Normal Distribution Probability Calculations and Table-Reading Techniques | 正态分布概率的计算方法与查表技巧

📚 Normal Distribution Probability Calculations and Table-Reading Techniques | 正态分布概率的计算方法与查表技巧

The normal distribution is one of the most important probability models in statistics and is heavily tested in A-level Mathematics. This guide explains how to calculate probabilities using the standard normal distribution table, with clear step-by-step methods and practical tips to avoid common mistakes.

正态分布是统计学中最重要的概率模型之一，也是A-level数学考试的高频考点。本文将系统讲解如何使用标准正态分布表计算概率，并提供清晰的步骤与实用技巧，帮助同学们避免常见错误。

1. The Normal Distribution and Its Notation | 正态分布与记号

A continuous random variable X follows a normal distribution with mean μ and variance σ², written as X ~ N(μ, σ²). The curve is symmetric about the mean, bell-shaped, and the total area under the curve equals 1.

连续型随机变量 X 若服从均值为 μ、方差为 σ² 的正态分布，则记作 X ~ N(μ, σ²)。其曲线关于均值对称，呈钟形，曲线下方的总面积为 1。

The probability that X lies between two values a and b is the area under the curve from a to b. Because X is continuous, P(X = a) = 0 for any single value a, so inequalities such as “less than” and “less than or equal to” give identical probabilities.

随机变量 X 落在 a 与 b 之间的概率，就是曲线在 a 到 b 区间下的面积。由于 X 是连续型的，任意单点 a 的概率 P(X = a) = 0，因此“小于”和“小于等于”所对应的概率是相同的。

X ~ N(μ, σ²) ⇒ Z = (X − μ) / σ ~ N(0, 1)

The standard normal distribution has mean 0 and variance 1, written Z ~ N(0, 1). Converting X to Z is called standardisation, and it allows us to use one single table for all normal distributions.

标准正态分布的均值为 0、方差为 1，记作 Z ~ N(0, 1)。将 X 转化为 Z 的过程称为标准化，它使我们能够用同一张表处理所有正态分布问题。

2. The Standard Normal Distribution Table | 标准正态分布表的结构

The standard normal table usually gives the cumulative probability Φ(z) = P(Z < z), which is the area under the standard normal curve to the left of z. The table is arranged with rows representing the integer part and first decimal of z, and columns representing the second decimal place.

标准正态分布表通常给出累积概率 Φ(z) = P(Z < z)，即标准正态曲线在 z 左侧的面积。表的行表示 z 的整数部分和第一位小数，列表示 z 的第二位小数。

For example, to find Φ(1.25), look down the left column to 1.2, then across to the column headed 0.05. The entry at that intersection is the required probability.

例如，要求 Φ(1.25)，先在左侧列找到 1.2，然后横向找到表头为 0.05 的列，行列交叉处的数值即为所求概率。

z	0.00	0.01	0.02	0.03	0.04	0.05
1.0	0.8413	0.8438	0.8461	0.8485	0.8508	0.8531
1.1	0.8643	0.8665	0.8686	0.8708	0.8729	0.8749
1.2	0.8849	0.8869	0.8888	0.8907	0.8925	0.8944
1.3	0.9032	0.9049	0.9066	0.9082	0.9099	0.9115
1.4	0.9192	0.9207	0.9222	0.9236	0.9251	0.9265

From this excerpt, Φ(1.25) = 0.8944. This means P(Z < 1.25) = 0.8944. Notice that the values in the table are always between 0 and 1, and they increase as z increases.

从表中可读出 Φ(1.25) = 0.8944，即 P(Z < 1.25) = 0.8944。注意表中数值始终介于 0 和 1 之间，且随着 z 增大而增大。

3. Finding P(Z < z) | 计算 P(Z < z)

If z is positive, simply read the table directly. For example, P(Z < 1.5) is found by looking at row 1.5 and column 0.00, giving 0.9332.

若 z 为正数，直接查表即可。例如，P(Z < 1.5)，查第 1.5 行与第 0.00 列，得到 0.9332。

If z is negative, use the symmetry of the normal curve. The property is:

若 z 为负数，则需利用正态曲线的对称性，其性质为：

Φ(−z) = 1 − Φ(z)

Therefore, to find P(Z < −1.5), first find Φ(1.5) = 0.9332, then compute 1 − 0.9332 = 0.0668. So P(Z < −1.5) = 0.0668.

因此，求 P(Z < −1.5) 时，先查出 Φ(1.5) = 0.9332，再计算 1 − 0.9332 = 0.0668，即 P(Z < −1.5) = 0.0668。

Always check whether z is positive or negative before reading the table.
For negative z, never try to read a negative row directly; use the complement rule.

查表前务必判断 z 的正负。
对于负 z，不要试图直接读负数行，而应使用互补规则。

4. Finding P(Z > z) | 计算 P(Z > z)

Because the total area under the curve is 1, the probability to the right of z is:

由于曲线下总面积为 1，z 右侧的概率为：

P(Z > z) = 1 − Φ(z)

For example, P(Z > 1.25) = 1 − 0.8944 = 0.1056. This is the upper-tail probability.

例如，P(Z > 1.25) = 1 − 0.8944 = 0.1056，这就是上尾概率。

For a negative z, the same formula works. P(Z > −1.25) = 1 − Φ(−1.25). Using symmetry, Φ(−1.25) = 1 − Φ(1.25) = 0.1056, so P(Z > −1.25) = 1 − 0.1056 = 0.8944.

对于负 z，公式同样适用。P(Z > −1.25) = 1 − Φ(−1.25)。由对称性，Φ(−1.25) = 1 − Φ(1.25) = 0.1056，所以 P(Z > −1.25) = 1 − 0.1056 = 0.8944。

It is often helpful to sketch the curve and shade the required area before doing any calculation. This reduces the chance of using the wrong tail.

计算前先画出曲线草图并标出所求区域，往往能有效避免用错尾端。

5. Finding P(a < Z < b) | 计算区间概率

The probability that Z lies between two values a and b is the area between them:

Z 落在 a 与 b 之间的概率为两者之间的面积：

P(a < Z < b) = Φ(b) − Φ(a)

This is always calculated by subtracting the lower cumulative probability from the upper cumulative probability. For example, P(0.5 < Z < 1.25) = Φ(1.25) − Φ(0.5) = 0.8944 − 0.6915 = 0.2029.

计算方法始终是“上端累积概率减去下端累积概率”。例如，P(0.5 < Z < 1.25) = Φ(1.25) − Φ(0.5) = 0.8944 − 0.6915 = 0.2029。

When one or both bounds are negative, the same subtraction formula still works. For example, P(−1.5 < Z < 0.5) = Φ(0.5) − Φ(−1.5). Using Φ(−1.5) = 0.0668, we get 0.6915 − 0.0668 = 0.6247.

当区间一端或两端为负数时，减法公式依然成立。例如，P(−1.5 < Z < 0.5) = Φ(0.5) − Φ(−1.5)。利用 Φ(−1.5) = 0.0668，得到 0.6915 − 0.0668 = 0.6247。

If a and b have the same sign, draw the curve to see whether the area is a middle region or a tail region.
Do not add the two tail probabilities unless the interval covers the whole distribution outside a middle range.

若 a 与 b 同号，画图判断所求区域是中间区域还是尾端区域。
除非区间是中间区域之外的整个尾部，否则不要将两个尾部概率相加。

6. Standardising X to Z | 将 X 标准化为 Z

For a general normal distribution X ~ N(μ, σ²), any probability statement about X can be converted into a statement about Z using:

对于一般正态分布 X ~ N(μ, σ²)，关于 X 的概率问题都可以通过下式转化为关于 Z 的问题：

Z = (X − μ) / σ

Suppose X ~ N(100, 15²). To find P(X < 120), standardise: Z = (120 − 100) / 15 = 1.3333. Then P(X < 120) = P(Z < 1.3333) ≈ Φ(1.33) = 0.9082.

设 X ~ N(100, 15²)，求 P(X < 120)。标准化：Z = (120 − 100) / 15 = 1.3333。于是 P(X < 120) = P(Z < 1.3333) ≈ Φ(1.33) = 0.9082。

When standardising, always use the population standard deviation σ, not σ². A common error is to divide by the variance instead of the standard deviation.

标准化时务必使用总体标准差 σ，而不是方差 σ²。一个常见错误是除以方差而不是标准差。

For example, if X ~ N(50, 9), then σ = 3. The correct z-score for X = 56 is (56 − 50) / 3 = 2, not (56 − 50) / 9.

例如，若 X ~ N(50, 9)，则 σ = 3。X = 56 对应的 z 分数应为 (56 − 50) / 3 = 2，而不是 (56 − 50) / 9。

7. Working Backwards: Finding X from a Probability | 逆查法：由概率反求 X

Sometimes the probability is given and the value of X or z is unknown. For example, find k such that P(Z < k) = 0.95.

有时题目给出概率，要求反求 X 或 z。例如，求 k 使得 P(Z < k) = 0.95。

Look inside the table for the probability closest to 0.9500. The closest entry is 0.9505 at z = 1.65, or 0.9495 at z = 1.64. Depending on the convention of your exam board, you may interpolate or choose the nearest value.

在表内查找与 0.9500 最接近的概率。最接近的值为 z = 1.65 对应的 0.9505，或 z = 1.64 对应的 0.9495。具体取值方式需依据你所在考试局的规定，可能需要插值或取最近值。

If P(Z < k) = p, then k = Φ⁻¹(p)

After finding z, convert back to X using X = μ + zσ. For instance, if X ~ N(100, 15²) and P(X < x) = 0.95, then z ≈ 1.645, so x = 100 + 1.645 × 15 = 124.675.

求出 z 后，再利用 X = μ + zσ 反推 X。例如，若 X ~ N(100, 15²) 且 P(X < x) = 0.95，则 z ≈ 1.645，所以 x = 100 + 1.645 × 15 = 124.675。

For upper-tail questions, remember that if P(Z > k) = p, then P(Z < k) = 1 − p. Always convert the statement to a cumulative probability before using the inverse table.

对于上尾问题，若 P(Z > k) = p，则 P(Z < k) = 1 − p。先转化为累积概率，再进行逆查表。

8. Using Symmetry to Simplify Calculations | 利用对称性简化计算

The standard normal curve is symmetric about 0, so the area to the left of −z equals the area to the right of z. This gives two key identities:

标准正态曲线关于 0 对称，因此 −z 左侧的面积等于 z 右侧的面积。由此得到两个关键恒等式：

Φ(−z) = 1 − Φ(z) and P(−1 < Z < 1) = 2Φ(1) − 1

For example, P(−2 < Z < 2) = 2Φ(2) − 1 = 2(0.9772) − 1 = 0.9544. This is the famous 95.4% rule for two standard deviations.

例如，P(−2 < Z < 2) = 2Φ(2) − 1 = 2(0.9772) − 1 = 0.9544，这就是著名的“两个标准差约 95.4%”规则。

Similarly, P(−1.96 < Z < 1.96) = 2Φ(1.96) − 1 = 0.95, which is the critical value used for 95% confidence intervals.

类似地，P(−1.96 < Z < 1.96) = 2Φ(1.96) − 1 = 0.95，这是 95% 置信区间所用的临界值。

When using symmetry, always write down the symmetry step explicitly in your exam solution. This shows the examiner that you understand why the property works.

在考试中使用对称性时，务必在解答中明确写出对称性步骤，这能向阅卷老师展示你理解该性质成立的原因。

9. Common Pitfalls and How to Avoid Them | 常见错误与规避方法

Below are frequent mistakes students make when using the normal table, along with corrected strategies.

以下是同学们在使用正态分布表时最常见的错误，以及对应的纠正策略。

Error / 错误	Why it happens / 原因	Correct approach / 正确做法
Dividing by σ² instead of σ	Confusing variance and standard deviation	Always take the square root of the variance first
Using the upper tail directly for negative z	Misreading the table direction	Draw a sketch and convert to Φ(z) before reading
Forgetting to subtract from 1 for P(Z > z)	Not noticing that the table gives only left-tail areas	Remember total area = 1, so right tail = 1 − left tail
Adding Φ(a) and Φ(b) for interval probabilities	Thinking both areas are separate tails	Always subtract: Φ(b) − Φ(a)
Rounding too early when standardising	Using 1.3 instead of 1.3333	Keep at least 4 decimal places until the final answer

Another common pitfall is forgetting that P(X = a) = 0. This means P(X ≤ a) and P(X < a) are identical. Do not waste time trying to distinguish "less than" from "less than or equal to" for continuous distributions.

另一个常见错误是忘记 P(X = a) = 0。这意味着 P(X ≤ a) 与 P(X < a) 完全相同。对于连续型分布，不必浪费时间区分“小于”和“小于等于”。

10. Worked Example with a Realistic Exam Question | 综合例题

A factory produces metal rods with lengths normally distributed, mean 10.0 cm and standard deviation 0.2 cm. Find the probability that a randomly chosen rod is longer than 10.4 cm.

某工厂生产的金属棒长度服从正态分布，均值为 10.0 cm，标准差为 0.2 cm。求随机抽取一根金属棒，其长度超过 10.4 cm 的概率。

Z = (10.4 − 10.0) / 0.2 = 0.4 / 0.2 = 2.0

P(X > 10.4) = P(Z > 2.0) = 1 − Φ(2.0) = 1 − 0.9772 = 0.0228

Therefore, the probability is 0.0228, or about 2.28%. This is a one-step standardisation followed by an upper-tail calculation.

因此，该概率为 0.0228，即约 2.28%。这是一个先标准化再做上尾计算的简单问题。

Now find the range of lengths that contains the middle 90% of the rods.

接着求包含中间 90% 金属棒的长度范围。

For the middle 90%, the lower tail and upper tail each contain 5%. Thus we need z such that Φ(z) = 0.95, giving z ≈ 1.645. By symmetry, the lower boundary is at z = −1.645.

中间 90% 意味着左右尾端各含 5%。因此需求 Φ(z) = 0.95 的 z，得到 z ≈ 1.645。由对称性，下边界对应 z = −1.645。

x₁ = 10.0 − 1.645 × 0.2 = 9.671 and x₂ = 10.0 + 1.645 × 0.2 = 10.329

The middle 90% of rods have lengths between 9.671 cm and 10.329 cm.

中间 90% 的金属棒长度介于 9.671 cm 与 10.329 cm 之间。

11. Quick Table-Reading Tips for the Exam | 考试查表小贴士

Use the following quick checklist whenever you face a normal distribution question.

遇到正态分布题目时，请使用以下快速检查清单。

Write down X ~ N(μ, σ²) and identify μ and σ clearly.
Sketch the curve and shade the required region.
Convert the boundary value(s) to z-scores.
Rewrite the probability in terms of Φ(z).
Use symmetry or the complement rule where needed.
Read the table carefully, matching row and column correctly.
Give the final answer as a decimal, usually to 4 decimal places unless the question states otherwise.

写出 X ~ N(μ, σ²)，并明确 μ 与 σ。
画出曲线草图，标出所求区域。
将边界值标准化为 z 分数。
用 Φ(z) 表示所求概率。
必要时利用对称性或互补规则。
仔细查表，正确匹配行与列。
最终答案通常保留 4 位小数，除非题目另有要求。

Also, always check whether the final probability is sensible. A probability must be between 0 and 1. If you compute a negative probability or a value greater than 1, you have made a sign or subtraction error.

另外，务必检查最终概率是否合理。概率必须介于 0 和 1 之间。如果你算出了负数或大于 1 的值，说明你在符号或减法上出了错。

12. Conclusion | 总结

Mastering normal distribution probability calculations requires understanding the standard normal table, the standardisation formula, and the symmetry properties of the curve. With consistent practice and careful sketching, this topic becomes one of the most reliable scoring areas in A-level Mathematics.

掌握正态分布概率计算，关键在于理解标准正态分布表、标准化公式以及曲线的对称性质。通过持续练习和认真画图，这个专题将成为A-level数学中最稳定的得分点之一。

Remember the three golden rules: standardise first, sketch always, and check the tail direction before reading the table.

请记住三条黄金法则：先标准化、勤画草图、查表前确认尾端方向。

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Normal Distribution Probability Calculations and Table-Reading Techniques | 正态分布概率的计算方法与查表技巧