Finding μ and σ from Probability Conditions for a Normal Distribution | 由概率条件求正态分布参数μ与σ

📚 Finding μ and σ from Probability Conditions for a Normal Distribution | 由概率条件求正态分布参数μ与σ

In A-level Mathematics, you are often given a normal distribution with unknown mean μ or unknown standard deviation σ, together with one or two probability conditions. The task is to recover these parameters by first standardising the normal variable and then using the standard normal distribution table or an inverse normal function.

在A-level数学中，经常遇到这样的问题：正态分布的均值μ或标准差σ未知，但给了一个或两个概率条件。我们的任务是通过标准化正态变量，再借助标准正态分布表或逆正态函数，求出这些参数。

1. The Standard Normal Distribution | 标准正态分布

If X ~ N(μ, σ²), then the standardised variable Z = (X − μ) / σ follows N(0, 1). This transformation allows us to convert any normal probability into a probability involving Z, which can be looked up in a table or computed using a calculator.

若 X ~ N(μ, σ²)，则标准化变量 Z = (X − μ) / σ 服从 N(0, 1)。这一变换使我们能将任意正态概率转化为关于 Z 的概率，从而通过查表或计算器求得结果。

Z = (X − μ) / σ

The key idea is that every statement like P(X < a) = p can be rewritten as P(Z < (a − μ)/σ) = p. Then (a − μ)/σ is a specific quantile of the standard normal distribution, often denoted z.

核心思想是：任何形如 P(X < a) = p 的条件都可以改写为 P(Z < (a − μ)/σ) = p。于是 (a − μ)/σ 就是标准正态分布的某个分位数，通常记作 z。

2. Finding the z-value from a probability | 由概率求z值

To solve for μ or σ, we first need to find the standard normal z-value that corresponds to the given probability. For example, if P(Z < z) = 0.95, then z ≈ 1.6449. If P(Z > z) = 0.05, then z ≈ 1.6449 as well, by symmetry.

要求解μ或σ，首先需要找到与给定概率对应的标准正态z值。例如，若 P(Z < z) = 0.95，则 z ≈ 1.6449。若 P(Z > z) = 0.05，由对称性同样有 z ≈ 1.6449。

Use the inverse normal function on your calculator: usually invNorm(area, μ, σ) with μ=0, σ=1.

使用计算器上的逆正态函数：通常为 invNorm(概率, μ, σ)，其中 μ=0, σ=1。
When using a table, remember that most tables give P(Z < z) for positive z. For negative z or right-tail probabilities, use symmetry.

查表时，大多数表给出正z对应的 P(Z < z)。负z或右尾概率需利用对称性转换。

3. Solving for μ when σ is known | 已知σ求μ

Suppose X ~ N(μ, σ²) with σ known, and we are told P(X < a) = p. We standardise:

设 X ~ N(μ, σ²)，σ 已知，并已知 P(X < a) = p。我们标准化：

P( Z < (a − μ)/σ ) = p

Let z_p be the standard normal quantile such that P(Z < z_p) = p. Then (a − μ)/σ = z_p, so μ = a − z_p σ.

令 z_p 满足 P(Z < z_p) = p 的标准正态分位数。则 (a − μ)/σ = z_p，因此 μ = a − z_p σ。

If the condition is P(X > a) = p, then by symmetry P(X < a) = 1 − p, so use the quantile z_{1−p} with a minus sign carefully.

若条件是 P(X > a) = p，则由对称性 P(X < a) = 1 − p，需谨慎使用分位数 z_{1−p} 和符号。

4. Solving for σ when μ is known | 已知μ求σ

Suppose X ~ N(μ, σ²) with μ known, and we are told P(X < a) = p. Standardising gives:

设 X ~ N(μ, σ²) 且 μ 已知，已知 P(X < a) = p。标准化得：

(a − μ)/σ = z_p

Therefore σ = (a − μ) / z_p. Always check that σ > 0. If the z-value is negative, the order of a and μ matters; the sign will automatically be correct if you solve the equation directly.

因此 σ = (a − μ) / z_p。始终检查 σ > 0。若z值为负，注意 a 与 μ 的先后顺序；直接解方程时符号会自动正确。

5. Solving for both μ and σ from two probability conditions | 由两个概率条件求μ和σ

In the most common exam question, both μ and σ are unknown, and two independent probability statements are given. For example:

在最常见的考试题中，μ和σ均未知，并给出两个独立概率条件。例如：

P(X < 60) = 0.2, P(X > 90) = 0.1

Each condition produces one linear equation in μ and σ:

每个条件产生一个关于μ和σ的线性方程：

60 = μ + z₁ σ, 90 = μ + z₂ σ

where z₁ is the z-value for the lower tail and z₂ is the z-value for the upper tail (with appropriate sign). Subtract the two equations to eliminate μ, then solve for σ, then substitute back to find μ.

其中 z₁ 为下尾对应的z值，z₂ 为上尾对应的z值（注意符号）。将两式相减消去μ，解出σ，再代回求μ。

6. Worked Example 1: Finding μ | 例题1：求μ

The mass X of a bag of flour is normally distributed with mean μ grams and standard deviation 8 grams. Given that P(X < 250) = 0.9, find μ.

某袋面粉的质量 X 服从正态分布，均值 μ 克，标准差 8 克。已知 P(X < 250) = 0.9，求 μ。

Standardise: P(Z < (250 − μ)/8) = 0.9. From the normal table, P(Z < 1.2816) = 0.9, so (250 − μ)/8 = 1.2816.

标准化：P(Z < (250 − μ)/8) = 0.9。查表得 P(Z < 1.2816) = 0.9，所以 (250 − μ)/8 = 1.2816。

250 − μ = 10.2528 ⇒ μ = 239.7472 ≈ 239.7

So the mean mass is approximately 239.7 g.

因此平均质量约为239.7克。

7. Worked Example 2: Finding σ | 例题2：求σ

The lifetime X of a battery, in hours, is normally distributed with mean 100 hours and unknown standard deviation σ. Given that P(X > 110) = 0.05, find σ.

某电池寿命 X（小时）服从正态分布，均值100小时，标准差σ未知。已知 P(X > 110) = 0.05，求 σ。

Since P(X > 110) = 0.05, we have P(X < 110) = 0.95. Thus P(Z < (110 − 100)/σ) = 0.95, and (110 − 100)/σ = 1.6449.

因为 P(X > 110) = 0.05，所以 P(X < 110) = 0.95。因此 P(Z < (110 − 100)/σ) = 0.95，即 (110 − 100)/σ = 1.6449。

10 / σ = 1.6449 ⇒ σ = 10 / 1.6449 ≈ 6.079

The standard deviation is about 6.08 hours.

标准差约为6.08小时。

8. Worked Example 3: Finding μ and σ together | 例题3：求μ和σ

Heights X of a plant species are normally distributed with mean μ cm and standard deviation σ cm. Given that P(X < 40) = 0.1 and P(X > 55) = 0.2, find μ and σ.

某植物高度 X 服从正态分布，均值 μ 厘米，标准差 σ 厘米。已知 P(X < 40) = 0.1 和 P(X > 55) = 0.2，求 μ 和 σ。

First find z-values: for P(Z < z₁) = 0.1, z₁ = −1.2816. For P(Z > z₂) = 0.2, P(Z < z₂) = 0.8, so z₂ = 0.8416.

先求z值：由 P(Z < z₁) = 0.1，得 z₁ = −1.2816。由 P(Z > z₂) = 0.2，得 P(Z < z₂) = 0.8，因此 z₂ = 0.8416。

Write two equations:

写出两个方程：

40 = μ − 1.2816 σ

55 = μ + 0.8416 σ

Subtract the first from the second: 15 = 2.1232 σ, so σ ≈ 7.066. Substitute back: μ = 40 + 1.2816 × 7.066 ≈ 49.06.

两式相减：15 = 2.1232 σ，所以 σ ≈ 7.066。代回：μ = 40 + 1.2816 × 7.066 ≈ 49.06。

9. Common pitfalls | 常见易错点

Using the wrong tail: always draw a diagram and check whether the given probability is less than or greater than 0.5 to decide the sign of the z-value.

用错尾端：务必画图，通过给定概率小于还是大于0.5来判断z值的正负。
Forgetting that the standard normal table may give P(Z < z) only for positive z; use symmetry: P(Z < −z) = 1 − P(Z < z).

忘记标准正态表仅给出正z的 P(Z < z)；利用对称性：P(Z < −z) = 1 − P(Z < z)。
Confusing σ and σ²: the normal distribution parameter is variance σ², but the standard deviation is σ. All z-formulas use σ.

混淆σ和σ²：正态分布的参数是方差σ²，但标准差是σ。所有z公式中使用σ。
Rounding too early: keep at least 4 decimal places for z-values until the final answer.

过早四舍五入：在最终答案之前，z值至少保留4位小数。

10. Summary | 总结

To find μ and/or σ from probability conditions:

由概率条件求μ和/或σ的步骤：

Step / 步骤	Action / 操作
1	Convert each probability statement into a standard normal quantile equation / 将每个概率条件转化为标准正态分位数方程
2	Write X = μ + zσ for each boundary, using the correct sign for z / 对每个边界写出 X = μ + zσ，注意z的符号
3	Solve the linear system for μ and σ / 解线性方程组求μ和σ
4	Check that σ > 0 and that the answers are plausible / 检验σ > 0且答案合理

Master this standard method, and you can handle any exam question of this type confidently.

掌握这一标准方法，就能自信应对此类所有考试题。

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Finding μ and σ from Probability Conditions for a Normal Distribution | 由概率条件求正态分布参数μ与σ