Year 12 SQA Statistics: Cross-disciplinary Comprehensive Question Training | SQA 高阶统计：跨学科综合题型训练

📚 Year 12 SQA Statistics: Cross-disciplinary Comprehensive Question Training | SQA 高阶统计：跨学科综合题型训练

In the SQA Higher Statistics syllabus, the ability to apply statistical methods to real-world, cross-disciplinary contexts is essential. This article provides integrated practice covering descriptive statistics, probability, distributions, regression, hypothesis testing, and experimental design. Each section presents a scenario-based question typical of Year 12 assessments, with step-by-step bilingual guidance.

在SQA高等统计大纲中，将统计方法应用于跨学科真实情境的能力至关重要。本文提供综合训练，涵盖描述统计、概率、分布、回归、假设检验和实验设计。每节提供一个基于情境的题目，配以中英双语的逐步指导。

1. Descriptive Statistics in Environmental Science | 描述性统计在环境科学中的应用

A researcher recorded monthly rainfall (mm) in a Scottish moorland over 12 months: 120, 98, 145, 67, 210, 88, 154, 113, 76, 132, 105, 188.

研究者记录了苏格兰某沼泽地12个月的月降雨量（毫米）：120, 98, 145, 67, 210, 88, 154, 113, 76, 132, 105, 188。

To analyse central tendency and spread, first sort the data: 67, 76, 88, 98, 105, 113, 120, 132, 145, 154, 188, 210.

为分析集中趋势与离散程度，首先将数据排序：67, 76, 88, 98, 105, 113, 120, 132, 145, 154, 188, 210。

The median is the average of the 6th and 7th values: (113+120)/2 = 116.5 mm.

中位数是第6和第7个值的平均数：(113+120)/2 = 116.5 mm。

The lower quartile Q1 is the median of the lower half: (88+98)/2 = 93 mm.

下四分位数Q1是下半部分的中位数：(88+98)/2 = 93 mm。

The upper quartile Q3 is the median of the upper half: (145+154)/2 = 149.5 mm.

上四分位数Q3是上半部分的中位数：(145+154)/2 = 149.5 mm。

Interquartile range IQR = Q3 – Q1 = 149.5 – 93 = 56.5 mm.

四分位距 IQR = 149.5 – 93 = 56.5 mm。

Outlier boundaries: lower fence = Q1 – 1.5 × IQR = 93 – 84.75 = 8.25 mm; upper fence = Q3 + 1.5 × IQR = 149.5 + 84.75 = 234.25 mm. No data points lie outside these fences, so no outliers.

异常值界限：下界 = 93 – 1.5 × 56.5 = 8.25 mm；上界 = 149.5 + 84.75 = 234.25 mm。没有数据点落在界限外，因此无异常值。

The mean x̄ = sum / n = (67+76+88+98+105+113+120+132+145+154+188+210) / 12 = 1496 / 12 ≈ 124.67 mm.

平均值 x̄ = 总和 / 样本量 = 1496 / 12 ≈ 124.67 mm。

Sample standard deviation uses the formula below, where n = 12:

样本标准差采用以下公式，其中 n = 12：

s = √[ Σ(xᵢ – x̄)² / (n – 1) ]

Calculating squared deviations yields s ≈ 43.2 mm. This indicates moderate variability around the mean.

计算平方偏差后得到 s ≈ 43.2 mm，表明降雨量在均值附近有中等程度的变异。

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2. Probability and Genetics | 概率与遗传学

A monohybrid cross of heterozygous parents (Aa × Aa) follows Mendelian inheritance. Construct a Punnett square to find the genotype probabilities.

杂合子亲本（Aa × Aa）单因子杂交遵循孟德尔遗传。绘制庞纳特方格求基因型概率。

Gametes	A	a
A	AA	Aa
a	Aa	aa

The probability of homozygous recessive (aa) is 1/4 = 0.25.

隐性纯合子 (aa) 的概率为 1/4 = 0.25。

Now consider a litter of 5 offspring. The number of aa pups follows a binomial distribution X ~ B(5, 0.25). The probability of at least 2 homozygous recessive is P(X ≥ 2) = 1 – P(X=0) – P(X=1).

现考虑一窝5只幼崽。aa 幼崽的数量服从二项分布 X ~ B(5, 0.25)。至少2只为隐性纯合子的概率为 P(X ≥ 2) = 1 – P(X=0) – P(X=1)。

Using the binomial formula P(X = k) = C(n,k) · pᵏ · (1 – p)ⁿ⁻ᵏ:

利用二项概率公式 P(X = k) = C(n,k) · pᵏ · (1 – p)ⁿ⁻ᵏ：

P(X = 0) = (0.75)⁵ ≈ 0.2373

P(X = 1) = 5 × 0.25 × (0.75)⁴ ≈ 0.3955

Hence P(X ≥ 2) = 1 – (0.2373 + 0.3955) = 0.3672. So there is about a 36.7% chance.

因此 P(X ≥ 2) = 1 – (0.2373 + 0.3955) = 0.3672，即约36.7%的可能性。

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3. Binomial Distribution in Quality Control | 二项分布在质量控制中的应用

A factory produces light bulbs with a 5% defect rate. A random sample of 20 bulbs is selected. Find the probability that exactly 2 are defective, and the probability of at least one defective.

某工厂生产的灯泡有5%不合格。随机抽取20个灯泡，求恰有2个不合格的概率，以及至少1个不合格的概率。

Let X ~ B(20, 0.05) be the number of defective bulbs. The probability of exactly 2 is:

设不合格灯泡数 X ~ B(20, 0.05)。恰有2个的概率为：

P(X = 2) = C(20,2) · (0.05)² · (0.95)¹⁸

C(20,2) = 190, (0.05)² = 0.0025, (0.95)¹⁸ ≈ 0.3972. Multiply to get P(X = 2) ≈ 190 × 0.0025 × 0.3972 ≈ 0.1887.

C(20,2) = 190，(0.05)² = 0.0025，(0.95)¹⁸ ≈ 0.3972。相乘得 P(X = 2) ≈ 0.1887。

The probability of at least one defective is simpler via complement: P(X ≥ 1) = 1 – P(X = 0) = 1 – (0.95)²⁰ ≈ 1 – 0.3585 = 0.6415.

至少1个不合格的概率利用补集计算：P(X ≥ 1) = 1 – (0.95)²⁰ ≈ 1 – 0.3585 = 0.6415。

Quality control engineers can use these probabilities to determine whether an unusually high defect count signals a process shift.

质量控制工程师可利用这些概率判断异常高的次品数是否意味着流程漂移。

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4. Normal Distribution in Psychology | 正态分布在心理学中的应用

IQ scores are modelled as N(100, 15²). Calculate the proportion of the population with IQ above 130, and determine the threshold for the lowest 5%.

智商分数服从正态分布 N(100, 15²)。计算智商超过130的人口比例，并找出最低5%的临界分数。

For X = 130, compute the Z-score: Z = (130 – 100) / 15 = 2.00.

对 X = 130，计算Z分数：Z = (130 – 100) / 15 = 2.00。

Using standard normal tables, P(Z < 2.00) ≈ 0.9772, so P(X > 130) = 1 – 0.9772 = 0.0228, i.e. 2.28%.

查标准正态表得 P(Z < 2.00) ≈ 0.9772，故 P(X > 130) = 1 – 0.9772 = 0.0228，即2.28%。

The lowest 5% corresponds to the Z-score for which P(Z < z) = 0.05. From tables, z₀.₀₅ ≈ –1.645.

最低5%对应概率0.05的Z分数。查表得 z₀.₀₅ ≈ –1.645。

Rearrange to find X: X = μ + z·σ = 100 + (–1.645) × 15 ≈ 75.3. Thus, an IQ of about 75 marks the lower 5% boundary.

解出 X = 100 + (–1.645) × 15 ≈ 75.3。因此智商约75分为最低5%的界限。

Such analysis helps psychologists interpret extreme scores and set diagnostic criteria.

此类分析有助于心理学家解释极端分数并设定诊断标准。

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5. Linear Regression in Economics | 线性回归在经济学中的应用

A firm records monthly advertising spend (x, in £1000) and sales revenue (y, in £1000) over 6 months:

某公司记录6个月的月度广告支出（x，千英镑）和销售收入（y，千英镑）：

x (spend)	2.0	3.0	4.5	5.5	7.0	8.0
y (revenue)	12	18	24	30	36	42

Calculate the least-squares regression line y = a + bx and interpret the slope. Then predict sales for x = 6.0.

计算最小二乘回归线 y = a + bx，解释斜率，并预测 x = 6.0 时的销售收入。

First compute sums: n = 6, Σx = 30.0, Σy = 162.0, Σxy = 953, Σx² = 176.5, Σy² = 4860.

先计算各项和：n = 6, Σx = 30.0, Σy = 162.0, Σxy = 953, Σx² = 176.5, Σy² = 4860。

The slope b = Sxy / Sxx, where Sxx = Σx² – (Σx)²/n = 176.5 – 30²/6 = 176.5 – 150 = 26.5. Sxy = Σxy – (Σx)(Σy)/n = 953 – (30×162)/6 = 953 – 810 = 143.

斜率 b = Sxy / Sxx，其中 Sxx = 176.5 – 900/6 = 26.5，Sxy = 953 – (30×162)/6 = 143。

b = 143 / 26.5 ≈ 5.396

Intercept a = (Σy/n) – b·(Σx/n) = (162/6) – 5.396 × (30/6) = 27 – 5.396 × 5 = 27 – 26.98 ≈ 0.02.

截距 a = 27 – 5.396 × 5 ≈ 0.02。

So the regression line is y = 0.02 + 5.396x. The slope 5.396 means that for each additional £1000 spent on advertising, sales revenue increases by roughly £5396 on average.

因此回归线为 y = 0.02 + 5.396x。斜率5.396表示广告支出每增加1000英镑，销售收入平均增加约5396英镑。

Prediction at x = 6.0: y = 0.02 + 5.396 × 6 = 0.02 + 32.376 = 32.396, so sales are predicted at about £32,400.

当 x = 6.0 时，y = 0.02 + 5.396 × 6 = 32.396，预测销售收入约为32,400英镑。

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6. Hypothesis Testing in Medicine | 假设检验在医学中的应用

A new drug is trialled to assess whether it significantly lowers diastolic blood pressure. Nine patients had their pressure measured before and after treatment; the mean reduction is 5.2 mmHg with standard deviation of the differences s_d = 4.1 mmHg. Test at 5% significance whether the drug is effective (one-tailed).

某新药试验评估其是否显著降低舒张压。9名患者治疗前后测得血压，减少量平均为5.2 mmHg，差值的标准差 s_d = 4.1 mmHg。在5%显著性水平下检验该药是否有效（单尾检验）。

Let μ_d be the population mean reduction. State hypotheses: H₀: μ_d = 0; H₁: μ_d > 0.

设 μ_d 为总体平均降低量。假设：H₀: μ_d = 0；H₁: μ_d > 0。

Since the sample size is small and population variance unknown, use a one-sample t-test on the differences.

由于样本量小且总体方差未知，对差值采用单样本t检验。

Test statistic: t = (x̄_d – 0) / (s_d / √n) = 5.2 / (4.1/3) = 5.2 / 1.3667 ≈ 3.804.

检验统计量：t = (x̄_d – 0) / (s_d / √n) = 5.2 / (4.1/3) = 5.2 / 1.3667 ≈ 3.804。

Degrees of freedom v = n – 1 = 8. The critical value for a one-tailed t-test at 5% level from tables is t₈,₀.₀₅ ≈ 1.860.

自由度 v = 8。查表得单尾5%临界值 t₈,₀.₀₅ ≈ 1.860。

Since 3.804 > 1.860, we reject H₀. There is sufficient evidence to conclude the drug significantly reduces diastolic blood pressure.

由于3.804 > 1.860，拒绝H₀。有充分证据表明该药显著降低舒张压。

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7. Chi-squared Test for Independence in Sociology | 独立性卡方检验在社会学中的应用

A survey explores the association between gender and voting preference. Observed frequencies are shown in the contingency table:

一项调查探讨性别与投票倾向之间的关系。观察频数列联表如下：

	Party A	Party B	Total
Male	40	30	70
Female	35	45	80
Total	75	75	150

Perform a χ² test for independence at the 5% significance level.

在5%显著性水平下进行χ²独立性检验。

Expected frequencies are computed by (row total × column total)/grand total. For Male, Party A: E = 70×75/150 = 35. Others similarly: Male, Party B = 35; Female, Party A = 40; Female, Party B = 40.

期望频数 = (行合计 × 列合计) / 总计。男性-党A：E = 70×

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