AS AQA Statistics: Interdisciplinary Mixed-Question Practice | AS AQA 统计：跨学科综合题型训练

📚 AS AQA Statistics: Interdisciplinary Mixed-Question Practice | AS AQA 统计：跨学科综合题型训练

In AS AQA Statistics, the ability to apply statistical methods in real-world contexts is a vital skill. This article presents a series of cross-disciplinary mixed questions, linking statistics with biology, psychology, geography, business, and sports science. Each section focuses on a key topic from the specification and translates it into an authentic scenario, helping you build confidence and exam technique.

在 AS AQA 统计课程中，将统计方法应用于真实情境是一项关键能力。本文提供了一系列跨学科综合题型，将统计与生物学、心理学、地理学、商学和运动科学联系起来。每个小节聚焦考纲中的一个核心主题，并将其转化为真实场景，帮助你建立信心并提升考试技巧。

1. Sampling Methods in Ecology | 生态学中的抽样方法

A team of ecologists wants to estimate the density of a rare plant species in a 200 m × 200 m heathland. They decide to use 1 m² quadrats. To avoid bias, they implement simple random sampling by generating pairs of random coordinates using a calculator. An alternative would be systematic sampling with quadrats placed at regular intervals along transects.

一组生态学家想要估计一片200米×200米石南荒原中某种稀有植物的密度。他们决定使用1平方米的样方。为了避免偏差，他们通过计算器生成随机坐标对来实施简单随机抽样。另一种替代方法是沿着样带等间距放置样方进行系统抽样。

In this context, simple random sampling gives every possible quadrat location an equal chance of selection, which is ideal for statistical inference. However, systematic sampling is often logistically easier and may capture environmental gradients if transects are carefully placed perpendicular to the slope.

在此背景下，简单随机抽样使每个可能的样方位置都有相同的被选概率，非常适合于统计推断。而系统抽样通常在操作上更简便，如果样带巧妙地被设置为垂直于坡度，就可能捕捉到环境梯度。

2. Descriptive Statistics in Psychology | 心理学中的描述统计

A psychologist recorded reaction times (in milliseconds) of 12 participants completing a cognitive task: 230, 245, 252, 260, 265, 272, 275, 285, 290, 310, 325, 350. She wishes to summarise the data using measures of central tendency and spread.

一位心理学家记录了12名参与者完成一项认知任务的反应时间（毫秒）：230, 245, 252, 260, 265, 272, 275, 285, 290, 310, 325, 350。她希望使用集中趋势和离散度量来汇总这些数据。

The median is (272 + 275) ÷ 2 = 273.5 ms. The interquartile range (IQR) is found by locating Q&sub1; (position 3.25: 253 ms) and Q&sub3; (position 9.75: 306.25 ms), giving IQR ≈ 53.25 ms. Because the data set contains an outlier-friendly value of 350, the IQR and median are more appropriate than the mean and standard deviation.

中位数为 (272+275)÷2=273.5 毫秒。通过定位第一四分位数Q&sub1;（第3.25位：253毫秒）和第三四分位数Q&sub3;（第9.75位：306.25毫秒），得到四分位距 IQR≈53.25 毫秒。由于数据集中含有被看作异常值的350，中位数和IQR比均值和标准差更合适。

3. Graphical Representation in Geography | 地理学中的图形表示

Geography students measure the daily discharge (m³/s) of a river over 20 days. They construct a box plot to compare low-flow and high-flow periods. The five-number summary is: minimum 8, Q&sub1; 14, median 21, Q&sub3; 30, maximum 48.

地理专业的学生测量了一条河流连续20天的日流量（立方米/秒）。他们绘制箱线图来对比枯水期和丰水期。五数概括为：最小值8，Q&sub1; 14，中位数21，Q&sub3; 30，最大值48。

The box plot reveals a slight positive skew, as the upper whisker (from 30 to 48) is longer than the lower whisker (from 8 to 14). This suggests that very high discharges occur occasionally, perhaps after heavy rainfall. Students are also asked to calculate the lower fence: Q&sub1; − 1.5 × IQR = 14 − 1.5 × 16 = −10, so no low-end outliers.

箱线图显示出轻微的正偏态，因为上须（30到48）比下须（8到14）更长。这表明偶尔会出现极高的流量，可能在大雨之后。还要求学生计算下内限：Q&sub1;−1.5×IQR = 14 − 1.5×16 = −10，因此没有低端异常值。

4. Probability: Mendelian Genetics | 概率：孟德尔遗传学

In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). Two heterozygous tall plants (Tt) are crossed. We can model the offspring’s genotype using a tree diagram or a Punnett square to find probabilities.

在豌豆植株中，高茎等位基因(T)对矮茎等位基因(t)为显性。将两棵杂合高茎植株(Tt)进行杂交。我们可以使用树状图或庞纳特方格来求出后代的基因型概率。

Each parent produces gametes T and t with probability 0.5. The offspring genotype probabilities: TT = 0.5×0.5 = 0.25, Tt = (0.5×0.5)+(0.5×0.5) = 0.5, tt = 0.25. Since T is dominant, the probability of a tall phenotype is P(TT or Tt) = 0.75. This is a simple application of the multiplication and addition rules of probability.

每个亲本产生配子T和t的概率均为0.5。后代基因型概率：TT=0.5×0.5=0.25，Tt=(0.5×0.5)+(0.5×0.5)=0.5，tt=0.25。由于T为显性，高茎表型的概率为P(TT或Tt)=0.75。这是概率乘法和加法规则的一个简单应用。

5. Binomial Distribution in Quality Control | 二项分布与质量控制

A factory produces light bulbs. Historically, 5% are defective. A random sample of 15 bulbs is selected. The number of defective bulbs, X, follows a binomial distribution: X ~ B(15, 0.05). The quality manager wants to know the probability of finding at most one defective bulb in the sample.

一家工厂生产灯泡。历史上，次品率为5%。现随机选取15个灯泡为样本。次品灯泡的数量X服从二项分布：X ~ B(15, 0.05)。质量经理想知道样本中最多个次品的概率。

P(X ≤ 1) = P(X=0) + P(X=1). Using the binomial formula: P(X=0) = ¹&sup5;C&sub0; × 0.05&sup0; × 0.95¹&sup5; ≈ 0.4633. P(X=1) = ¹&sup5;C&sub1; × 0.05¹ × 0.95¹&sup4; ≈ 0.3658. So P(X ≤ 1) ≈ 0.8291. This means there is about an 83% chance that the sample contains no more than one defective bulb, which assists in setting inspection rules.

P(X ≤ 1) = P(X=0) + P(X=1)。使用二项概率公式：P(X=0) = ¹&sup5;C&sub0; × 0.05&sup0; × 0.95¹&sup5; ≈ 0.4633。P(X=1) = ¹&sup5;C&sub1; × 0.05¹ × 0.95¹&sup4; ≈ 0.3658。因此P(X ≤ 1) ≈ 0.8291。这意味着样本里不超过一个次品的概率约83%，有助于制定抽检规则。

6. Normal Distribution in Sports Science | 运动科学中的正态分布

The heights of 16-year-old male basketball players in a national programme are normally distributed with mean 187 cm and standard deviation 8 cm. A coach wants to know the proportion of players taller than 195 cm, and also the height that 90% of players exceed.

国家训练计划中16岁男子篮球运动员的身高服从正态分布，均值为187厘米，标准差为8厘米。一位教练想知道身高超过195厘米的比例，以及90%的球员都超过的那个身高值。

For X ~ N(187, 8²), P(X > 195) = P(Z > (195−187)/8) = P(Z > 1). From statistical tables, Φ(1) = 0.8413, so P(Z > 1) = 1 − 0.8413 = 0.1587. About 15.9% are taller than 195 cm. For the 90^th percentile reversed, we use P(Z > −z) = 0.90, so z ≈ −1.2816. Height = 187 − 1.2816×8 ≈ 176.7 cm. Thus 90% of players are taller than about 177 cm.

对于 X ~ N(187, 8²)，P(X > 195) = P(Z > (195−187)/8)=P(Z > 1)。查表得Φ(1)=0.8413，故P(Z > 1)=1−0.8413=0.1587。约15.9%的球员身高超过195厘米。对于反向求第90百分位数，利用P(Z > −z)=0.90，得z≈−1.2816。身高 = 187 − 1.2816×8 ≈ 176.7厘米。因此90%的球员身高超过约177厘米。

7. Hypothesis Testing (Binomial) in Medicine | 医学中的假设检验（二项分布）

A pharmaceutical company claims that a new migraine drug is effective in at least 70% of cases. In a trial with 20 patients, 11 report relief. Test the claim at the 5% significance level. Let p be the true proportion of patients who gain relief from the drug. H&sub0;: p = 0.7, H&sub1;: p < 0.7.

一家制药公司声称其新型偏头痛药物在至少70%的病例中有效。在一项20名患者的试验中，11名报告缓解。在5%显著性水平下检验该声明。令p为药物真正见效的患者比例。H&sub0;: p = 0.7，H&sub1;: p < 0.7。

Under H&sub0;, X ~ B(20, 0.7). The observed test statistic is x = 11. We need P(X ≤ 11) ≈ 0.0103 (from table or calculator). Since 0.0103 < 0.05, we reject H&sub0;. There is sufficient evidence, at the 5% level, to suggest that the proportion of patients who gain relief is less than 70%. This could be due to the drug being less effective than claimed, or due to the small sample size.

在H&sub0;下，X ~ B(20, 0.7)。观测检验统计量x=11。需要计算P(X ≤ 11)≈0.0103（查表或计算器）。由于0.0103<0.05，我们拒绝H&sub0;。在5%显著性水平下，有足够证据表明见效比例小于70%。这可能因为药物不如宣称的有效，或由于样本量较小。

8. Hypothesis Testing (Normal) in Manufacturing | 制造业中的假设检验（正态分布）

A machine cuts steel rods whose lengths are normally distributed. The target mean is 150.0 mm. From past data, the population standard deviation is known to be 0.8 mm. A quality inspector takes a random sample of 25 rods and obtains a mean length of 149.5 mm. Test at the 2% significance level whether the machine needs adjustment.

一台机器切割钢棒，其长度服从正态分布。目标均值为150.0毫米。根据历史数据，总体标准差已知为0.8毫米。一名质检员随机抽取25根钢棒，得到平均长度为149.5毫米。在2%显著性水平下检验机器是否需要调整。

H&sub0;: μ = 150, H&sub1;: μ ≠ 150, two-tailed test. Test statistic: Z = (149.5 − 150) / (0.8/√25) = −0.5 / 0.16 = −3.125. Critical values at 2%: ±2.326. Since |−3.125| > 2.326, we reject H&sub0;. There is evidence that the mean length has shifted from 150 mm. The machine should be recalibrated.

H&sub0;: μ = 150，H&sub1;: μ ≠ 150，双尾检验。检验统计量：Z = (149.5−150)/(0.8/√25) = −0.5/0.16 = −3.125。2%显著性水平下的临界值为±2.326。由于|−3.125|>2.326，我们拒绝H&sub0;。有证据表明平均长度已偏离150毫米。该机器需要重新校准。

9. Correlation and Regression in Business | 商学中的相关与回归

A marketing analyst studied the relationship between monthly advertising spend (£100s) and sales revenue (£1000s) for 10 small businesses. The data produced a product moment correlation coefficient (PMCC) r = 0.892. Test H&sub0;: ρ = 0 against H&sub1;: ρ > 0 at the 5% level. Then calculate the equation of the regression line given that the means are x¯ = 15, y¯ = 22, Sxx = 640, Sxy = 512.

一位市场分析员研究10家小企业的月广告支出（百英镑）与销售收入（千英镑）之间的关系。数据得到积矩相关系数 r = 0.892。在5%水平下检验 H&sub0;: ρ = 0 对 H&sub1;: ρ > 0。然后根据已知均值 x¯ = 15, y¯ = 22, Sxx = 640, Sxy = 512，计算回归线方程。

Critical value for n = 10 (one-tailed, 0.05) is 0.549. Since 0.892 > 0.549, reject H&sub0;. There is strong evidence of positive correlation. The gradient b = Sxy / Sxx = 512 / 640 = 0.8. Intercept a = y¯ − b x¯ = 22 − 0.8×15 = 10. Regression line: y = 10 + 0.8x. For every additional £100 spent on advertising, sales are predicted to increase by about £800.

对于n=10（单尾，0.05），临界值为0.549。由于0.892 > 0.549，拒绝H&sub0;。有强有力的证据表明存在正相关。斜率 b = Sxy/Sxx = 512/640 = 0.8。截距 a = y¯ − b x¯ = 22 − 0.8×15 = 10。回归线：y = 10 + 0.8x。广告支出每增加100英镑，销售收入预计增加约800英镑。

10. Integrated Mixed Challenge: Customer Satisfaction Survey | 综合挑战：顾客满意度调查

A supermarket chain conducted a survey on 150 customers. They recorded waiting times at checkouts (seconds), overall satisfaction scores (out of 10), and whether the customer would recommend the store. The tasks blend sampling, descriptive statistics, normal distribution, and hypothesis testing.

一家连锁超市对150名顾客进行了调查。他们记录了收银台等待时间（秒）、总体满意度评分（满分10分）以及顾客是否会推荐该店。这些任务融合了抽样、描述统计、正态分布和假设检验。

(a) The survey used stratified sampling by store section (fresh food, ambient, non-food). Explain one advantage of this method compared to simple random sampling in this context.

(a) 该调查按区域（生鲜食品、常温商品、非食品）采用了分层抽样。请说明在此情境下，与简单随机抽样相比，该方法的一个优势。

Stratified sampling ensures proportional representation of customers from different sections, which may have distinct waiting time patterns. This reduces variability in estimates and increases precision compared to simple random sampling, where one section might be under- or over-represented by chance.

分层抽样能确保不同区域的顾客都有按比例的代表，而各区域的等待时间模式可能有差异。与简单随机抽样相比，这减少了估计值的变异性并提高了精度，简单随机抽样可能因偶然原因导致某个区域的代表性不足或过度。

(b) The waiting times were approximately normal with mean μ = 250 s and standard deviation σ = 40 s. Find the probability that a randomly selected customer waits more than 5 minutes (300 s). Hence estimate how many customers in the sample of 150 would be expected to wait over 5 minutes.

(b) 等待时间近似服从均值为 μ = 250秒、标准差 σ = 40秒的正态分布。求随机选取的一名顾客等待时间超过5分钟（300秒）的概率。并由此估计150名样本中预期有多少人会等待超过5分钟。

Z = (300 − 250) / 40 = 1.25. P(Z > 1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056. Expected number = 150 × 0.1056 ≈ 15.8, so about 16 customers.

Z = (300−250)/40=1.25。P(Z > 1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056。预期人数 = 150×0.1056 ≈ 15.8，即大约16名顾客。

(c) The manager believes that the true proportion p of customers who would recommend the store has increased from 0.65. In the sample, 108 out of 150 would recommend. Conduct a hypothesis test at the 10% significance level.

H&sub0;: p = 0.65, H&sub1;: p > 0.65. Test statistic: under H&sub0;, X ~ B(150, 0.65), normal approximation justified since n is large. Mean = 150×0.65 = 97.5, variance = 150×0.65×0.35 = 34.125, standard deviation ≈ 5.842. Observed x = 108, Z = (108 − 97.5) / 5.842 ≈ 1.797. Critical value at 10% one-tailed is 1.282. Since 1.797 > 1.282, reject H&sub0;. There is evidence of an increase in the recommendation rate.

H&sub0;: p = 0.65, H&sub1;: p > 0.65。检验统计量：在H&sub0;下，X ~ B(150, 0.65)，因n大

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AS AQA Statistics: Interdisciplinary Mixed-Question Practice | AS AQA 统计：跨学科综合题型训练