即可。用等宽的条形表示每个区间，且条形之间不留空隙，因为数据是连续的。

The distribution appears slightly positively skewed: there is a longer tail towards higher waiting times. The modal class is 70–89 seconds, and most customers wait between 50 and 109 seconds.

分布呈现出轻微的正偏态：较高等待时间一侧有一条较长的尾巴。众数所在组是 70–89 秒，大多数顾客的等待时间集中在 50 到 109 秒之间。

4. Measures of Central Tendency | 集中趋势的度量

Using the raw data, the exact mean waiting time is 2343 ÷ 30 = 78.1 seconds. The median lies between the 15th and 16th ordered values: 75 and 78, giving a median of 76.5 seconds.

利用原始数据，精确的平均等待时间为 2343 ÷ 30 = 78.1 秒。中位数位于第 15 和第 16 个排序值之间：75 和 78，因此中位数为 76.5 秒。

For grouped data, we estimate the mean with midpoints (40, 60, 80, 100, 120): (3×40 + 8×60 + 9×80 + 7×100 + 3×120) ÷ 30 = 2380 ÷ 30 ≈ 79.3 s. The median from grouped data uses 69.5 + (15−11)/9 × 20 ≈ 78.4 s. The two sets of results are close, demonstrating how grouping works in practice.

对于分组数据，我们用组中值（40、60、80、100、120）来估计均值：(3×40 + 8×60 + 9×80 + 7×100 + 3×120) ÷ 30 = 2380 ÷ 30 ≈ 79.3 秒。分组数据的中位数则为 69.5 + (15−11)/9 × 20 ≈ 78.4 秒。两组结果很接近，这展示了分组在实际中是如何运作的。

5. Measures of Spread | 离散程度的度量

The exact variance is calculated as s² = [Σx² − (Σx)²/n] ÷ (n−1). Here Σx² = 197283, Σx = 2343, n = 30. This gives s² ≈ 492.9 and standard deviation s ≈ 22.2 seconds. The range is 120 − 38 = 82 seconds, and the interquartile range can be found from the ordered list: Q1 = 62.5, Q3 = 95, so IQR = 32.5 seconds.

精确方差的计算公式为 s² = [Σx² − (Σx)²/n] ÷ (n−1)。此处 Σx² = 197283，Σx = 2343，n = 30。由此得出 s² ≈ 492.9，标准差 s ≈ 22.2 秒。极差为 120 − 38 = 82 秒，四分位距可从排序列表中得出：Q1 = 62.5，Q

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