📚 Finding the Mean and Standard Deviation from a Normal Distribution | 由正态分布反求均值和标准差
In A-Level Mathematics, you are often given a normal distribution with known mean and standard deviation, and asked to calculate probabilities. However, a common and slightly harder type of question requires you to work backwards: given one or more probability statements, find the unknown mean μ or standard deviation σ. This article explains the key steps, with clear worked examples and common pitfalls.
在 A-Level 数学中，你通常已知正态分布的均值和标准差，然后计算概率。但有一类常见且稍难的问题要求你反向思考：给定一个或多个概率条件，反求未知的均值 μ 或标准差 σ。本文将解释关键步骤，并给出清晰的例题与常见误区。
1. Review of the Normal Distribution and Standard Normal | 回顾正态分布与标准正态分布
If a random variable X follows a normal distribution with mean μ and standard deviation σ, we write X ~ N(μ, σ²). The probability that X is less than a particular value x is found from the cumulative distribution function, usually via the standard normal variable Z = (X − μ) / σ, where Z ~ N(0, 1).
若随机变量 X 服从均值为 μ、标准差为 σ 的正态分布，记作 X ~ N(μ, σ²)。X 小于某个值 x 的概率由累积分布函数给出，通常通过标准正态变量 Z = (X − μ) / σ 计算，其中 Z ~ N(0, 1)。
The standard normal distribution has a mean of 0 and a standard deviation of 1. Tables or calculators give us P(Z < z) for any z. Conversely, they can give us the z-value corresponding to a given probability; this is called the inverse normal function.
标准正态分布的均值为 0，标准差为 1。查表或使用计算器可以得到任意 z 对应的 P(Z < z)。反过来，它们也能给出某个概率所对应的 z 值，这就是所谓的“逆正态函数”。
z = (x − μ) / σ
2. The Role of z-Scores | 标准分数 z 的作用
A z-score measures how many standard deviations a value x lies from the mean μ. For any normal distribution, a given probability corresponds to a unique z-score in the standard normal distribution. Therefore, to find unknown μ or σ, we first convert the probability statement into an equation involving z.
标准分数 z 衡量的是数值 x 相对于均值 μ 偏离了多少个标准差。对于任何正态分布，一个给定的概率对应标准正态分布中唯一的 z 值。因此，要反求未知的 μ 或 σ，我们先把概率条件转化为关于 z 的方程。
For example, if P(X < a) = 0.8, then we can find the value z₀ such that P(Z < z₀) = 0.8. This gives z₀ = (a − μ) / σ, an equation linking μ and σ.
例如，若 P(X < a) = 0.8，则可找到使 P(Z < z₀) = 0.8 的 z₀，于是有 z₀ = (a − μ) / σ，这是一个关联 μ 和 σ 的方程。
It is essential to draw a sketch of the normal curve. Mark the mean, the given value x, and shade the area described by the probability. This helps you decide whether the z-score is positive or negative.
务必画一张正态曲线草图，标出均值、给定值 x，并涂出概率对应的面积。这有助于判断 z 值是正还是负。
3. Finding z-Values from Probabilities | 从概率反求 z 值
Using a calculator (e.g., invNorm) or a standard normal table, you can find z such that P(Z < z) = p. For example, P(Z < 1.2816) ≈ 0.9, so if P(X < a) = 0.9 then z = 1.2816.
使用计算器（例如 invNorm 功能）或标准正态表，可以找到满足 P(Z < z) = p 的 z。例如 P(Z < 1.2816) ≈ 0.9，因此若 P(X < a) = 0.9，则 z = 1.2816。
For probabilities smaller than 0.5, the z-value is negative. For instance, P(Z < −1.6449) ≈ 0.05. If the given probability is a right-tail probability, subtract it from 1 first because tables usually give left-tail probabilities.
当概率小于 0.5 时，z 值为负。例如 P(Z < −1.6449) ≈ 0.05。如果给定的是右尾概率，需要先用 1 减去它，因为表通常给出的是左尾概率。
P(Z < z) = p ⇔ z = Φ⁻¹(p)
When using a table, always check whether the table gives the area to the left or to the right of z. Exam boards often specify the format; if unsure, draw the curve.
查表时，务必确认表给出的是 z 左侧还是右侧的面积。考试局通常会注明格式；如果不确定，画图辅助判断。
4. Single Unknown: Solving for the Mean μ | 单个未知数：反求均值 μ
If σ is known and μ is unknown, a single probability statement is enough. Suppose X ~ N(μ, 25) and P(X < 20) = 0.3. Find μ.
如果 σ 已知而 μ 未知，一个概率条件就足够了。设 X ~ N(μ, 25)，且 P(X < 20) = 0.3，求 μ。
First, find z such that P(Z < z) = 0.3. Since 0.3 < 0.5, z is negative. Using a calculator, z ≈ −0.5244. Then set:
首先，找到满足 P(Z < z) = 0.3 的 z。因为 0.3 < 0.5，所以 z 为负。使用计算器得 z ≈ −0.5244。于是有：
−0.5244 = (20 − μ) / 5
Solving gives μ = 20 + 5 × 0.5244 = 22.622. Notice that the mean is larger than 20 because the probability of being less than 20 is only 0.3, so 20 lies to the left of the mean.
解得 μ = 20 + 5 × 0.5244 = 22.622。注意均值大于 20，因为小于 20 的概率只有 0.3，说明 20 位于均值的左侧。
Always check the sign of the z-score: if the given value is greater than μ, then z is positive; if less than μ, z is negative.
始终检查 z 的符号：若给定值大于 μ，则 z 为正；若小于 μ，则 z 为负。
5. Single Unknown: Solving for the Standard Deviation σ | 单个未知数：反求标准差 σ
If μ is known and σ is unknown, we can also solve from one probability statement. Suppose X ~ N(50, σ²) and P(X > 60) = 0.1. Find σ.
如果 μ 已知而 σ 未知，同样可以用一个概率条件求解。设 X ~ N(50, σ²)，且 P(X > 60) = 0.1，求 σ。
Since P(X > 60) = 0.1, we have P(X < 60) = 0.9. The z-value for 0.9 is 1.2816. Thus:
因为 P(X > 60) = 0.1，所以 P(X < 60) = 0.9。0.9 对应的 z 值为 1.2816。因此：
1.2816 = (60 − 50) / σ
Rearranging gives σ = 10 / 1.2816 ≈ 7.803. Here x = 60 is above the mean, so z is positive and the equation is straightforward.
整理得 σ = 10 / 1.2816 ≈ 7.803。这里 x = 60 高于均值，因此 z 为正，方程直接可解。
Remember that σ must be positive. If your algebraic result gives a negative value, check your signs or the direction of the inequality.
记住 σ 必须为正。如果代数结果得到负值，请检查符号或不等号的方向。
6. Simultaneous Equations for μ and σ | 用联立方程求 μ 和 σ
When both μ and σ are unknown, we need two independent probability statements. Each statement produces one linear equation in μ and σ after substituting the appropriate z-score.
当 μ 和 σ 都未知时，需要两个独立的概率条件。将相应的 z 值代入后，每个条件都会产生一个关于 μ 和 σ 的线性方程。
For example, X ~ N(μ, σ²), with P(X < 8) = 0.25 and P(X < 12) = 0.75. From tables, z₀.₂₅ ≈ −0.6745 and z₀.₇₅ ≈ 0.6745. We write:
例如，X ~ N(μ, σ²)，已知 P(X < 8) = 0.25，P(X < 12) = 0.75。查表得 z₀.₂₅ ≈ −0.6745，z₀.₇₅ ≈ 0.6745。于是：
−0.6745 = (8 − μ) / σ
0.6745 = (12 − μ) / σ
These are two linear equations. Multiply the first by −1: 0.6745 = (μ − 8) / σ. Equating the two expressions gives (μ − 8)/σ = (12 − μ)/σ, so μ − 8 = 12 − μ, hence μ = 10.
这是两个线性方程。将第一个方程两边乘以 −1：0.6745 = (μ − 8) / σ。令两式相等得 (μ − 8)/σ = (12 − μ)/σ，因此 μ − 8 = 12 − μ，解得 μ = 10。
Then substitute μ = 10 into either equation: 0.6745 = (12 − 10)/σ, so σ = 2 / 0.6745 ≈ 2.965. The solution is μ = 10, σ ≈ 2.965.
再将 μ = 10 代入任一方程：0.6745 = (12 − 10)/σ，得 σ = 2 / 0.6745 ≈ 2.965。最终 μ = 10，σ ≈ 2.965。
In general, for two probabilities p₁ and p₂, you get:
一般地，对于两个概率 p₁ 和 p₂，你可以得到：
σ = (x₂ − x₁) / (z₂ − z₁), μ = x₁ − z₁σ
This formula works because both z₁ and z₂ are linear transforms of x₁ and x₂.
这个公式成立，是因为 z₁ 和 z₂ 都是 x₁、x₂ 的线性变换。
7. Using Symmetry to Simplify | 利用对称性简化问题
The normal distribution is symmetric about its mean μ. Therefore, if P(X < a) = P(X > b), then a and b are symmetric about μ, meaning μ = (a + b) / 2. This can give μ directly without needing σ.
正态分布关于均值 μ 对称。因此，若 P(X < a) = P(X > b)，则 a 和 b 关于 μ 对称，即 μ = (a + b) / 2。这可以直接求出 μ，无需知道 σ。
Similarly, if P(X < μ − k) = P(X > μ + k) for any k, the symmetric relationship is obvious from the sketch of the density curve.
类似地，如果 P(X < μ − k) = P(X > μ + k) 对任意 k 成立，这种对称关系从密度曲线图中就一目了然。
Another common symmetric statement is P(X < a) = 0.025 and P(X > b) = 0.025. The values a and b are equally distant from μ, and the z-scores are opposites: z_a = −1.96 and z_b = 1.96.
另一种常见的对称表述是 P(X < a) = 0.025 且 P(X > b) = 0.025。此时 a 和 b 到 μ 的距离相等，对应的 z 值互为相反数：z_a = −1.96，z_b = 1.96。
Whenever you see equal tail probabilities, check whether the values are symmetric; this often halves the algebra.
当你看到尾部概率相等时，先检查数值是否对称；这通常能减半代数运算量。
8. Common Pitfalls: Interval and Tail Probabilities | 常见误区：区间概率与尾概率
Many students forget to convert a right-tail probability to a left-tail probability. If you are given P(X > c) = p, the z-score you find must correspond to P(Z > z) = p. Since tables often give P(Z < z), you should look up 1 − p.
许多学生忘记将右尾概率转换为左尾概率。如果给定 P(X > c) = p，找到的 z 值必须对应 P(Z > z) = p。由于表格通常给出 P(Z < z)，你需要查 1 − p 对应的 z 值。
Another pitfall is misreading an interval statement. For example, P(a < X < b) = 0.6 does not directly give a single z-value. You must split it into two equations: P(X < a) and P(X < b), using the fact that the total area between them is 0.6.
另一个误区是误读区间条件。例如 P(a < X < b) = 0.6 并不能直接给出一个 z 值。你必须将其拆分为两个方程：P(X < a) 和 P(X < b)，利用两者之间的总面积为 0.6 这一事实。
When the question says “the middle 80% of the distribution lies between a and b”, this means P(X < a) = 0.1 and P(X < b) = 0.9. The z-values are −1.2816 and 1.2816, not ±0.8.
当题目说“分布中间 80% 位于 a 和 b 之间”时，这意味着 P(X < a) = 0.1，P(X < b) = 0.9。对应的 z 值为 −1.2816 和 1.2816，而不是 ±0.8。
Always write down the probability relationship before substituting numbers. For instance, “P(X > m) = 0.9” means m is below the mean; the z-score will be negative.
在代入数值之前，总是先写下概率关系。例如，“P(X > m) = 0.9” 意味着 m 低于均值；此时 z 值为负。
9. Using Calculators and Tables Correctly | 正确使用计算器与正态分布表
On modern calculators, the inverse normal function is often labelled invNorm or similar. You input the left-tail probability, the mean (usually 0), and the standard deviation (usually 1), and it returns the z-value. Check your calculator model for the exact menu.
在现代计算器上，逆正态函数通常标记为 invNorm 或类似名称。你输入左尾概率、均值（通常为 0）和标准差（通常为 1），它会返回 z 值。请查阅你的计算器型号的具体菜单。
When using printed tables, note whether they list cumulative probabilities from −∞ to z or from z to +∞. Some tables give P(Z < z) for positive z only; for negative z you may need to use symmetry: P(Z < −z) = 1 − P(Z < z).
使用印刷表格时，注意它列出的是从 −∞ 到 z 的累积概率，还是从 z 到 +∞ 的概率。有些表只给出正 z 对应的 P(Z < z)；对于负 z，你可能需要利用对称性：P(Z < −z) = 1 − P(Z < z)。
Always round z to at least 4 decimal places when performing intermediate calculations to avoid errors in the final answer. If your exam expects a specific rounding (e.g., 3 significant figures), apply it only at the end.
在中间计算中，z 值至少保留 4 位小数，以避免最终答案产生误差。如果考试要求特定修约（例如 3 位有效数字），请只在最后一步修约。
10. Full Worked Example | 完整例题演练
Problem: A machine produces nails whose lengths, in mm, follow a normal distribution with mean μ and standard deviation σ. It is found that 5% of nails are shorter than 48.1 mm, and 15% are longer than 52.3 mm. Find μ and σ.
问题：某机器生产的钉子长度（单位 mm）服从均值为 μ、标准差为 σ 的正态分布。已知 5% 的钉子短于 48.1 mm，15% 的钉子长于 52.3 mm。求 μ 和 σ。
We have P(X < 48.1) = 0.05 and P(X > 52.3) = 0.15, so P(X < 52.3) = 0.85. Let z₁ correspond to 0.05 and z₂ correspond to 0.85.
已知 P(X < 48.1) = 0.05，P(X > 52.3) = 0.15，因此 P(X < 52.3) = 0.85。设 z₁ 对应 0.05，z₂ 对应 0.85。
Using inverse normal: z₁ = Φ⁻¹(0.05) ≈ −1.6449, z₂ = Φ⁻¹(0.85) ≈ 1.0364. The equations are:
使用逆正态函数：z₁ = Φ⁻¹(0.05) ≈ −1.6449，z₂ = Φ⁻¹(0.85) ≈ 1.0364。方程为：
−1.6449 = (48.1 − μ) / σ
1.0364 = (52.3 − μ) / σ
Subtract the first equation from the second:
用第二个方程减去第一个方程：
1.0364 − (−1.6449) = (52.3 − 48.1) / σ
So 2.6813 = 4.2 / σ, giving σ = 4.2 / 2.6813 ≈ 1.566. Then use the second equation: 1.0364 = (52.3 − μ) / 1.566, so 52.3 − μ = 1.0364 × 1.566 ≈ 1.623, hence μ = 52.3 − 1.623 = 50.677.
于是 2.6813 = 4.2 / σ，得 σ = 4.2 / 2.6813 ≈ 1.566。再利用第二个方程：1.0364 = (52.3 − μ) / 1.566，所以 52.3 − μ = 1.0364 × 1.566 ≈ 1.623，因此 μ = 52.3 − 1.623 = 50.677。
So the nail lengths have mean approximately 50.7 mm and standard deviation approximately 1.57 mm. Always check: 48.1 and 52.3 are roughly symmetric about 50.7, and the probabilities 0.05 and 0.15 are plausible.
因此钉子长度均值约为 50.7 mm，标准差约为 1.57 mm。务必检查：48.1 和 52.3 大致关于 50.7 对称，概率 0.05 和 0.15 也合理。
11. Summary | 总结
To find an unknown mean or standard deviation from a normal distribution, follow these steps:
要从正态分布反求未知均值或标准差，请遵循以下步骤：
- Write down the probability statement in terms of X and convert it to a left-tail probability if necessary.
- 把概率条件用 X 表示，如果必要，先转换为左尾概率。
- Use the inverse normal function or tables to find the corresponding z-score. Draw a sketch to determine its sign.
- 使用逆正态函数或查表得到对应的 z 值。画草图判断其符号。
- Substitute z into the formula z = (x − μ) / σ. If both μ and σ are unknown, set up two equations and solve them simultaneously.
- 将 z 代入公式 z = (x − μ) / σ。如果 μ 和 σ 都未知，建立两个方程并联立求解。
- Check your answers by substituting back into the original probability statements, using your calculator to verify the probabilities.
- 将答案代回原概率条件进行检验，用计算器验证概率。
Mastering this reverse process is essential for exam success in statistics. With practice, you will learn to recognise symmetric patterns, handle left and right tails confidently, and avoid the common sign errors that cost marks.
掌握这种反向求解过程对于统计部分的考试成功至关重要。通过练习，你将学会识别对称模式，自信地处理左尾和右尾，并避免因符号错误而失分。
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