📚 Exercise 24J: Solving for Unknown Mean and Standard Deviation in Normal Distribution | 正态分布未知均值与标准差的求解
Exercise 24J in the IB Mathematics HL Core course typically deals with a challenging class of normal distribution problems: finding the population mean (μ) and/or standard deviation (σ) when they are not given directly. Instead of plugging into a known distribution, you are presented with one or more probability statements that allow you to work backwards. Mastering these inverse normal problems is essential for high marks on IB exams, as they test your ability to link standardised scores, probability tables, and algebraic reasoning.
IB 数学 HL 核心课程中的练习 24J 通常涉及一类具有挑战性的正态分布问题：在总体均值 (μ) 和/或标准差 (σ) 未知的情况下求解这些参数。题目不会直接给出参数，而是提供一个或多个概率条件，让你逆向推导。掌握这类逆向正态问题对于在 IB 考试中取得高分至关重要，因为它们考察你将标准化分数、概率表与代数推理联系起来的能力。
1. Quick Recap: Standardisation and Z-Scores | 快速回顾：标准化与 Z 分数
a random variable X follows a normal distribution with mean μ and variance σ², we write X ~ ew variable Z follows the standard normal distribution N(0, 1), for which probabilities are tabulated or computed with a calculator. This standardisation step is the key to unlocking the unknown parameters.
如果随机变量 X 服从均值为 μ、方差为 σ² 的正态分布，记作 X ~ N(μ, σ²)。为了计算概率，我们使用公式 Z = (X - μ) / σ 将任意 X 值转化为 Z 分数。新变量 Z 服从标准正态分布 N(0, 1)，其概率可以通过表格或计算器获得。这一标准化步骤是解开未知参数的关键。
e equation we can solve for one unknown parameter if the other is given. When both μ and σ are unknown, we need at least two independent probability conditions to form a system of two equations.
当给出一个概率，例如 P(X < a) = p，我们首先找出满足 P(Z < z) = p 的 Z 值，然后建立方程 z = (a - μ) / σ。如果另一个参数已知，单个方程可以解出一个未知参数。当 μ 和 σ 都未知时，我们至少需要两个独立的概率条件，以形成包含两个方程的方程组。
2. Reading the Inverse Normal Table or Using InvNorm | 读取逆向正态表或使用 InvNorm 功能
0.05, you first convert it to a left-tail: P(X < a) = 0.95, then find the Z-value for 0.95, which is approximately 1.6449. Remember that the standard normal table always gives cumulative probabilities from the left.
要找出对应左尾概率 p 的 Z 值，你可以使用标准正态分布表或图形计算器上的逆正态函数。例如，若 P(Z < z) = 0.8，则 z ≈ 0.8416。如果给出的概率是右尾概率，比如 P(X > a) = 0.05，你首先要将其转化为左尾：P(X < a) = 0.95，然后找出对应 0.95 的 Z 值，约为 1.6449。请记住，标准正态表始终给出从左侧累积的概率。
On TI-Nspire or Casio calculators, the invNorm function allows you to type in the area (left-tail), the mean, and the standard deviation. When solving for μ or σ, you will typically use invNorm(area, 0, 1) to get the Z-score, and then set up your algebraic equation.
在 TI-Nspire 或 Casio 计算器上，invNorm 函数允许你输入左侧面积、均值和标准差。在求解 μ 或 σ 时，你通常会使用 invNorm(面积, 0, 1) 来获得 Z 分数，然后建立代数方程。
3. One Unknown Parameter: μ or σ Given | 单个未知参数：已知 μ 或 σ
Sometimes Exercise 24J includes problems where only one parameter is missing. Suppose X ~ imilarly, if μ were known and σ unknown, a single equation would allow you to solve for σ.
有时练习 24J 中会出现只有一个参数缺失的题目。假设 X ~ N(μ, 2²) 且 P(X < 20) = 0.3。我们已知 σ = 2。使用 invNorm(0.3, 0, 1) 得到 z = -0.5244。然后建立方程 (20 - μ)/2 = -0.5244，可解得 μ = 20 + 2(0.5244) = 21.0488。类似地，如果已知 μ 而 σ 未知，单一方程也能让你求出 σ。
This type of problem reinforces the direct link between standardised values and raw data. Always check the sign of the Z-score: if the probability is less than 0.5, the Z-score will be negative, indicating the raw value is below the mean.
这类问题强化了标准化值与原始数据之间的直接联系。务必检查 Z 分数的符号：如果概率小于 0.5，Z 分数将为负，表明原始数值低于均值。
4. Two Unknown Parameters: Building a System | 两个未知参数：建立方程组
44. Thus (55 - μ)/σ = -0.5244. Second, P(X > 75) = 0.1 means P(X < 75) = 0.9, giving z₂ ≈ 1.2816, and (75 - μ)/σ = 1.2816. Solving these two equations simultaneously yields μ and σ.
练习 24J 中最常见且难度较大的题目是使用两个概率条件同时求出 μ 和 σ。例如，已知 P(X < 55) = 0.3 且 P(X > 75) = 0.1，我们可以将每个条件转化为标准化方程。首先，P(X < 55) = 0.3 对应 z₁，使得 P(Z < z₁) = 0.3，故 z₁ ≈ -0.5244，于是 (55 - μ)/σ = -0.5244。其次，P(X > 75) = 0.1 意味着 P(X < 75) = 0.9，得到 z₂ ≈ 1.2816，且 (75 - μ)/σ = 1.2816。联立这两个方程即可解出 μ 和 σ。
The system can be solved by subtraction or substitution. From the two equations:
55 - μ = -0.5244σ ... (1)
75 - μ = 1.2816σ ... (2)
44 × 11.07 ≈ 60.80. Such problems require careful handling of signs and accurate Z-values.
该方程组可以通过消元法或代入法求解。从两个方程：
55 - μ = -0.5244σ ... (1)
75 - μ = 1.2816σ ... (2)
用 (2) 减去 (1) 消去 μ：(75 - μ) - (55 - μ) = 1.2816σ - (-0.5244σ) → 20 = 1.806σ → σ = 20 / 1.806 ≈ 11.07。然后代入 σ 可求得 μ = 55 + 0.5244 × 11.07 ≈ 60.80。这类题目需要小心处理符号并保证 Z 值精确。
5. Symmetric Interval Probabilities | 对称区间概率
Another classic problem type asks: find μ or σ given an interval centred at the mean. If X ~ echnique helps when the interval is symmetric about the mean, even if the mean is unknown.
另一类经典问题是：给定以均值为中心的区间，求 μ 或 σ。若 X ~ N(μ, σ²) 且 P(μ - d < X < μ + d) = 0.90，正态曲线的对称性意味着两侧尾部的面积各为 0.05。因此上界 μ + d 满足 P(X < μ + d) = 0.95。对应的 Z 值为 1.6449，可得 d/σ = 1.6449 → d = 1.6449σ。当区间关于均值对称时，即使均值未知，此方法也很有用。
must standardise both boundaries and equate the probability area to 0.95. This usually results in two equations with the same unknowns; however, you may need the central area property: the Z-scores will be symmetric only if the interval is centred on μ. More often you solve by standardising both limits and using the difference of cumulative probabilities.
当 μ 和 σ 都未知，而给出的区间不以均值为中心时，比如 P(a < X < b) = 0.95，你必须对两个边界都进行标准化，并令概率面积等于 0.95。这通常会产生两个含有相同未知数的方程；但你可能需要用到中心面积的性质：只有当区间关于 μ 对称时，Z 分数才会互为相反数。更常见的是对两个界限分别标准化，并利用累积概率之差来求解。
6. Step-by-Step Strategy for Exercise 24J | 应对练习 24J 的逐步策略
A systematic approach drastically reduces errors in these multistep problems. Follow these steps:
一个系统的方法可以大幅减少这类多步骤问题中的错误。请遵循以下步骤：
- Step 1: ven probability as a left-tail cumulative probability (e.g., P(X > c) → P(X < c) = 1 - p).
- Step 2: Use the inverse normal function on your GDC (or standard normal table) to find the Z-value(s) corresponding to each cumulative area, keeping at least four decimal places.
- Step 3: Write an equation of the form (Xvalue - μ) / σ = Z for each condition.
- Step 4: Rearrange to isolate μ or express both equations in terms of μ and σ.
- Step 5: Solve the system by elimination or substitution, solving for one parameter first.
- Step 6: Substitute back to find the remaining parameter, and check that your answers produce the original probabilities using a quick normal calculation.
- 第 1 步：将每个给定的概率改写为左尾累积概率（例如，P(X > c) → P(X < c) = 1 - p）。
- 第 2 步：使用 GDC 的逆正态函数（或标准正态表）找出每个累积面积对应的 Z 值，至少保留四位小数。
- 第 3 步：为每个条件写出形如 (X_值 - μ) / σ = Z 的方程。
- 第 4 步：对方程进行整理，分离 μ，或者用 μ 和 σ 表达每个方程。
- 第 5 步：通过消元法或代入法求解方程组，先求出一个参数。
- 第 6 步：代回求出另一个参数，并用计算器进行快速的正态计算，验证所求答案能否还原原始概率。
This structured plan mirrors what markers expect to see in IB 'show that' and 'find' questions. Writing down the standardised equations clearly often gains method marks even if arithmetic slips occur.
这一结构化的计划与阅卷人期望在 IB "证明" 和 "求值" 题中看到的步骤一致。清晰地写出标准化方程往往能获得方法分，即使在算术上出现小错误。
7. Worked Example 1: Two Tail Probabilities | 例题 1：两个尾部概率
The daily rainfall in a region follows a normal distribution with unknown mean μ mm and standard deviation σ mm. It is known that 5% of days have rainfall below 12 mm, and 10% of days have rainfall above 45 mm. Find μ and σ.
某地区的日降雨量服从正态分布，均值 μ mm 和标准差 σ mm 未知。已知 5% 的日降雨量低于 12 mm，10% 的日降雨量高于 45 mm。求 μ 和 σ。
ond, P(X > 45) = 0.10 means P(X < 45) = 0.90, with Z ≈ 1.2816, giving (45 - μ)/σ = 1.2816.
首先，P(X < 12) = 0.05 对应的 Z 分数约为 -1.6449（因为 P(Z < -1.6449) = 0.05）。于是 (12 - μ)/σ = -1.6449。其次，P(X > 45) = 0.10 表示 P(X < 45) = 0.90，对应的 Z ≈ 1.2816，得到 (45 - μ)/σ = 1.2816。
tion with calculator shows P(X < 12) ≈ 0.05 and P(X > 45) ≈ 0.10, confirming the solution.
整理：12 - μ = -1.6449σ 且 45 - μ = 1.2816σ。用第二个方程减去第一个方程：(45 - μ) - (12 - μ) = 33 = (1.2816 + 1.6449)σ = 2.9265σ，因此 σ ≈ 11.28 mm。代回可得 μ = 12 + 1.6449 × 11.28 ≈ 30.56 mm。用计算器快速验证可知 P(X < 12) ≈ 0.05 且 P(X > 45) ≈ 0.10，结果正确。
8. Worked Example 2: Given a Central Probability Interval | 例题 2：给定中心概率区间
The lengths of leaves from a tree are normally distributed. 80% of the leaves measure between 6.2 cm and 8.6 cm. The interval is symmetrical about the mean. Find the mean length and standard deviation.
某种树木的叶片长度服从正态分布。80% 的叶片长度在 6.2 cm 和 8.6 cm 之间，且该区间关于均值对称。求平均长度和标准差。
Since the interval is symmetric, the mean μ = (6.2 + 8.6)/2 = 7.4 cm. The central 80% leaves tail areas of 10% proximately 1.2816. With X = 8.6, (8.6 - 7.4)/σ = 1.2816 → 1.2/σ = 1.2816 → σ ≈ 0.9365 cm.
由于区间是对称的，均值 μ = (6.2 + 8.6)/2 = 7.4 cm。中间的 80% 意味着两侧尾部的面积各为 10%，因此 P(X < 8.6) = 0.90。对应 0.90 的 Z 值为 1.2816。代入 X = 8.6，(8.6 - 7.4)/σ = 1.2816 → 1.2/σ = 1.2816 → σ ≈ 0.9365 cm。
ndardise both using (a - μ)/σ and (b - μ)/σ. Then we would have two equations with two unknowns, requiring simultaneous solution as in Example 1.
如果区间不对称，我们会将边界记为 a 和 b，写出 P(X < b) - P(X < a) = 0.80，并利用 (a - μ)/σ 和 (b - μ)/σ 对两者进行标准化。这样会得到含有两个未知数的两个方程，需要像例题 1 那样联立求解。
9. Handling Probabilities Involving the Mean Exactly | 处理恰好涉及均值的概率
a must equal μ, because the median equals the mean in a normal distribution. Using this property can reduce a two-parameter problem to a single-parameter one.
当你遇到像 P(X < μ) 或 P(X > μ) 这样的概率时，正态曲线的对称性会立即告诉你 P(X < μ) = 0.5 且 P(X > μ) = 0.5。这可能是一个强大的捷径：如果你的条件之一是 P(X > a) = 0.5，那么 a 必然等于 μ，因为在正态分布中中位数等于均值。利用这一性质可以将双参数问题简化为单参数问题。
Z = 1.0364, so (30 - 20)/σ = 1.0364 → σ ≈ 9.65. This is much faster than solving a system.
例如，若已知 P(X < 20) = 0.5 且 P(X > 30) = 0.15，第一个条件会立即给出 μ = 20。然后第二个条件变为 P(X > 30) = 0.15 → P(X < 30) = 0.85，Z = 1.0364，因此 (30 - 20)/σ = 1.0364 → σ ≈ 9.65。这比解方程组要快得多。
10. Common Mistakes to Watch Out For | 常见错误与注意事项
The most frequent error is misidentifying the Z-score's sign. If the observed X is below the mean, the Z-score will be negative, and vice versa. Many students forget to attach the negative sign when the left-tail area is less than 0.5. Another common slip is treating a right-tail probability as a left-tail one without subtraction from 1. Always convert right-tail statements before finding Z.
最常见的错误是弄错 Z 分数的符号。如果观测值 X 低于均值，Z 分数将为负值，反之亦然。许多学生会在左尾面积小于 0.5 时忘记添加负号。另一个常见的疏忽是没有从 1 中减去右尾概率，就直接将其当作左尾概率使用。务必在查找 Z 值之前将右尾表述转换过来。
Rounding too early can lead to inaccurate final results, especially when the Z-values are used in systems where errors propagate. Keep at least four decimal places throughout the calculation, and only round your final answers to an appropriate degree of accuracy, typically three significant figures as per IB convention.
过早舍入会导致最终结果不准确，尤其是在方程组中使用 Z 值时会放大误差。在整个计算过程中至少保留四位小数，只有在最终答案时才依据 IB 惯例舍入到适当精度，通常是三位有效数字。
Another pitfall is assuming symmetry blindly. Unless the problem explicitly states that an interval is symmetric about the mean, do not assume the Z-scores are opposites. Stick to the standardisation procedure and solve the system.
另一个陷阱是盲目假设对称性。除非题目明确说明区间关于均值对称，否则不要假设 Z 分数互为相反数。坚持使用标准化程序并求解方程组。
11. Calculator Tips for Efficiency | 提高效率的计算器技巧
Modern graphic display calculators can do much of the heavy lifting. Use the invNorm function with mean 0 and standard deviation 1 to get accurate Z-scores without a table. On the TI-84, for instance, invNorm(0.05,0,1) returns -1.644853626. You can store these values in variables for later use, reducing rounding errors.
现代的图形计算器可以承担大量运算工作。使用 invNorm 函数，均值和标准差分别设为 0 和 1，即可在没有表格的情况下获得精确的 Z 分数。例如，在 TI-84 上，invNorm(0.05,0,1) 返回 -1.644853626。你可以将这些值存储在变量中以便后续使用，从而减少舍入误差。
For solving systems, you can use the calculator’s equation solver or set up two equations in terms of μ and σ, then solve simultaneously. Some IB exam questions expect you to show algebraic working, so use the calculator for checking rather than as a black box.
对于解方程组，你可以使用计算器的方程求解器，或者列出关于 μ 和 σ 的两个方程，然后联立求解。某些 IB 考试题要求展示代数推理过程，因此应将计算器作为检验工具，而不是一个黑箱。
Also, after finding μ and σ, always test your answers by inputting the original X values and the obtained parameters into the normalcdf function to see if the given probabilities are reproduced. This verification step takes only a few seconds and can catch algebraic mistakes.
此外，在求出 μ 和 σ 之后，始终将原始的 X 值和求得的参数输入 normalcdf 函数进行检验，看是否能重现给定的概率。这一验证步骤仅需几秒钟，却能帮你发现代数错误。
12. Connecting to IB Exam Questions and Beyond | 联系 IB 考题与拓展
Exercise 24J problems appear frequently in IB Paper 1 and Paper 2, sometimes as standalone short questions and other times as part of larger statistical inference tasks. In topics like calibration of equipment or biological measurements, you need to determine the underlying distribution parameters from observed percentages. This inverse thinking also lays the groundwork for confidence intervals and hypothesis testing, where you repeatedly go from tail probabilities back to critical values.
练习 24J 中的题目经常出现在 IB 试卷一和试卷二中，有时是独立的简答题，有时则作为较大的统计推断任务的一部分。在诸如设备校准或生物测量等课题中，你需要从观察到的百分比确定潜在的分布参数。这种逆向思维也为置信区间和假设检验奠定了基础，在那些内容中，你需要反复从尾部概率回到
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图