Case Study Practice: Statistical Analysis in Action | 案例分析实战演练：统计学实际应用

📚 Case Study Practice: Statistical Analysis in Action | 案例分析实战演练：统计学实际应用

In CIE A-Level Statistics, real-world case studies blend data handling, probability, and distributions. This article walks you through a comprehensive scenario — a beverage company investigating consumer taste ratings — from sampling to normal approximation, reinforcing key S1 concepts.

在CIE A-Level统计学中，真实案例研究融合了数据处理、概率与分布。本文将带领你深入一个综合情景——一家饮料公司调查消费者口味评分——从抽样到正态近似，巩固S1的核心概念。

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1. The Scenario and Sampling | 案例情景与抽样方法

A beverage firm plans to launch a new fruit-flavoured drink targeting young adults aged 18–25. To gauge taste preferences, the company designs a survey. Because the target consumers are mainly university students, a stratified sample of 120 individuals is drawn from three local universities, with strata sizes proportional to each campus’s student population.

一家饮料公司计划推出一款新水果口味饮料，目标群体为18–25岁的年轻消费者。为了解口味偏好，公司设计了一项调查。由于目标消费者主要是大学生，他们从当地三所大学中按各校园学生人数比例分层抽取了120人作为样本。

University	Population	Sample Size
Campus A	1000	40
Campus B	1500	60
Campus C	500	20
Total	3000	120

Each selected participant is asked to taste a sample of the drink and rate it on a scale from 1 (dislike extremely) to 10 (like extremely). The data are then recorded anonymously.

每位被选中的参与者被要求品尝该饮料样品，并从1（非常不喜欢）到10（非常喜欢）进行评分。数据随后被匿名记录。

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2. Organising and Displaying Data | 数据的整理与展示

After collecting the ratings, the raw data are organised into a frequency table. Grouping is not needed because the ratings are integers from 1 to 10. The table below summarises the scores of all 120 participants.

收集评分后，原始数据被整理成一个频数分布表。由于评分是从1到10的整数，无需进行分组。下表汇总了全部120名参与者的评分。

Score (x)	1	2	3	4	5	6	7	8	9	10
Frequency (f)	2	4	6	10	15	25	20	18	12	8

To visualise the distribution, we can construct a stem-and-leaf diagram. Since scores are single digits, stems are the integer parts and leaves are the last digit (here just the score itself). For a data set this small, a bar chart or histogram with equal class widths would also clearly show the shape — slightly left‑skewed, peaking around score 6.

为了可视化分布，我们可以绘制茎叶图。由于评分是个位数，茎即为整数部分，叶即为最后一位数字（此处即为评分本身）。对于这样的小型数据集，等宽条形图或直方图也能清晰显示分布形状——略微左偏，在6分附近出现峰值。

Another powerful graphical summary is the box‑and‑whisker plot. Using cumulative frequencies, we find the median at the 60.5th value, which lies in score 6; the lower quartile Q₁ at the 30.5th value falls in score 5; and the upper quartile Q₃ at the 90.5th value falls in score 8. The interquartile range IQR = Q₃ − Q₁ gives a robust measure of spread.

另一种强有力的图形摘要是箱线图。通过累积频数，我们找到中位数位于第60.5个值，落在6分；下四分位数Q₁位于第30.5个值，落在5分；上四分位数Q₃位于第90.5个值，落在8分。四分位距IQR = Q₃ − Q₁提供了一种稳健的离散度量。

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3. Measures of Central Tendency and Spread | 集中趋势与离散程度的度量

Let us compute the sample mean x̄ and variance s² for ungrouped data using the frequency table. Summations are performed as follows:

让我们利用频数表计算未分组数据的样本均值x̄和方差s²。计算过程如下：

Σf = 120, Σfx = 1×2 + 2×4 + 3×6 + 4×10 + 5×15 + 6×25 + 7×20 + 8×18 + 9×12 + 10×8 = 765

x̄ = Σfx / Σf = 765 / 120 = 6.375

For variance we need Σfx²:

为计算方差需要Σfx²：

Σfx² = 1²×2 + 2²×4 + 3²×6 + 4²×10 + 5²×15 + 6²×25 + 7²×20 + 8²×18 + 9²×12 + 10²×8 = 5411

s² = (Σfx² / Σf) − x̄² = 5411/120 − (6.375)² ≈ 45.0917 − 40.6406 = 4.4511

Hence the standard deviation s = √4.4511 ≈ 2.110. This tells us that typical ratings deviate from the mean by about 2.1 points, indicating moderate variability around the central value of 6.4.

因此标准差s = √4.4511 ≈ 2.110。这表明典型评分偏离均值约2.1分，说明围绕中心值6.4存在着中等程度的变异。

The five‑number summary (minimum=1, Q₁≈5.2, median=6, Q₃≈7.8, maximum=10) complements the box plot, giving a clear picture of data location and spread without assuming symmetry.

五数概括（最小值=1，Q₁≈5.2，中位数=6，Q₃≈7.8，最大值=10）补充了箱线图，在不假设对称性的条件下清晰呈现了数据位置与分散情况。

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4. Probability and Combined Events | 概率与复合事件

From the survey, we estimate that a randomly chosen consumer gives a rating of 8 or higher with probability p ≈ 38/120 = 0.3167. For simplicity, we adopt p = 0.3 as the baseline ‘high‑score’ probability in further analysis.

根据调查，我们估计随机挑选的消费者给出8分及以上评分的概率为p ≈ 38/120 = 0.3167。为简单起见，在后续分析中我们采用p = 0.3作为基准’高分’概率。

Now suppose two consumers are selected independently at random. What is the probability that at least one of them gives a high score? Using the complement rule,

现在假设随机且独立地选取两名消费者。至少一人给出高分的概率是多少？利用补集法则，

P(at least one) = 1 − P(none high) = 1 − (0.7)² = 1 − 0.49 = 0.51

This simple probability shows how quickly the chance of finding a favourable opinion increases when multiple consumers are consulted, an important point when designing a sample size.

这一简单概率表明，在咨询多位消费者时，找到好评的机会迅速增加，这正是在设计样本量时的一个重要考量。

Similarly, we can calculate the probability that exactly one of two consumers rates the drink highly using the binomial formula (which will be formalised next) or by recognising the two sequences: (high, low) or (low, high).

类似地，我们可以用二项公式（稍后将正式化）或通过识别（高，低）和（低，高）两种序列，计算恰好有一名消费者给出高分的概率。

P(exactly one) = 2 × 0.3 × 0.7 = 0.42

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5. Discrete Random Variables: Binomial Distribution | 离散随机变量：二项分布

In a follow‑up tasting session, the company randomly selects 10 consumers and records how many of them give a score ≥ 8. Let X be the number of high‑score ratings. Under the baseline assumption, X follows the binomial distribution B(10, 0.3).

在一次后续品尝会中，公司随机选取10名消费者，记录其中给出≥8分的人数。设X为高分评分的个数。在基准假设下，X服从二项分布B(10, 0.3)。

The probability mass function is

概率质量函数为

P(X = k) = ¹⁰Cₖ × 0.3^k × 0.7^10−k, for k = 0, 1, … , 10

For example, the probability that exactly 3 consumers rate the drink highly is

例如，恰好有3名消费者给出高分的概率为

P(X = 3) = ¹⁰C₃ × (0.3)³ × (0.7)⁷

¹⁰C₃ = 120 → P(X=3) = 120 × 0.027 × 0.0823543 ≈ 0.2668

Often exam questions ask for P(X ≥ 2) or P(X ≤ 5). Using cumulative binomial tables or the formula, we find P(X ≥ 2) = 1 − P(X=0) − P(X=1). P(X=0) = 0.7¹⁰ ≈ 0.0282, P(X=1) = 10 × 0.3 × 0.7⁹ ≈ 0.1211, so P(X ≥ 2) ≈ 0.8507. Such a high probability suggests that at least 2 favourable ratings are almost certain in a group of 10.

考试常会要求求解P(X ≥ 2)或P(X ≤ 5)。利用累积二项表或公式，可求得P(X ≥ 2) = 1 − P(X=0) − P(X=1)。P(X=0)=0.7¹⁰ ≈ 0.0282，P(X=1)=10×0.3×0.7⁹ ≈ 0.1211，因此P(X ≥ 2) ≈ 0.8507。如此高的概率意味着在10人小组里几乎必然会出现至少2个好评。

It is also important to recall the mean and variance of a binomial random variable: E(X) = np = 10 × 0.3 = 3, Var(X) = np(1−p) = 10 × 0.3 × 0.7 = 2.1.

同样重要的是记住二项随机变量的均值和方差：E(X) = np = 10×0.3 = 3，Var(X) = np(1−p) = 10×0.3×0.7 = 2.1。

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6. Normal Approximation to Binomial | 二项分布的正态近似

When the company scales up the survey to n = 200 consumers with the same p = 0.3, using the binomial formula directly becomes impractical. Provided that np = 60 and nq = 140 are both greater than 5, the normal approximation N(μ, σ²) with continuity correction can be used.

当公司将调查规模扩大至n = 200名消费者且仍假设p = 0.3时，直接使用二项公式变得不切实际。由于np = 60和nq = 140均大于5，可以使用带连续性校正的正态近似N(μ, σ²)。

μ = np = 200 × 0.3 = 60

σ² = npq = 200 × 0.3 × 0.7 = 42 ⇒ σ = √42 ≈ 6.4807

Suppose we wish to find the probability that at least 70 out of 200 consumers rate the drink highly, i.e. P(X ≥ 70). Applying the continuity correction, we compute P(X ≥ 69.5) under N(60, 42).

假设我们希望求出200名消费者中至少有70人给出高分的概率，即P(X ≥ 70)。运用连续性校正，我们在N(60, 42)下计算P(X ≥ 69.5)。

z = (69.5 − 60) / 6.4807 ≈ 1.466

Using standard normal tables, Φ(1.47) ≈ 0.9292, so P(Z ≥ 1.466) ≈ 1 − 0.928 = 0.072 (approximately). This small probability (about 7.2%) indicates that observing at least 70 high scores would be somewhat surprising if the true p is only 0.3.

利用标准正态表，Φ(1.47) ≈ 0.9292，因此P(Z ≥ 1.466) ≈ 1 − 0.928 = 0.072（约）。这一小概率（约7.2%）表明，如果真实p仅为0.3，观察到至少70个高分多少会有些意外。

Always remember to state the continuity correction and to check the np and nq conditions before applying the normal approximation. These conditions are frequently rewarded in mark schemes.

务必记住陈述连续性校正，并在应用正态

Published by TutorHao | Year 12 统计 Revision Series | aleveler.com

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