📚 Exercise 26E.2: Normal Distribution – Finding Unknown Parameters | 练习 26E.2:正态分布——求解未知参数
Exercise 26E.2 in the IB Mathematics course challenges students to apply the normal distribution in reverse – using given probabilities to determine the unknown mean μ or standard deviation σ. This type of problem frequently appears in both Analysis & Approaches (AA) and Applications & Interpretation (AI) syllabuses, testing a deep understanding of standardisation and the inverse normal function.
IB 数学课程中的练习 26E.2 要求学生反向应用正态分布——利用给定概率确定未知的均值 μ 或标准差 σ。这类问题经常出现在分析与方法 (AA) 以及应用与解释 (AI) 的大纲中,考查学生对标准化和逆正态函数的深入理解。
1. What Makes Exercise 26E.2 Unique? | 练习 26E.2 的独特之处
ckward to find either μ or σ. This reverses the usual flow and forces you to think algebraically about the relationship between X, Z, μ and σ.
与那些通过查表 Φ(z) 求 P(X < x) 的基本概率题不同,练习 26E.2 要求反过来:已知 P(X < x) = p,逆向求出 μ 或 σ。这颠倒了常规思路,迫使你用代数思维处理 X、Z、μ 和 σ 之间的关系。
Often the exercise presents two separate probability statements involving the same distribution, requiring you to set up and solve simultaneous equations. This combination of normal probability and algebra makes these questions some of the most rewarding on the IB exam.
通常该练习会给出关于同一个分布的两个独立概率陈述,需要你建立并求解联立方程。这种正态概率与代数的结合,使得这些题目成为 IB 考试中最有挑战性也最有成就感的题目之一。
2. Recap of the Standard Normal Distribution | 标准正态分布回顾
lities given by Φ(z) = P(Z < z). In IB examinations you will use a GDC function such as normalcdf to find probabilities, but the core principle remains: any normal distribution X ~ N(μ, σ²) can be transformed to Z.
标准正态变量 Z 服从 N(0, 1²),其累积概率由 Φ(z) = P(Z < z) 给出。在 IB 考试中,你将使用 GDC 函数如 normalcdf 求概率,但核心原理不变:任意正态分布 X ~ N(μ, σ²) 都可以转换为 Z。
≈ 1.64485 and Φ⁻¹(0.10) ≈ −1.28155 appear repeatedly when working with tail probabilities.
这种转换之所以关键,是因为标准正态临界值是固定的;例如,处理尾部概率时,Φ⁻¹(0.95) ≈ 1.64485 和 Φ⁻¹(0.10) ≈ −1.28155 会反复出现。
3. The Inverse Normal Function | 逆正态函数
Your GDC provides an invNorm inverse normal) function that returns the z-value such that P(Z < z) = p. For instance, invNorm(0.975) returns about 1.960. This is the engine behind all reverse-lookup problems in Exercise 26E.2.
你的 GDC 提供 invNorm(或逆正态)函数,返回满足 P(Z < z) = p 的 z 值。例如,invNorm(0.975) 返回约 1.960。这就是练习 26E.2 中所有反向查找问题的动力来源。
before applying the inverse normal. This small step prevents sign errors that frequently ruin otherwise correct work.
务必检查尾部方向:如果给定的是 P(X > a) = 0.05,你必须先重写为 P(X < a) = 0.95,再使用逆正态函数。这一小步可以避免经常破坏正确解答的符号错误。
4. The Standardisation Formula as a Bridge | 作为桥梁的标准化公式
The link between X and Z is the centring and scaling formula
Z = (X − μ) / σ
X 和 Z 之间的联系是中心化和尺度化公式
Z = (X − μ) / σ
Rearranging gives X = μ + Zσ, which is especially useful for setting up linear equations.
当 μ 或 σ 未知时,该方程便成为将已知 z 值和给定 X 边界翻译成含未知参数等式的工具。重新整理可得 X = μ + Zσ,这对于建立线性方程特别有用。
6E.2 you will frequently substitute z = Φ⁻¹(p) into this formula and then solve for μ or σ – a clear, algebraic pathway from probability to parameter.
在练习 26E.2 中,你将频繁地把 z = Φ⁻¹(p) 代入该公式,并随后求出 μ 或 σ——这是一条从概率到参数的清晰代数路径。
5. Finding μ When σ Is Known | 当 σ 已知时求 μ
Suppose X ~ N(μ, 5²) and P(X < 30) = 0.15. Standardising gives
(30 − μ) / 5 = Φ⁻¹(0.15) ≈ −1.03643
假设 X ~ N(μ, 5²) 且 P(X < 30) = 0.15。标准化可得
(30 − μ) / 5 = Φ⁻¹(0.15) ≈ −1.03643
f.). This simple linear solution is typical when only one parameter is unknown and σ is given.
两边乘以 5 并求解:30 − μ = −5.18215,因此 μ ≈ 35.2(保留三位有效数字)。这种简单的线性求解在只有一个参数未知且 σ 已知时非常典型。
Watch out for negative z-values: they indicate the boundary is below the mean. Keeping the algebraic signs consistent is crucial for obtaining a sensible answer.
注意负的 z 值:它们表示边界低于均值。保持代数符号的一致性对于得到合理答案至关重要。
6. Finding σ When μ Is Known | 当 μ 已知时求 σ
If X ~ Φ⁻¹(0.92) ≈ 1.40507. Then (110 − 100) / σ = 1.40507, which yields σ = 10 / 1.40507 ≈ 7.12.
若 X ~ N(100, σ²) 且 P(X > 110) = 0.08,改写为 P(X < 110) = 0.92,并求 z = Φ⁻¹(0.92) ≈ 1.40507。则 (110 − 100) / σ = 1.40507,得出 σ = 10 / 1.40507 ≈ 7.12。
The same algebraic approach works, but always remember that σ must be positive. If your method returns a negative σ, you have either misidentified the tail or mishandled the sign of the Z-score.
同样的代数方法可行,但切记 σ 必须为正。若你的方法得到一个负的 σ,要么是尾部弄错了,要么是 Z 分数的符号处理有误。
7. When Both μ and σ Are Unknown | 当 μ 和 σ 均未知时
reate a system of two equations. For example: P(X < 20) = 0.10 and P(X > 50) = 0.05.
这正是练习 26E.2 的核心。关于同一个正态变量的两个独立概率陈述构成一个二元方程组。例如:P(X < 20) = 0.10 且 P(X > 50) = 0.05。
t; 50) = 0.95 → z₂ = Φ⁻¹(0.95) ≈ 1.64485. The standardisation yields two linear equations:
(20 − μ) / σ = −1.28155 and (50 − μ) / σ = 1.64485
转换为左尾概率:P(X < 20) = 0.10 → z₁ = Φ⁻¹(0.10) ≈ −1.28155。第二个,P(X < 50) = 0.95 → z₂ = Φ⁻¹(0.95) ≈ 1.64485。标准化得到两个线性方程:
(20 − μ) / σ = −1.28155 且 (50 − μ) / σ = 1.64485
) = 1.64485σ − (−1.28155σ) → 30 = 2.92640σ, so σ ≈ 10.25. Substitute back to get μ ≈ 33.1.
两边乘以 σ,然后用第二个方程减去第一个:(50−μ) − (20−μ) = 1.64485σ − (−1.28155σ) → 30 = 2.92640σ,故 σ ≈ 10.25。回代求得 μ ≈ 33.1。
This subtraction method neatly cancels μ, making it the most efficient technique when the two X-values are given. The same logic applies if one statement gives an upper tail and another a lower tail – always convert to a consistent left-tail form.
这种相减法巧妙地消去了 μ,当两个 X 值均已知时,这是最高效的技巧。如果其中一个为右尾、另一个为左尾,同样的逻辑也适用——始终转换为一致的左尾形式。
8. Using the GDC Efficiently | 高效使用图形计算器
Your GDC can store inverse normal results and solve the system directly. On a TI-84: use invNorm( to get z₁ and z₂, then assign them
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