📚 IB Math: Quantiles and Solving for k | IB数学：分位数与k值求解的常见题型
Quantiles are a fundamental concept in statistics, frequently tested in IB Mathematics. Many students struggle with questions that ask them to find the value of k given a quantile condition. In this article, we will systematically explore common question types and solution methods, helping you build confidence for both Paper 2 and the internal assessment.
分位数是统计学中的核心概念，也是 IB 数学考试中常见的考点。许多同学在遇到“已知分位数条件求 k 值”这类题目时常常感到困惑。本文将系统梳理常见题型与解法，帮助你在 Paper 2 和内部评估中都能从容应对。
1. What is a Quantile? | 什么是分位数？
A quantile divides a dataset into equal parts. The most common quantiles are quartiles (dividing data into four parts), percentiles (dividing data into one hundred parts), and deciles (dividing data into ten parts). In IB Mathematics: Analysis and Approaches (AI) and Applications and Interpretation (AI), quantiles are often used in box-and-whisker plots and cumulative frequency graphs.
分位数将一组数据划分为相等的部分。最常见的分位数包括四分位数（将数据分成四部分）、百分位数（分成一百部分）和十分位数（分成十部分）。在 IB 数学中，无论是分析类还是应用类课程，分位数常用于箱线图和累积频率图。
2. Understanding the k Value in Quantile Problems | 理解分位数问题中的 k 值
In IB statistics problems, k often appears in two contexts. First, k may represent a data value that corresponds to a given quantile, such as “Find k such that the 70th percentile is k.” Second, k may appear inside a function that defines a probability distribution, and you must solve for k using quantile information. Both types require careful interpretation of the word “quantile.”
在 IB 统计题中，k 通常出现在两种情境中。第一种：k 代表某个数据值，例如“已知第 70 百分位数为 k，求 k”。第二种：k 出现在概率密度函数或累积分布函数中，我们需要利用分位数条件反解 k。两种题型都需要准确理解“分位数”的含义。
3. Quantiles from a Frequency Table | 由频率表求分位数
When data is presented in a grouped frequency table, we use interpolation to estimate a quantile. The key formula involves the cumulative frequency, class width, and the position of the quantile. Suppose the quantile position is L = q × n, where q is the quantile fraction and n is the total frequency. Then we locate the class interval containing L and apply linear interpolation.
当数据以分组频率表给出时，我们使用插值法估计分位数。关键公式涉及累积频率、组距和分位数位置。假设分位数位置为 L = q × n，其中 q 是分位数比例，n 是总频数。然后找到包含 L 的组区间，并应用线性插值。
Estimated quantile = lower boundary + [(L − cumulative frequency before class) / class frequency] × class width
For example, consider the following frequency table:
例如，考虑以下频率表：
| Class Interval | Frequency | Cumulative Frequency |
| 0–10 | 5 | 5 |
| 10–20 | 8 | 13 |
| 20–30 | 12 | 25 |
| 30–40 | 9 | 34 |
To find the 70th percentile, first compute L = 0.70 × 34 = 23.8. The class interval is 20–30 because the cumulative frequency reaches 25 after this class. Using linear interpolation:
要求第 70 百分位数，先计算 L = 0.70 × 34 = 23.8。该位置落在 20–30 这一组，因为该组累积频率达到 25。使用线性插值：
k = 20 + [(23.8 − 13) / 12] × 10 = 20 + 9 = 29
Therefore, the 70th percentile is 29. This method is consistently taught in IB and is a common Paper 2 question.
因此，第 70 百分位数为 29。这一方法在 IB 中反复出现，是 Paper 2 的常见考题。
4. Quantiles from a Cumulative Frequency Graph | 由累积频率图求分位数
When a cumulative frequency curve (ogive) is provided, the steps are straightforward. First, calculate L = q × n. Second, draw a horizontal line from L on the cumulative frequency axis to the curve. Then draw a vertical line down to the x-axis. The x-coordinate of the intersection is the required quantile. For example, finding the median (q = 0.5) is done exactly this way.
当题目给出累积频率曲线（ogive）时，步骤如下：先计算 L = q × n；然后从累积频率轴上的 L 点画水平线至曲线；再垂直向 x 轴画线，交点处的 x 坐标即为所需分位数。例如求中位数（q = 0.5）时正是使用这一方法。
When the graph is drawn accurately on graph paper, the answer can be read directly. However, IB exams often require students to show the lines on the graph. Communication of method is rewarded in Paper 2, where a correct answer without working may lose marks.
当图形在坐标纸上精确绘制时，可以直接读出答案。然而，IB 考试常常要求学生在图上画出辅助线。Paper 2 中十分重视过程展示，只有答案而没有过程可能无法获得满分。
5. Solving for k in a Normal Distribution | 正态分布中求解 k
In IB Mathematics: Applications and Interpretation, normal distribution problems are extremely common. A typical question: “The random variable X is normally distributed with mean 50 and standard deviation 8. Find k such that P(X < k) = 0.85."
在 IB 数学：应用与解释课程中，正态分布问题非常常见。典型题目如：“随机变量 X 服从正态分布，均值为 50，标准差为 8。已知 P(X < k) = 0.85，求 k。”
To solve this, we convert X into a standard normal variable Z. We have P(Z < (k − 50)/8) = 0.85. From the inverse normal table or a GDC, we know the z-score for 0.85 is approximately 1.0364. Therefore:
解法是将 X 标准化为标准正态变量 Z。我们有 P(Z < (k − 50)/8) = 0.85。通过逆正态表或图形计算器，知 0.85 对应的 z 分数约为 1.0364。因此：
(k − 50) / 8 = 1.0364 → k = 50 + 8 × 1.0364 = 58.2912
Thus k ≈ 58.3. On the GDC, we would use the inverse normal function with tail = left, area = 0.85, and the given mean and standard deviation.
因此 k ≈ 58.3。在图形计算器上，我们使用逆正态功能，选择左尾、面积 0.85、以及给定的均值和标准差。
6. Solving for k in a Discrete Probability Distribution | 离散概率分布中求解 k
Sometimes the probability distribution itself contains k. For example, a random variable Y has the following probability mass function:
有时概率分布本身包含 k。例如，随机变量 Y 的概率质量函数如下：
| y | 1 | 2 | 3 | 4 |
| P(Y=y) | 0.1 | 0.3 | k | 0.2 |
First, use the total probability law: the sum of all probabilities is 1. Thus 0.1 + 0.3 + k + 0.2 = 1, giving k = 0.4. Then, if the question asks for the median, we find the smallest value y for which P(Y ≤ y) ≥ 0.5.
首先利用总概率为 1 的性质：0.1 + 0.3 + k + 0.2 = 1，得到 k = 0.4。若题目进一步要求中位数，我们需要找到使 P(Y ≤ y) ≥ 0.5 的最小 y 值。
Here P(Y ≤ 1) = 0.1, P(Y ≤ 2) = 0.4, P(Y ≤ 3) = 0.8. Since 0.8 ≥ 0.5, the median is y = 3. This kind of combined question tests both the concept of probability distributions and quantiles.
这里 P(Y ≤ 1) = 0.1，P(Y ≤ 2) = 0.4，P(Y ≤ 3) = 0.8。因为 0.8 ≥ 0.5，中位数为 y = 3。这类综合题同时考察了概率分布和分位数的理解。
7. Solving for k in Continuous Distributions | 连续分布中求解 k
In IB AI, continuous distributions other than normal are occasionally tested. For example, a continuous random variable X has a probability density function f(x) = kx for 0 ≤ x ≤ 2, and 0 otherwise. To find k, integrate the pdf over its support and set the integral to 1.
在 IB 应用类课程中，偶尔会考察正态分布以外的连续分布。例如，连续变量 X 的概率密度函数为 f(x) = kx（0 ≤ x ≤ 2），其余为 0。要求 k，可以对其定义域积分并令积分为 1。
∫₀² kx dx = k × [x²/2]₀² = k × 2 = 1 → k = 1/2
Then the median m satisfies ∫₀ᵐ (1/2)x dx = 0.5. This gives m²/4 = 0.5, so m = √2. A common mistake is forgetting that the integral from the lower bound to the median equals 0.5.
然后中位数 m 满足 ∫₀ᵐ (1/2)x dx = 0.5。于是 m²/4 = 0.5，解得 m = √2。常见的错误是忘记中位数定义中从下界到中位数的积分为 0.5。
8. Percentiles and Quartiles with GDC | 使用图形计算器求百分位数与四分位数
In IB exams, the GDC is essential
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