📚 Exercise 22F: Inverse Normal Problems – Finding μ and σ | 练习22F：逆正态问题——求解未知的均值与标准差
Exercise 22F in many IB Mathematics textbooks focuses on a particularly challenging area of the normal distribution: working backwards. Instead of finding a probability for a given value, you are given a probability and must find an unknown mean (μ) or standard deviation (σ). These inverse normal problems require a solid understanding of standardisation, the use of the GDC’s invNorm function and careful equation solving. This article breaks down the methods, common mistakes and worked examples to help you master the skills needed for both SL and HL exams.
在IB数学教材中，练习22F通常聚焦于正态分布中一个特别有挑战性的内容：逆向求解。你不再是根据给定数值求概率，而是已知概率，去求未知的均值（μ）或标准差（σ）。这类逆正态问题需要你对标准化方法、计算器的invNorm功能以及方程求解有扎实的掌握。本文将拆解解题方法、常见错误，并提供实例讲解，帮助你掌握SL和高阶HL考试所需的技巧。
1. Understanding Inverse Normal Problems | 理解逆正态问题
In a typical normal distribution question, you are given a mean μ and standard deviation σ, and you calculate something like P(X < a). Exercise 22F reverses this: you know the probability, but one or both of the population parameters are missing. For example, you might be told that the top 5% of scores in a test are above a certain mark, and you need to find the mean if the standard deviation is known, or find the standard deviation if the mean is known. The key is to link the given probability to a Z-score, then write an equation involving μ and σ.
在一个典型的正态分布问题中，你已知均值 μ 和标准差 σ，然后计算诸如 P(X < a) 这样的概率。练习22F将这一过程颠倒：你知道概率，但总体参数中的一个或两个却未知。例如，题目可能告诉你一次考试中前5%的分数高于某个分数，要求你在已知标准差时求均值，或在已知均值时求标准差。关键在于将给定的概率与 Z 分数联系起来，然后写出含有 μ 和 σ 的方程。
2. The Standard Normal Distribution and Z-scores | 标准正态分布与Z分数
All normal distributions can be standardised using the formula: Z = (X – dard normal distribution has μ = 0 and σ = 1. When you are given a probability involving X ~ N(μ, σ²), you can convert the boundary value x into a Z-score, so P(X < x) = P(Z < (x – ndard normal table or, more efficiently, by using the inverse normal function on your GDC.
所有正态分布都可以用公式 Z = (X – μ) / σ 进行标准化。标准正态分布的均值为 0，标准差为 1。当你遇到一个涉及 X ~ N(μ, σ²) 的概率时，你可以将边界值 x 转换为 Z 分数，因此 P(X < x) = P(Z < (x – μ)/σ)。与给定累计概率相对应的 Z 分数可以通过标准正态分布表查得，或者更高效地，使用图形计算器的逆正态功能得到。
3. Solving for μ When σ is Known | 已知标准差求均值
isation formula. Rearrange to solve for μ. For example, if P(X > 70) = 0.1, σ = 8, you find z such that P(Z > z) = 0.1. This is equivalent to P(Z < z) = 0.9, so z ≈ 1.2816 (from invNorm). Then set (70 – μ)/8 = 1.2816 and solve: μ = 70 – 8×1.2816 ≈ 59.75.
如果已知标准差 σ 和一个特定的概率，你就可以求出未知的均值 μ。首先使用逆正态功能求出与给定概率相对应的 Z 值。然后将这个 Z 值、已知的边界 x 值和 σ 一起代入标准化公式。重新整理后解出 μ。例如，假设 P(X > 70) = 0.1，σ = 8，先找到满足 P(Z > z) = 0.1 的 z。这等价于 P(Z < z) = 0.9，因此 z ≈ 1.2816（通过 invNorm）。然后建立方程 (70 – μ)/8 = 1.2816，解得 μ = 70 – 8×1.2816 ≈ 59.75。
4. Solving for σ When μ is Known | 已知均值求标准差
method is very similar. Find the Z-value for the stated probability, then plug it into (x – on. For instance, if P(X < 50) = 0.025, μ = 60, and lower-tail probability 0.025 gives z = –1.9600 (invNorm(0.025)). Then (50 – 60)/σ = –1.96, which simplifies to –10/σ = –1.96, so σ = 10/1.96 ≈ 5.102.
当均值 μ 已知而标准差 σ 未知时，方法非常类似。求出已知概率对应的 Z 值，然后将其代入 (x – μ)/σ = z。重新整理方程以分离 σ。注意 σ 在分式中的位置。例如，若 P(X < 50) = 0.025，μ = 60，下尾概率 0.025 对应的 z = –1.9600（invNorm(0.025)）。于是 (50 – 60)/σ = –1.96，化简为 –10/σ = –1.96，解得 σ = 10/1.96 ≈ 5.102。
5. Solving for Both μ and σ Simultaneously | 同时求解均值和标准差
requiring you to find both μ and σ. This creates a system of two equations in the form (x₁ – μ)/σ = z₁ and (x₂ – ores with their correct probabilities (e.g. upper tail vs lower tail) and to keep signs consistent. A GDC can be used to solve these simultaneous equations numerically, but IB exams often expect you to show the algebraic manipulation as well.
练习22F中更难的题目会给出两个不同 x 值对应的概率，要求你同时求出 μ 和 σ。这就构成了一个方程组：(x₁ – μ)/σ = z₁ 和 (x₂ – μ)/σ = z₂。你可以通过将两个方程相减，或者通过消元法解出 μ 和 σ。注意要使 Z 分数与正确的概率相匹配（例如，上尾概率还是下尾概率），并保持符号一致。图形计算器可以用来对这个方程组进行数值求解，但IB考试通常也期望你展示代数变换的过程。
6. Using GDC to Solve Efficiently | 利用图形计算器高效求解
Your graphical display calculator is essential in inverse normal problems. The invNorm (or inverse normal) function directly gives the Z-value for any cumulative probability from a standard normal distribution. For example, invNorm(0.95, 0, 1) returns 1.64485. When solving for μ or σ, you can set up an equation and use the numerical solver (Nspire: menu > Algebra > Solve; TI-84: Math > Solver). However, you must still write down the standardisation equation and the correct Z-value in your working to earn method marks. Never just write the final answer without showing the link to Z-scores.
图形计算器在求解逆正态问题时不可或缺。invNorm（或逆正态）函数可以直接给出标准正态分布下任意累积概率对应的 Z 值。例如，invNorm(0.95, 0, 1) 返回 1.64485。在解 μ 或 σ 时，你可以建立方程并使用数值求解器（Nspire：menu > Algebra > Solve；TI-84：Math > Solver）。但是，你仍然必须在解题步骤中写出标准化方程和正确的 Z 值，才能获得方法分。绝不要只写最终答案，而不展示与 Z 分数相关联的过程。
7. Common Pitfalls and How to Avoid Them | 常见错误与避免方法
) = p, you must convert this to a left-tail probability before using invNorm: P(X < k) = 1 – s forget to square σ or take square roots when transforming between variance and standard deviation. Always check whether the question asks for the standard deviation or the variance. Finally, rounding too early can lead to inaccurate final answers; keep intermediate Z-values to at least 3 significant figures.
一个常见错误是处理分布的尾部不当。如果题目给出的是 P(X > k) = p，在使用 invNorm 之前，你必须将其转换为左尾概率：P(X < k) = 1 – p。另一个陷阱是当边界值位于均值左侧时，忽略了 Z 分数的负号，此时 Z 为负值。此外，学生们有时会在方差和标准差的转换中忘记对方差开平方或对标准差平方。一定要看清楚题目要求的是标准差还是方差。最后，过早地舍入会导致最终答案不准确；中间步骤中的 Z 值应至少保留三位有效数字。
8. Worked Example 1: Unknown Mean | 例题1：未知均值
Question: The weights of apples from an orchard are normally distributed with a standard deviation of 15 grams. It is known that 12% of apples weigh more than 200 grams. Find the mean weight μ.
题目：某果园苹果的重量服从正态分布，标准差为15克。已知12%的苹果重量超过200克。求平均重量 μ。
Step 1 – Identify the tail: 12. Convert to left-tail: P(X < 200) = 0.88. Use invNorm(0.88, 0, 1) to obtain z ≈ 1.1750.
第一步——确定尾部：我们有 P(X > 200) = 0.12。转换为左尾：P(X < 200) = 0.88。使用 invNorm(0.88, 0, 1) 得到 z ≈ 1.1750。
Step 2 – Set up the equation: (200 – μ)/15 = 1.1750. Multiply both sides: 200 – μ = 15 × 1.1750 = 17.625. Therefore μ = 200 – 17.625 = 182.375.
第二步——建立方程：(200 – μ)/15 = 1.1750。两边同乘：200 – μ = 15 × 1.1750 = 17.625。因此 μ = 200 – 17.625 = 182.375。
Answer: The mean weight is approximately 182 grams (3 s.f.). Always state the conclusion in context.
答案：平均重量约为182克（三位有效数字）。始终要在情境中给出结论。
9. Worked Example 2: Unknown Standard Deviation | 例题2：未知标准差
Question: The lengths of a type of metal rod are normally distributed with mean 50.0 cm. Only 3% of rods are shorter than 48.5 cm. Find the standard deviation σ.
题目：某种金属棒的长度服从正态分布，均值为50.0 cm。只有3%的金属棒短于48.5 cm。求标准差 σ。
P(X < 48.5) = 0.03 directly gives a left-tail probability. Use invNorm(0.03, 0, 1) → z = –1.8808 (negative because 48.5 < 50). Now (48.5 – 50)/σ = –1.8808, which simplifies to –1.5/σ = – is 0.798 cm (3 s.f.). Note: if a question asks for variance, you would square this result.
P(X < 48.5) = 0.03 直接就是左尾概率。使用 invNorm(0.03, 0, 1) → z = –1.8808（为负值，因为 48.5 < 50）。现在得到 (48.5 – 50)/σ = –1.8808，化简为 –1.5/σ = –1.8808。解得 σ = 1.5 / 1.8808 ≈ 0.7975 cm。标准差为 0.798 cm（三位有效数字）。注意：如果题目问的是方差，你需要对这个结果进行平方。
10. Worked Example 3: Two Unknowns | 例题3：两个未知参数
Question: The scores on a test are normally distributed. It is found that 10% of students scored below 45 marks, and 20% scored above 72 marks. Find the mean μ and standard deviation σ of the scores.
题目：某测试的分数服从正态分布。已知10%的学生得分低于45分，20%的学生得分高于72分。求分数的均值 μ 和标准差 σ。
For the first condition: P(X < 45) = 0.10 → z₁ = invNorm(0.10) = –1.2816. Equation (45 – μ)/σ = –1.2816 → 45 – μ = –1.2816σ condition: P(X > 72) = 0.20 → P(X < 72) = 0.80 → z₂ = invNorm(0.80) = 0.8416. Equation (72 – μ)/σ = 0.8416 → 72 – μ = 0.8416σ (2). Subtract (1) from (2): (72 – μ) – (45 – μ) = 0.8416σ – (–1.2816σ). This gives 27 = 2.1232σ, so σ ≈ 12.72. Substitute back into (2): 72 – 416 × 12.72 ≈ 10.70 → μ = 61.30. The mean is 61.3 and standard deviation is 12.7 (3 s.f.).
对于第一个条件：P(X < 45) = 0.10 → z₁ = invNorm(0.10) = –1.2816。方程 (45 – μ)/σ = –1.2816 → 45 – μ = –1.2816σ (1)。第二个条件：P(X > 72) = 0.20 → P(X < 72) = 0.80 → z₂ = invNorm(0.80) = 0.8416。方程 (72 – μ)/σ = 0.8416 → 72 – μ = 0.8416σ (2)。用 (2) 减去 (1)：(72 – μ) – (45 – μ) = 0.8416σ – (–1.2816σ)。得到 27 = 2.1232σ，因此 σ ≈ 12.72。回代到 (2)：72 – μ = 0.8416 × 12.72 ≈ 10.70 → μ = 61.30。均值为61.3，标准差为12.7（三位有效数字）。
11. Exercise 22F: Overview and Practice Strategies | 练习22F：概要与练习策略
Exercise 22F typically contains a mix of these three types of problems, building in difficulty. Start by clearly labelling every given piece of information and drawing a normal curve with the probability area shaded. This helps you immediately see whether you are dealing with a left-tail, right-tail or central probability. If two conditions are given, create a system of equations. Always check your final answers by plugging them back into the original distribution to verify that the stated probabilities are reproduced. Practise exam-style questions where context (weights, volumes, test scores) is given, as IB marks are awarded for correct interpretations, not just calculations.
练习22F通常包含这三种类型的题目，难度逐步递增。解题时，先清楚地标注每一条已知信息，并画出一条正态曲线，涂上对应的概率区域。这能帮助你立刻判断处理的是左尾、右尾还是中心概率。如果给出了两个条件，就要建立方程组。始终通过将最终答案代回原始分布来检验，以验证给定的概率能否重现。练习在特定语境（重量、体积、考试分数）下的考题，因为IB考试不仅给计算过程打分，还会对正确的解释给予分数。
12. Final Tips for Exam Success | 考试成功最终建议
Memorise the standardisation formula Z = (X – ur GDC’s invNorm function and the equation solver, but never rely solely on the calculator – ith μ and σ quickly. If a problem asks for the variance, remember to square the standard deviation only at the final step. Lastly, stay calm and systematic; drawing a diagram and listing equations will keep you from making sign errors and will make Exercise 22F one of your strengths in the IB Mathematics exam.
熟记标准化公式 Z = (X – μ)/σ，并理解它的各种变形。熟练使用计算器的 invNorm 功能和方程求解器，但绝不能完全依赖计算器——解题步骤中一定要写出 Z 分数的方程。合理安排时间：这类题目代数运算量较大，所以要多加练习快速求解含有 μ 和 σ 的线性方程。如果题目问的是方差，切记要到最后一步才对方差进行平方。最后，保持冷静和系统性；画图并列出方程能避免符号错误，并使练习22F成为你IB数学考试中的强项之一。
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