📚 Exercise 24C.1: Normal Probability Calculations with Your GDC | 练习24C.1：用GDC计算正态概率
The normal distribution forms the backbone of statistical analysis in the IB Mathematics curriculum. Exercise 24C.1 is designed to build your confidence in using a graphical display calculator (GDC) to compute probabilities and critical values for normal distributions. Whether you are working with the standard normal Z ~ N(0,1) or a general normal variable X ~ N(μ, σ²), the GDC provides fast and accurate results, eliminating the need for statistical tables. This article walks you through every command, concept, and common pitfall associated with this exercise, ensuring you can tackle any normal probability question with ease.
正态分布是IB数学课程中统计分析的基石。练习24C.1旨在帮助你建立信心，熟练使用图形显示计算器（GDC）来计算正态分布的概率和临界值。无论你是在处理标准正态Z ~ N(0,1)，还是一般正态变量X ~ N(μ, σ²)，GDC都能提供快速而准确的结果，从而无需查阅统计表。本文将带你逐一了解与该练习相关的每条命令、每个概念以及常见误区，确保你能轻松应对任何正态概率问题。
1. The Normal Distribution Family | 正态分布族
A normal distribution is a continuous probability distribution characterised by a symmetric, bell-shaped curve. It is fully described by two parameters: the mean μ (mu) and the standard deviation σ (sigma). The peak of the curve is located at the mean, and the spread of the data is controlled by σ. The total area under the curve equals 1, and the curve is asymptotic to the horizontal axis, meaning it never actually touches it.
正态分布是一种连续概率分布，其特点是呈对称的钟形曲线。该分布完全由两个参数描述：均值μ（mu）和标准差σ（sigma）。曲线的峰值位于均值处，数据的离散程度由σ控制。曲线下方的总面积为1，且曲线以横轴为渐近线，意味着它永远不会真正与横轴相交。
The probability density function of a general normal variable is given by:
一般正态变量的概率密度函数为：
f(x) = 1 / (σ√2π) × e−(x − μ)² / (2σ²)
You are not required to integrate this function by hand. Instead, the GDC evaluates the area numerically, which is precisely why Exercise 24C.1 focuses on calculator technique rather than algebraic derivation. Approximately 68% of the data lies within one standard deviation of the mean, 95% within two, and 99.7% within three. Understanding these properties helps you verify whether your GDC output is reasonable before finalising an answer.
你不需要手工对这个函数积分。相反，GDC会以数值方式计算面积，这正是练习24C.1侧重于计算器技巧而非代数推导的原因。约有68%的数据落在均值的一个标准差范围内，95%落在两个标准差范围内，99.7%落在三个标准差范围内。理解这些特性有助于你在最终确定答案之前，先判断GDC输出是否合理。
2. Standard Normal Distribution Z ~ N(0,1) | 标准正态分布 Z ~ N(0,1)
eviously, students used Z-tables to look up probabilities, but GDC technology now performs the same task instantly and to far greater precision. Any normal variable X ~ ²) can be converted into the standard normal variable Z using the standardisation formula:
标准正态分布是均值为μ = 0、标准差为σ = 1的特殊情形。过去学生使用Z表来查概率，但如今的GDC技术可以瞬间完成同样的任务，且精度高得多。任何正态变量X ~ N(μ, σ²)都可以通过标准化公式转换为标准正态变量Z：
z = (x − μ) / σ
ond, P(Z > z) = 1 − P(Z < z). These identities let you check a GDC answer or convert a 'greater than' question into a 'less than' question when a particular command only handles left-tail areas.
标准正态曲线的对称性产生了两个在练习24C.1中极其有用的恒等式。第一，P(Z < −z) = P(Z > z)。第二，P(Z > z) = 1 − P(Z < z)。这些恒等式让你能够检验GDC的答案，或在某条命令只能处理左尾面积时，把"大于"问题转换为"小于"问题。
5) ≈ 0.0668. Recognising this pattern saves time and reduces keystroke errors in the exam.
例如，由于P(Z < 1.5) ≈ 0.9332，利用对称性可立即得到P(Z < −1.5) ≈ 0.0668以及P(Z > 1.5) ≈ 0.0668。识别出这一规律可以节省时间，并减少考试中的按键错误。
3. Reading the Question Before Touching the GDC | 先读题，再动计算器
Most lost marks in Exercise 24C.1 come from misreading the question, not from pressing the wrong button. Before entering any command, extract four pieces of information: the distribution and its parameters, the variable of interest, the direction of the inequality, and whether the question asks for a probability or a value of x.
练习24C.1中丢失的分数大多源于误读题目，而非按错按钮。在输入任何命令之前，先提取四条信息：分布及其参数、所关注的变量、不等号的方向，以及题目问的是概率还是x的取值。
Watch out for parameters disguised as variances. A question that states X ~ N(40, 9) usually means μ = 40 and σ² = 9, so σ = 3. If instead the question says X ~ vention your textbook is using and write it down explicitly at the start of your solution.
注意那些伪装成方差的参数。题目写X ~ N(40, 9)通常意味着μ = 40而σ² = 9，所以σ = 3。若题目写X ~ N(40, 3²)，则σ同样等于3。务必判断你的教材使用的是哪种约定，并在解题开头明确写下来。
- Underline the words 'less than', 'greater than', 'between', 'at least', 'at most'. | 划出'小于'、'大于'、'介于'、'至少'、'至多'等词。
- Circle the sample size if the question involves a mean of n observations. | 若题目涉及n个观测值的均值，圈出样本容量。
- Mark whether the final answer should be a probability or a value with units. | 标明最终答案应是概率还是带单位的数值。
4. The Normal CDF Command | 正态累积分布函数命令
The cumulative distribution function (CDF) command returns the area under the normal curve between two limits. Its general syntax is normalcdf(lower, upper, μ, σ) on Texas Instruments calculators, and Ncd (Normal cumulative distribution) on Casio models. The result is always a probability between 0 and 1.
累积分布函数（CDF）命令返回正态曲线下两个界限之间的面积。在德州仪器（TI）计算器上，其通用语法为normalcdf(lower, upper, μ, σ)；在卡西欧机型上则为Ncd（正态累积分布）。结果始终是介于0和1之间的概率。
| Calculator | 计算器 | Path | 路径 | Syntax | 语法 |
| TI-84 Plus | TI-84 Plus | 2nd VARS → 2:normalcdf( | normalcdf(lower, upper, μ, σ) |
| TI-Nspire | TI-Nspire | Menu → 6 Statistics → 5 Distributions → 2 Normal Cdf | normCdf(lower, upper, μ, σ) |
| Casio fx-CG50 | 卡西欧 fx-CG50 | MENU → 7 Distribution → NORM → Ncd | Lower, Upper, σ, μ |
| Casio fx-9860 | 卡西欧 fx-9860 | STAT → DIST → NORM → Ncd | Lower, Upper, σ, μ |
Note carefully that the Casio interface asks for σ before μ, the reverse of the TI order. Mixing these up is one of the most common errors in Exercise 24C.1 and produces answers that look plausible but are completely wrong.
请特别注意，卡西欧界面先要求输入σ再输入μ，与TI的顺序相反。混淆这两者正是练习24C.1中最常见的错误之一，会得出看似合理却完全错误的答案。
5. Handling Infinite Limits | 处理无限界限
large number instead. On TI models use −1E99 for negative infinity and 1E99 for positive infinity. On Casio models the built-in Ncd screen offers a '−∞' and '∞' toggle, usually accessed with the F1/F2 function keys.
许多题目要求计算P(X < a)或P(X > a)，这需要无限界限。由于计算器无法输入∞符号，可改用一个非常大的数。在TI机型上，用−1E99表示负无穷，用1E99表示正无穷。在卡西欧机型上，内置的Ncd界面提供了'−∞'与'∞'的切换选项，通常通过F1/F2功能键实现。
- P(X < a) → lower = −1E99, upper = a | P(X < a) → 下限 = −1E99，上限 = a
- P(X > a) → lower = a, upper = 1E99 | P(X > a) → 下限 = a，上限 = 1E99
- P(a < X < b) → lower = a, upper = b | P(a < X < b) → 下限 = a，上限 = b
a) equals P(X < a), because the probability of any single exact value is zero. You therefore never need to worry about whether the inequality is strict or inclusive in a normal probability question.
对于连续分布，P(X ≤ a)等于P(X < a)，因为任何单一精确取值的概率都为零。因此在正态概率问题中，你完全不必担心不等号是严格还是非严格。
6. The Inverse Normal Command | 逆正态命令
When a question gives you a probability and asks for the corresponding value of x, you need the inverse normal function. On TI calculators this is invNorm(area, μ, σ), while on Casio it is InvN. The critical rule is that the 'area' you enter is always the left-tail area, that is, P(X < x).
当题目给出概率并要求求出对应的x值时，就需要使用逆正态函数。在TI计算器上这是invNorm(area, μ, σ)，在卡西欧上则是InvN。关键规则是：你输入的'面积'始终是左尾面积，即P(X < x)。
If a question refers to the 'top 10%' or 'the highest 10%', the left-tail area is 1 − 0.10 = 0.90. If it refers to the 'bottom 10%', the area is simply 0.10. Casio' ng of Left, Right or Central, so you can enter the area exactly as the question states it.
如果题目提到'最高的10%'，则左尾面积为1 − 0.10 = 0.90；如果提到'最低的10%'，面积就是0.10。卡西欧的InvN界面通过提供Left、Right或Central的Tail设置，能更优雅地处理这一点，让你按题目原文直接输入面积。
invNorm(0.90, μ, σ) gives x such that P(X < x) = 0.90
Always round your final answer appropriately. For a value such as a mass or a length, three significant figures is normally sufficient; give the answer in the context of the question and never leave it as a raw calculator output with ten decimal places.
务必对最终答案进行适当取整。对于质量或长度之类的数值，通常保留三位有效数字即可；请结合题目情境作答，绝不要留下带十位小数的原始计算器输出。
7. Standardising by Hand Versus Using the GDC | 手工标准化与使用GDC
Both routes are acceptable in Exercise 24C.1, and knowing both gives you a powerful cross-check. The standardising route converts x into z and then evaluates a standard normal probability; the direct route enters μ and σ straight into the calculator.
在练习24C.1中，两种途径都可接受，而掌握两者能提供强有力的交叉检验。标准化途径把x转换为z，再计算标准正态概率；直接途径则把μ和σ直接输入计算器。
| Method | 方法 | Command | 命令 | Result | 结果 |
| Standardise then CDF | 先标准化再用CDF | normalcdf(−1E99, 1.6, 0, 1) | 0.9452 |
| Direct CDF | 直接使用CDF | normalcdf(−1E99, 68, 60, 5) | 0.9452 |
If the two methods disagree, you have made an arithmetic slip in the standardisation step or entered a wrong parameter. Since IB mark schemes often award a method mark for correct standardisation, showing the z-value in your written solution is usually worthwhile even when you use the direct calculator route.
如果两种方法结果不一致，说明你在标准化步骤中出现了算术失误，或输入了错误的参数。由于IB评分方案常为正确的标准化给出方法分，即使你使用计算器的直接途径，在书面解答中写出z值通常也是值得的。
8. Using the Empirical Rule as a Sanity Check | 用经验法则做合理性检验
The 68–95–99.7 rule provides a rapid mental estimate that can catch gross errors before you commit to an answer. If a question asks for P(μ − σ < X < μ + σ), the answer should be close to 0.68; for two standard deviations, close to 0.95; for three, close to 0.997.
68–95–99.7法则提供了一种快速的脑内估算，能在你确定答案之前捕捉到严重错误。如果题目要求P(μ − σ < X < μ + σ)，答案应接近0.68；两个标准差应接近0.95；三个标准差应接近0.997。
- P(μ − σ < X < μ + σ) ≈ 0.6827 | P(μ − σ < X < μ + σ) ≈ 0.6827
- P(μ − 2σ < X < μ + 2σ) ≈ 0.9545 | P(μ − 2σ < X < μ + 2σ) ≈ 0.9545
- P(μ − 3σ < X < μ + 3σ) ≈ 0.9973 | P(μ − 3σ < X < μ + 3σ) ≈ 0.9973
Suppose your calculator returns 0.42 for P(μ − σ < X < μ + σ). Immediately you know something is wrong: either a parameter was entered incorrectly or the limits were swapped. This habit of estimating first is exactly what distinguishes a confident student from one who blindly trusts the screen.
假设你的计算器对P(μ − σ < X < μ + σ)返回0.42。你立刻就知道出错了：要么参数输入有误，要么上下限被颠倒。这种先估算的习惯，正是自信的学生与盲目相信屏幕的学生之间的区别。
9. Eight Common Pitfalls in Exercise 24C.1 | 练习24C.1中的八个常见陷阱
Examiners report the same mistakes year after year in normal probability questions. Reviewing them now will save you marks in the examination, because most are completely avoidable once you know they exist.
考官年复一年地报告正态概率题中相同的错误。现在回顾这些错误能帮你在考试中保住分数，因为一旦你意识到它们的存在，大多数都是完全可以避免的。
- Confusing σ with σ². If the question states the variance, take the square root before entering σ. | 混淆σ与σ²。若题目给出的是方差，输入σ之前要先开平方。
- Reversing the limits. (a < X < b) needs lower = a and upper = b; a negative result means you swapped them. | 颠倒上下限。P(a < X < b)需要下限 = a、上限 = b；若结果为负，说明你颠倒了它们。
- Using the wrong tail for InvN. The calculator always uses the left tail unless you change the Tail setting. | InvN使用了错误的尾部。除非更改Tail设置，计算器始终使用左尾。
- Forgetting to use −1E99 instead of leaving the lower limit blank. | 忘记使用−1E99，而把下限留空。
- Entering σ and μ in the wrong order on Casio. | 在卡西欧上把σ和μ的顺序输错。
- Rounding intermediate values. Keep full precision until the final line. | 对中间值过早取整。请保持完整精度直到最后一行。
- Leaving the calculator in a non-default mode after a trigonometry question. | 做完三角题后未把计算器恢复默认模式。
- Writing only the final number without the probability statement P(X < 68) = 0.945. | 只写最终数字，而没有写出概率陈述P(X < 68) = 0.945。
10. Worked Example 1: A Direct Probability | 例题一：直接求概率
The masses of apples from an orchard are normally distributed with mean 120 g and standard deviation 8 g. Find the probability that a randomly chosen apple has a mass greater than 132 g.
某果园苹果的质量服从正态分布，均值为120 g，标准差为8 g。求随机选取的一个苹果质量大于132 g的概率。
Solution. Let X be the mass of an apple, so X ~ N(120, 8²). Standardising: z = (132 − 120) / 8 = 1.5. Then:
解答。设X为苹果的质量，则X ~ N(120, 8²)。标准化：z = (132 − 120) / 8 = 1.5。于是：
P(X > 132) = P(Z > 1.5) = 1 − 0.9332 = 0.0668
Using the GDC directly: normalcdf(132, 1E99, 120, 8) = 0.0668072… tions above the mean and 1.5 is close to 2, so a small upper-tail probability is expected.
直接使用GDC：normalcdf(132, 1E99, 120, 8) = 0.0668072…，取整为0.0668。经验法则证实了其合理性：132 g恰好位于均值上方1.5个标准差处，而1.5接近2，因此预期一个小概率的上尾是合理的。
11. Worked Example 2: A Between Probability and an Inverse Problem | 例题二：区间概率与逆向问题
Exam scores are normally distributed with mean 65 and standard deviation 8. (a) Find the probability that a score lies between 55 and 75. (b) Find the score k such that only 5% of candidates score above k.
考试成绩服从正态分布，均值为65，标准差为8。(a) 求分数位于55到75之间的概率。(b) 求分数k，使得仅有5%的考生得分高于k。
ause the limits are symmetric about the mean, the answer is twice the area from 0 to 1.25:
对于(a)部分，对两个界限标准化：z₁ = (55 − 65) / 8 = −1.25，z₂ = (75 − 65) / 8 = 1.25。由于界限关于均值对称，答案为0到1.25之间面积的两倍：
P(55 < X < 75) = 2 × 0.39435 = 0.7887 ≈ 0.789
For part (b), 'above k' s is well over two standard deviations above the mean, entirely consistent with the empirical rule.
对于(b)部分，'高于k'意味着右尾面积为0.05，因此左尾面积为0.95。使用invNorm(0.95, 65, 8)得到k = 65 + 1.6449 × 8 = 78.16，故k ≈ 78.2。注意这已远超均值上方两个标准差，与经验法则完全一致。
12. Writing Solutions That Earn
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