📚 Finding μ and σ | 求均值与标准差
In Edexcel A-Level Mathematics, one of the most common and most rewarding question types in the Statistics section involves working backwards: instead of being given the mean (μ) and standard deviation (σ) and asked for a probability, you are given one or two probabilities and asked to find μ and σ. This skill combines the standard normal distribution, z-scores, the inverse normal function and simultaneous equations, and it appears regularly in Paper 3 (Statistics and Mechanics) as well as in the S1-style legacy papers. Once you understand the logic, these questions become almost mechanical and are a reliable source of full marks.
在 Edexcel A-Level 数学中，统计部分最常见、也最"划算"的一类题型是"反向求解"：题目不给概率让你算，而是给出一个或两个概率，让你反求均值 μ 与标准差 σ。这种题型综合了标准正态分布、Z 分数、逆正态函数以及联立方程，在 Paper 3（Statistics and Mechanics）中反复出现，在旧版 S1 试卷中也是常客。一旦掌握了其中的逻辑，这类题目几乎变成"机械操作"，是稳定拿满分的得分点。
1. Why Finding μ and σ Matters | 为什么要反求 μ 与 σ
Forward normal questions ask: given a distribution, what is the probability? Backward questions ask: given the probabilities, what is the distribution? Real-world modelling rarely hands you the parameters. A factory knows what proportion of bottles are under-filled; a biologist knows what percentage of plants are taller than a threshold. From that information, they must infer the underlying μ and σ. This is the reason examiners love the topic: it is the point where pure calculation meets genuine statistical modelling.
正向的正态题目问的是：已知分布，概率是多少？反向题目问的是：已知概率，分布是什么？现实建模中，参数几乎不会直接告诉你。工厂只知道有多少比例的瓶子灌装不足，生物学家只知道有多少百分比的植株高于某个阈值。他们必须从这些信息反推背后的 μ 与 σ。这正是考官偏爱这个考点的原因：它是纯计算与真实统计建模的交汇处。
The examination skill being tested is translation: turning a sentence written in words into one or two equations written in z-values, then solving those equations. Every mark in the mark scheme is attached to a visible step, so showing your standardisation clearly is worth real marks even if the final arithmetic slips.
这里考查的核心技能是"翻译"：把一句文字描述翻译成一个或两个用 z 值写出的方程，然后求解。评分标准中每一分都挂在可见的步骤上，所以即使最后算错，把标准化过程写清楚也能拿到实打实的步骤分。
2. The Normal Distribution Recap | 正态分布回顾
ll shape, and the probability that the random variable X falls within a given interval equals the area under that curve over the interval. Every normal curve is symmetric about the line x = μ, and the total area under the curve is exactly 1.
正态分布由两个参数定义：μ（均值）和 σ（标准差）。它的概率密度曲线就是大家熟悉的钟形；随机变量 X 落在某个区间内的概率，等于曲线在该区间下方的面积。每条正态曲线都关于直线 x = μ 对称，曲线下总面积为 1。
A crucial notational trap: we write X ~ n states the second parameter is 25 directly, it is still the variance. Many lost marks in this topic come from this single confusion.
一个关键的符号陷阱：我们写作 X ~ N(μ, σ²)，其中第二个参数是方差，而不是标准差。如果题目说方差是 25，那么 σ = 5；如果题目直接写 N(μ, 25)，那个 25 依然是方差。这个考点上丢的分，很大一部分就来自这一处混淆。
Useful landmarks for checking your answers: about 68.27% of the distribution lies within one standard deviation of the mean, about 95.45% within two, and about 99.73% within three.
用来检查答案的常用地标：大约 68.27% 的数据落在均值左右一个标准差内，约 95.45% 落在两个标准差内，约 99.73% 落在三个标准差内。
3. Standardising: The Z-Formula | 标准化：Z 公式
For any X ~ N(μ, σ²), we can convert to the standard normal variable Z ~ N(0, 1) using the standardisation formula below. The z-score tells you how many standard deviations the observed value is above (positive z) or below (negative z) the mean.
对于任意 X ~ N(μ, σ²)，都可以用下面的标准化公式转化为标准正态变量 Z ~ N(0, 1)。z 分数表示该观测值高于（z 为正）或低于（z 为负）均值多少个标准差。
Z = (X − μ) / σ
Rearranged, the same relationship gives the value of X in terms of μ, σ and z. This rear-ranged form is the one you actually use when the unknown is μ or σ.
把这个关系式变形，就能用 μ、σ 和 z 表示 X。当未知量是 μ 或 σ 时，真正好用的正是这个变形后的形式。
X = μ + Zσ
In a backward question, z is a known number (found from the inverse normal function), X is a known number (given in the question), and μ and σ are the unknowns. Each probability statement therefore generates one linear equation linking μ and σ. Two statements give two equations, which can be solved simultaneously.
在反向题目中，z 是已知数（由逆正态函数求出），X 是已知数（题目给出），未知的是 μ 和 σ。因此每一条概率信息都能生成一个联系 μ 与 σ 的线性方程。两条信息就得到两个方程，可以联立求解。
Notice that the equation is linear in μ and σ rather than in σ alone, which is exactly why a pair of them can be solved with straightforward algebra: no quadratic manipulation is required in standard Edexcel questions.
注意，这个方程关于 μ 和 σ 是线性的（而不是关于 σ 的二次式），这正是两个方程能用普通代数轻松解出的原因：标准 Edexcel 题目中不需要处理二次方程。
4. The Inverse Normal Function | 逆正态函数
tion of the standard normal distribution. We then set up the equation z = (a − μ) / σ and rearrange.
如果给出的是概率，例如 P(X < a) = p，我们首先找出对应的 z 值 z = Φ⁻¹(p)，其中 Φ 是标准正态分布的累积分布函数。然后建立方程 z = (a − μ) / σ 并变形。
The inverse normal function is available on every approved calculator, as well as in printed tables. In tables, you search the body of the table for the probability p closest to the one required, then read the z-value from the row and column margins. For probabilities not in the table, use the calculator and give your answer to three or four significant figures.
逆正态函数在所有被允许的计算器上都能调用，纸质表格上也有。用表格时，在表体中找到最接近所需概率的 p，再从行、列边缘读出 z 值。如果概率不在表中，就用计算器求，并把答案保留三到四位有效数字。
er-tail probabilities without any extra work. This single identity saves more time in the exam than almost any other trick in the topic.
由于标准正态曲线关于 0 对称，恒等式 Φ⁻¹(1 − p) = −Φ⁻¹(p) 让你无需额外操作就能处理右尾概率。这个恒等式在考试中节省的时间，几乎超过本考点其他任何技巧。
5. Handling Right-Tail and Two-Tail Probabilities | 处理右尾与双尾概率
ail probability: P(X < a) = 1 − p, then find the appropriate z. The resulting z will be positive only if p < 0.5; for p > 0.5 the z-value is negative, which correctly places the boundary value a below the mean.
计算器和表格通常要求输入左尾概率。如果给出的是右尾概率 P(X > a) = p，先转换为左尾概率 P(X < a) = 1 − p，再求对应的 z。只有当 p < 0.5 时求出的 z 才是正的；若 p > 0.5，z 为负值，这说明分界值 a 位于均值左侧，完全合理。
Two-tail information needs care as well. If the question says "5% of values lie above a", then P(X < a) = 0.95. If instead it says "the middle 90% lies between two values", then 5% corresponds to P(X < upper) = 0.95. Reading the sentence correctly is half the battle.
双尾信息同样需要小心。如果说"有 5% 的值在 a 之上"，则 P(X < a) = 0.95；如果说"中间 90% 落在两个值之间"，则每一尾各占 5%，因此上界对应 P(X < 上界) = 0.95。把句子读准，仗就打了一半。
- Left tail given: use p directly. | 给出左尾概率：直接使用 p。
- Right tail given: use 1 − p. | 给出右尾概率：使用 1 − p。
- Central interval given: each tail is (1 − central)/2. | 给出中心区间：每一尾为 (1 − 中心概率)/2。
6. Setting Up Two Equations: The General Method | 建立双方程：通用方法
The general method for finding two unknown parameters from two probability statements can be written as a fixed five-step routine. Follow it in order every time and the algebra will look after itself.
由两条概率信息求两个未知参数，可以总结为固定的五步流程。每次都按顺序执行，代数部分自然顺畅。
- tep 1: Convert every statement into the form P(X < a) = p, using 1 − p where necessary. | 第 1 步：把每条信息都化成 P(X < a) = p 的形式，必要时用 1 − p。
- Step 2: Find each z-value with the inverse normal function. | 第 2 步：用逆正态函数求出每个 z 值。
- Step 3: Write each statement as (a − μ)/σ = z, or equivalently μ = a − zσ. | 第 3 步：把每条信息写成 (a − μ)/σ = z，或等价地 μ = a − zσ。
- Step 4: Eliminate μ by subtracting the two equations, then solve for σ. | 第 4 步：两式相减消去 μ，先解出 σ。
- Step 5: Substitute σ back to obtain μ, then check both original statements. | 第 5 步：代回求出 μ，再代回检验两条原始信息。
atement and then subtracting gives a₁ − a₂ = (z₁ − z₂)σ, from which σ follows immediately.
两式相减是最简洁的路线。把每式写成 a − μ = zσ 再相减，得 a₁ − a₂ = (z₁ − z₂)σ，σ 立刻可得。
σ = (a₁ − a₂) / (z₁ − z₂)
Then substitute back into either equation to get μ. Always finish by substituting both μ and σ into the original probability statements as a sanity check, since the examination report regularly notes that candidates skip this step and fail to notice a sign error.
然后把 σ 代回任一方程求 μ。最后一定要把 μ 和 σ 代回原始概率条件做一次检验，因为考试报告反复指出，考生跳过这一步而未能发现符号错误。
7. Worked Example 1: Two Tail Probabilities | 例题一：两个尾部概率
(X < 20) = 0.25 and P(X > 45) = 0.10, find the mean and the standard deviation of X.
连续随机变量 X 服从正态分布。已知 P(X < 20) = 0.25 且 P(X > 45) = 0.10，求 X 的均值与标准差。
: Using the inverse normal function, z₁ = Φ⁻¹(0.25) = −0.6745 and z₂ = Φ⁻¹(0.90) = 1.2816.
第 1 步：第一条已是左尾概率；第二条化为 P(X < 45) = 1 − 0.10 = 0.90。第 2 步：用逆正态函数得 z₁ = Φ⁻¹(0.25) = −0.6745，z₂ = Φ⁻¹(0.90) = 1.2816。
first equation from the second eliminates μ and gives 25 = 1.9561σ, so σ = 12.78 (3 s.f.).
第 3 步：代入 X = μ + Zσ。第一个点：20 = μ − 0.6745σ；第二个点：45 = μ + 1.2816σ。第 4 步：第二式减第一式消去 μ，得 25 = 1.9561σ，所以 σ = 12.78（三位有效数字）。
− 28.62) / 12.78 = 1.2817, which matches z₂ to four decimal places, confirming the answer.
第 5 步：代回得 μ = 20 + 0.6745 × 12.78 = 28.62。检验第二条： (45 − 28.62) / 12.78 = 1.2817，与 z₂ 在四位小数上吻合，答案得到验证。
μ = 28.6, σ = 12.8 (3 s.f.)
Notice how the negative z-value was handled: it is simply carried through the algebra, producing a plus sign when rearranged for μ. There is no need to treat it as a special case.
注意负 z 值的处理方式：直接带着它做代数，整理成 μ 的表达式时自然变成加号，完全不需要当作特例处理。
8. Worked Example 2: A Symmetric Central Interval | 例题二：对称中心区间
The masses of a certain type of fruit, in grams, are normally distributed. It is found that 90% of the fruit have masses between 30 g and 70 g, and the interval is symmetric about the mean. Find μ and σ.
某种水果的质量（单位：克）服从正态分布。已知 90% 的水果质量在 30 克到 70 克之间，且该区间关于均值对称。求 μ 与 σ。
probability is 0.90, so each tail contains 0.05, and the upper boundary corresponds to P(X < 70) = 0.95.
由于正态分布对称且所给区间以均值为中心，均值即中点：μ = (30 + 70) / 2 = 50。中心概率为 0.90，故每一尾含 0.05，上界对应 P(X < 70) = 0.95。
e lower boundary: (30 − 50) / 12.16 = −1.6447, which is the negative of the same z-value, as symmetry requires.
逆正态给出 z = Φ⁻¹(0.95) = 1.6449。对上界标准化：(70 − 50) / σ = 1.6449，所以 σ = 20 / 1.6449 = 12.16（四位有效数字）。检验下界：(30 − 50) / 12.16 = −1.6447，正好是同一 z 值的相反数，符合对称性要求。
μ = 50, σ = 12.2 (3 s.f.)
This type of question is easier than it looks because symmetry supplies one of the two equations for free. Whenever an interval is described as symmetric about the mean, or as "the middle p%", start by writing down μ as the midpoint of the two stated values.
这类题目比看上去简单，因为对称性替你免费提供了一个方程。凡是题目说区间关于均值对称，或说"中间 p%"，第一步就先把 μ 写成两个给定值的中点。
9. Worked Example 3: A Quality Control Context | 例题三：质检情境
A machine fills packets of coffee. The filled mass X, in grams, is normally distributed. 30% of packets contain less than 40 g, and 20% contain more than 70 g. Estimate μ and σ to three significant figures.
一台机器灌装咖啡包，灌装质量 X（克）服从正态分布。30% 的包装少于 40 克，20% 的包装多于 70 克。求 μ 与 σ（保留三位有效数字）。
0.80. The corresponding z-values are z₁ = Φ⁻¹(0.30) = −0.5244 and z₂ = Φ⁻¹(0.80) = 0.8416.
化为左尾概率：P(X < 40) = 0.30，P(X < 70) = 1 − 0.20 = 0.80。对应的 z 值为 z₁ = Φ⁻¹(0.30) = −0.5244，z₂ = Φ⁻¹(0.80) = 0.8416。
0σ, hence σ = 21.96 (4 s.f.), and substituting back gives μ = 40 + 0.5244 × 21.96 = 51.52.
两个方程为 40 = μ − 0.5244σ 与 70 = μ + 0.8416σ。相减得 30 = 1.3660σ，故 σ = 21.96（四位有效数字），代回得 μ = 40 + 0.5244 × 21.96 = 51.52。
μ = 51.5, σ = 22.0 (3 s.f.)
s the interval from 40 to 70 spans roughly 1.4 standard deviations, which is consistent with capturing the middle 50% of the distribution. Sanity checks like this cost five seconds and catch most sign errors.
快速合理性检验：μ = 51.5 落在 40 与 70 之间；σ = 22.0 意味着 40 到 70 的区间大约跨越 1.4 个标准差，与"包含中间约 50%"是一致的。这样的检验只花五秒钟，却能抓出绝大多数符号错误。
10. Calculator Workflow | 计算器操作流程
. The calculator returns the z-value. On the fx-CG50 graphical model, go through DIST, then NORM, then InvN and enter the same information.
在卡西欧 fx
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