Applications and Interpretation SL: Practice Paper 1 with Worked Solutions | 应用与解释SL模拟试卷1及详细解答

📚 Applications and Interpretation SL: Practice Paper 1 with Worked Solutions | 应用与解释SL模拟试卷1及详细解答

This practice Paper 1 for IB Mathematics: Applications and Interpretation SL mirrors the style and demand of the real examination. You may use a graphic display calculator (GDC) throughout. The questions cover descriptive statistics, probability, bivariate data, financial mathematics, normal distributions, Voronoi diagrams, trigonometric modelling, Spearman's rank hypothesis testing, differentiation, and integration. For each question, a full worked solution is provided step by step, highlighting the reasoning and GDC use.

这份IB数学应用与解释标准水平模拟试卷1贴合真实考试的题型与难度。整场考试均可使用图形计算器。题目涵盖描述统计、概率、双变量数据、金融数学、正态分布、沃罗诺伊图、三角模型、斯皮尔曼等级假设检验、微分和积分。每道题都提供了完整的分步解答，突出推理过程和计算器的运用。

1. Test Overview and Instructions | 试卷概述与指令

This Paper 1 is designed for Applications and Interpretation SL. It lasts 1 hour 30 minutes and contains both short-response and extended-response questions. Marks are indicated in square brackets. A clean copy of the formula booklet is provided, and a GDC is required. All working must be shown to earn full marks.

本试卷专为应用与解释SL设计，时长1小时30分钟，包含简答题与拓展题。分数标示于方括号内。考场提供干净的公式手册，并要求使用图形计算器。必须写出所有解题步骤才能获得满分。

2. Question 1: Descriptive Statistics | 问题1：描述统计

Question: The frequency table shows the number of books read by 27 students in one month. Find the mean and the sample standard deviation. [4 marks]

问题：频数表给出了27名学生在一个月内阅读的书籍数量。求平均值与样本标准差。[4分]

Solution: ating fx and fx². The total frequency n = 27, Σf·x = 50, Σf·x² = 124. The sample mean is x̄ = 50 ÷ 27 ≈ 1.85.

解答：首先，扩展表格，计算 fx 与 fx²。总频数 n = 27，Σf·x = 50，Σf·x² = 124。样本均值 x̄ = 50 ÷ 27 ≈ 1.85。

s = √[ (Σfx² − (Σfx)²/n) ÷ (n − 1) ] = √[ (124 − 50²/27) ÷ 26 ]

Substituting the values: s = √[ (124 − 92.5926) ÷ 26 ] = √[ 31.4074 ÷ 26 ] ≈ √1.208 ≈ 1.099. The sample standard deviation is 1.10 (3 s.f.).

代入数值：s = √[ (124 − 92.5926) ÷ 26 ] = √[ 31.4074 ÷ 26 ] ≈ √1.208 ≈ 1.099。样本标准差为1.10（三位有效数字）。

3. Question 2: Probability without Replacement | 问题2：不放回概率

Question: A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. Find the probability that the two balls are of the same colour. [3 marks]

问题：一个袋子装有5个红球和3个蓝球。随机不放回地抽出两个球。求抽出的两个球颜色相同的概率。[3分]

Solution: : P(BB) = (3/8) × (2/7) = 6/56. Therefore, P(same) = 20/56 + 6/56 = 26/56 = 13/28 ≈ 0.464.

解答：同色即均为红球或均为蓝球。红球：P(RR) = (5/8) × (4/7) = 20/56。蓝球：P(BB) = (3/8) × (2/7) = 6/56。因此，P(同色) = 20/56 + 6/56 = 26/56 = 13/28 ≈ 0.464。

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4. Question 3: Bivariate Statistics and Linear Regression | 问题3：双变量统计与线性回归

Question: The table shows hours of study (x) and exam scores (y) for five students. (a) Calculate Pearson's correlation coefficient r. (b) Find the equation of the regression line of y on x. (c) Use the equation to estimate the score for 15 hours of study. [6 marks]

问题：表格显示5名学生的学习时间（x）与考试成绩（y）。(a) 计算皮尔逊相关系数 r。(b) 求 y 对 x 的回归直线方程。(c) 利用方程估计学习15小时的分数。[6分]

Solution: Enter the data into GDC lists and use the 2-Var Stats function. We obtain x̄ = 6, ȳ ives r = 0.9995. For the regression line, the slope b = Sxy / Sxx = 3.1, and intercept a = ȳ − b x̄ = 9.2. Hence the equation is y = 9.2 + 3.1x. When x = 15, y = 9.2 + 3.1(15) = 55.7 ≈ 56.

解答：将数据输入图形计算器列表，使用双变量统计功能。得到 x̄ = 6，ȳ = 27.8，Sx = 3.162，Sy = 9.808。计算器得 r = 0.9995。回归直线的斜率 b = Sxy / Sxx = 3.1，截距 a = ȳ − b x̄ = 9.2。因此方程为 y = 9.2 + 3.1x。当 x = 15，y = 9.2 + 3.1(15) = 55.7 ≈ 56。

5. Question 4: Financial Mathematics – Compound Interest | 问题4：金融数学 – 复利

Question: An investment of $2000 earns interest at 4.5% per annum compounded quarterly. (a) Find the value of the investment after 6 years. (b) Determine the time required for the investment to double in value. [4 marks]

问题：一笔2000美元的投资以年利率4.5%按季度复利计息。(a) 求6年后的投资价值。(b) 计算投资翻倍所需的时间。[4分]

Solution: Use the compound interest formula A = P(1 + r/n)^nt. Here P = 2000, r = 0.045, n = 4, t = 6. (a) A = 2000(1 + 0.045/4)^4×6 = 2000(1.01125)²⁴ ≈ 2612.03. The value after 6 years is approximately $2612. (b) For doubling, set 4000 = 2000(1.01125)^4t. Divide by 2000: 2 = (1.01125)^4t ln2 = 4t × ln1.01125 → t = ln2 / (4 × ln1.01125) ≈ 0.6931 / (4 × 0.011187) ≈ 15.49 years.

解答：使用复利公式 A = P(1 + r/n)^nt。其中 P = 2000，r = 0.045，n = 4，t = 6。(a) A = 2000(1 + 0.045/4)^4×6 = 2000(1.01125)²⁴ ≈ 2612.03。6年后的价值约为2612美元。(b) 翻倍时，4000 = 2000(1.01125)^4t。两边除以2000：2 = (1.01125)^4t。取自然对数：ln2 = 4t × ln1.01125 → t = ln2 / (4 × ln1.01125) ≈ 0.6931 / (4 × 0.011187) ≈ 15.49年。

6. Question 5: Normal Distribution | 问题5：正态分布

Question: The masses of apples from an orchard are normally distributed with mean 120 g and standard deviation 15 g. (a) Find the probability that a randomly selected apple weighs more than 140 g. (b) Determine the weight exceeded by the heaviest 10% of apples. [4 marks]

问题：某果园苹果的质量服从正态分布，均值120 g，标准差15 g。(a) 求随机选取的一个苹果质量大于140 g的概率。(b) 求被最重的10%苹果超过的质量阈值。[4分]

Solution: (a) X ~ 0.90. The inverse normal gives z = 1.2816. Then k = μ + zσ = 120 + 1.2816 × 15 ≈ 139.2 g.

解答：(a) X ~ N(120, 15²)。P(X > 140) = 1 − P(X ≤ 140)。标准化：z = (140 − 120) / 15 = 1.333。由计算器，Φ(1.333) = 0.9088。所以 P(X > 140) = 0.0912。(b) 求

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