AS Cambridge Statistics: Case Study Practical Exercise | AS剑桥统计：案例分析实战演练

📚 AS Cambridge Statistics: Case Study Practical Exercise | AS剑桥统计：案例分析实战演练

This article presents a comprehensive case study that mirrors the style of Cambridge AS Statistics examination questions. Through the analysis of real-world data on light bulb lifetimes, you will practise essential techniques such as data organisation, calculation of summary statistics, construction of cumulative frequency graphs, box plots, linear transformations, and applications of the binomial and geometric distributions. Each section integrates bilingual explanations to reinforce your understanding of both statistical concepts and the examination format.

本文提供一个贴近剑桥AS统计考试风格的完整案例分析。通过分析灯泡寿命的真实数据，你将练习数据整理、概括统计量计算、累积频率图构建、箱线图绘制、线性变换以及在二项分布和几何分布中的应用等核心技巧。每个小节均配有中英双语解析，帮助你在掌握统计概念的同时熟悉考试题型。

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1. Case Study Background | 案例背景介绍

A quality control department at BrightLux manufacturing randomly sampled 50 light bulbs and recorded their lifetimes in hours. The data were grouped into the following frequency distribution.

BrightLux制造公司的质量控制部门随机抽取了50只灯泡，记录下各自的使用寿命（以小时计），并将数据整理成如下频率分布表。

Lifetime (hours)	Number of bulbs (f)
100–150	5
150–200	12
200–250	18
250–300	10
300–350	3
350–400	2

The company defines a bulb as ‘defective’ if its lifetime is less than 150 hours. You will use this dataset to perform a series of statistical investigations.

该公司将寿命低于150小时的灯泡定义为“次品”。你将基于这组数据开展一系列统计分析。

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2. Organising Data | 数据整理

To compute summary statistics efficiently we first find the midpoint of each class interval, denoted by x.

为高效计算概括统计量，我们首先确定每个组区间的中点，记为x。

Lifetime (h)	Midpoint (x)	f	fx
100–150	125	5	625
150–200	175	12	2100
200–250	225	18	4050
250–300	275	10	2750
300–350	325	3	975
350–400	375	2	750
Total		Σf = 50	Σfx = 11250

The sum of frequencies confirms the sample size n = 50, and the total Σfx will be used to calculate the mean.

频率之和确认样本容量n = 50，总和Σfx将用于计算均值。

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3. Calculating Mean and Variance | 计算均值和方差

The sample mean is obtained by dividing Σfx by n.

样本均值等于Σfx除以n。

Mean, x̄ = Σfx / n = 11250 / 50 = 225 hours

To measure spread we compute the variance using the formula s² = (Σfx² / n) – x̄². First we determine fx² for each class.

为衡量分散程度，我们用公式s² = (Σfx² / n) – x̄²计算方差。首先计算每一组的fx²。

x	f	x²	fx²
125	5	15625	78125
175	12	30625	367500
225	18	50625	911250
275	10	75625	756250
325	3	105625	316875
375	2	140625	281250
Total			Σfx² = 2,711,250

Now substitute into the variance formula.

现在代入方差公式。

s² = (2,711,250 / 50) – (225)² = 54,225 – 50,625 = 3,600 hours²

The standard deviation is the positive square root of the variance.

标准差是方差的正平方根。

s = √3600 = 60 hours

Therefore, on average the bulbs last 225 hours with a standard deviation of 60 hours.

因此，灯泡的平均使用寿命为225小时，标准差为60小时。

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4. Cumulative Frequency and Percentiles | 累积频率与百分位数

We now build a cumulative frequency table to determine the median and quartiles.

现在我们构建累积频率表以确定中位数和四分位数。

Upper class boundary	Cumulative frequency
150	5
200	17
250	35
300	45
350	48
400	50

The median position is at n/2 = 25. This falls in the 200–250 class. Using linear interpolation:

中位数位置为n/2 = 25，落在200–250组内。采用线性插值法：

Median = 200 + [(25 – 17) / 18] × 50 = 200 + (8/18)×50 ≈ 222.2 hours

The lower quartile Q₁ is at n/4 = 12.5, lying in the 150–200 class.

下四分位数Q₁位于n/4 = 12.5，落在150–200组内。

Q₁ = 150 + [(12.5 – 5) / 12] × 50 = 150 + (7.5/12)×50 ≈ 181.3 hours

The upper quartile Q₃ is at 3n/4 = 37.5, in the 250–300 class.

上四分位数Q₃位于3n/4 = 37.5，落在250–300组内。

Q₃ = 250 + [(37.5 – 35) / 10] × 50 = 250 + (2.5/10)×50 = 262.5 hours

Thus the interquartile range IQR = Q₃ – Q₁ = 262.5 – 181.3 = 81.2 hours.

因此四分位距IQR = Q₃ – Q₁ = 262.5 – 181.3 = 81.2小时。

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5. Box Plot and Outliers | 箱线图与异常值

Outlier boundaries are defined using Q₁ – 1.5×IQR and Q₃ + 1.5×IQR.

异常值界限由Q₁ – 1.5×IQR和Q₃ + 1.5×IQR确定。

Lower boundary = 181.3 – 1.5×81.2 = 59.5 hours (below the minimum value, so no low outliers)

Upper boundary = 262.5 + 1.5×81.2 = 384.3 hours

The maximum recorded lifetime is 400 hours, which exceeds 384.3; therefore the value 400 is a high outlier.

记录的最大寿命为400小时，超过了384.3，因此数值400为一个高异常值。

A box plot can be sketched with the whisker extending to the next largest non-outlier value. The adjacent upper value from the data is in the 350–400 class, but the next largest non-outlier is taken as 350 (the upper boundary of the previous class). Thus the upper whisker ends at 350, and the point at 400 is marked as an outlier.

可以画出箱线图，其上须线延伸至相邻的最大非异常值。数据中次大非异常值取350（前一组的上界），因此上须线止于350，而400标为异常点。

The five‑number summary is: minimum ≈ 100, Q₁ = 181.3, median = 222.2, Q₃ = 262.5, maximum adjacent = 350, with an outlier at 400.

五数概括为：最小值约100，Q₁ = 181.3，中位数 = 222.2，Q₃ = 262.5，相邻最大值 = 350，异常值为400。

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6. Applying Linear Transformations | 线性变换应用

The quality team decides to express lifetimes in minutes instead of hours (1 hour = 60 minutes). If Y = 60X, find the mean and standard deviation of Y.

质量团队决定用分钟而非小时表示寿命（1小时 = 60分钟）。若Y = 60X，求Y的均值和标准差。

For a linear transformation Y = aX + b, the mean changes by the same rule: E(Y) = aE(X) + b, and the variance transforms as Var(Y) = a² Var(X).

对于线性变换Y = aX + b，均值遵循相同规则：E(Y) = aE(X) + b，方差变换为Var(Y) = a² Var(X)。

Here a = 60, b = 0. Using the previously calculated values:

此处a = 60，b = 0。代入之前计算的值：

Mean of Y = 60 × 225 = 13,500 minutes

Variance of Y = 60² × 3600 = 12,960,000 minutes²

Standard deviation of Y = √12,960,000 = 3,600 minutes

This illustrates how multiplying by a constant scales both the mean and the standard deviation.

这说明了乘以常数会等比例缩放均值和标准差。

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7. Probability of Selecting Bulbs | 选取灯泡的概率

Recall that a bulb is defective if its lifetime is below 150 hours. From the original data, the proportion of defectives p̂ = 5/50 = 0.1. Assume this proportion holds for the entire production and that bulbs are selected independently.

回顾次品定义为寿命低于150小时。从原始数据可知，次品比例p̂ = 5/50 = 0.1。假设该比例适用于整个生产批次，且灯泡的选择相互独立。

A retailer purchases a box of 8 bulbs. Find the probability that exactly 2 are defective.

某零售商购买了一盒8只灯泡。求恰好有2只是次品的概率。

Let X be the number of defective bulbs in the box. X follows a binomial distribution: X ~ B(8, 0.1).

设X为盒中次品个数。X服从二项分布：X ~ B(8, 0.1)。

P(X = 2) = ⁸C₂ × (0.1)² × (0.9)⁶ = 28 × 0.01 × 0.531441 ≈ 0.1488

So the probability is approximately 0.149.

因此概率约为0.149。

What is the probability that at least one bulb is defective? This is efficiently found via the complement rule.

至少有一只次品的概率？这可以通过补集规则高效计算。

P(X ≥ 1) = 1 – P(X = 0) = 1 – (0.9)⁸ = 1 – 0.430467 ≈ 0.5695

We can also calculate the expected number of defectives in a box of 8: E(X) = np = 8 × 0.1 = 0.8, with variance Var(X) = np(1-p) = 8 × 0.1 × 0.9 = 0.72.

我们还可以计算一盒8只中的期望次品数：E(X) = np = 8 × 0.1 = 0.8，方差Var(X) = np(1-p) = 8 × 0.1 × 0.9 = 0.72。

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8. Geometric Distribution Application | 几何分布应用

A separate quality check involves inspecting bulbs one by one until the first defective is found. Assuming the probability of a defective bulb remains 0.1 and inspections are independent, the number of bulbs inspected Y follows a geometric distribution: Y ~ Geo(0.1).

另一项质量检查是对灯泡逐个检测，直到发现第一个次品。假设次品概率保持0.1且各次检测独立，则所检灯泡数Y服从几何分布：Y ~ Geo(0.1)。

Find the probability that the first defective is the 5th bulb tested.

求第一个次品恰好是第5只受检灯泡的概率。

P(Y = 5) = (0.9)⁴ × 0.1 = 0.6561 × 0.1 = 0.06561

How many bulbs does the inspector expect to check before finding a defect? For a geometric distribution, the mean is 1/p.

检查员期望在找到次品前检查多少只灯泡？对于几何分布，均值为1/p。

E(Y) = 1 / 0.1 = 10 bulbs

The probability that more than 10 bulbs are tested before a defective appears is P(Y > 10). Using the formula for geometric tail probability, P(Y > r) = (1-p)^r.

在发现次品之前检测超过10只灯泡的概率为P(Y > 10)。使用几何分布的尾部概率公式，P(Y > r) = (1-p)^r。

P(Y > 10) = (0.9)¹⁰ ≈ 0.3487

These calculations highlight how the geometric distribution models the waiting time until the first success in a sequence of independent trials.

这些计算突显了几何分布如何模拟在一系列独立试验中直到首次成功所需的等待时间。

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9. Connecting the Statistical Ideas | 统计概念的综合串联

This case study demonstrates how descriptive statistics and probability distributions work together in quality control. The mean and variance gave us a summary of the typical lifetime and variation. The cumulative frequency graph and box plot visually displayed the distribution and identified an outlier. The binomial distribution allowed us to model the number of defectives in a fixed sample, while the geometric distribution described the sampling effort to find a defect.

本案例展示了描述性统计与概率分布在质量控制中如何协同作用。均值和方差给出了典型寿命和变异的概括；累积频率图和箱线图直观展示了分布并识别出异常值；二项分布使得我们能够建模固定样本中次品的数量；而几何分布则描述了为发现次品所需的抽样工作量。

In the Cambridge AS Statistics examination you can expect integrated problems that require you to move fluently between data representation, summary measures, and probability models. Practising with a full dataset like this will strengthen your ability to interpret information and choose the correct statistical techniques.

在剑桥AS统计考试中，你会遇到需要融合数据表示、概括量数和概率模型的综合问题。使用类似本案例的完整数据集进行练习，将增强你解读信息并选择正确统计方法的能力。

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Published by TutorHao | Statistics Revision Series | aleveler.com

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