Mastering Normal Distribution: Exercise 15J Explained | 掌握正态分布：练习15J详解

📚 Mastering Normal Distribution: Exercise 15J Explained | 掌握正态分布：练习15J详解

In IB Mathematics, normal distribution calculations form a cornerstone of statistical analysis. Exercise 15J is typically designed to strengthen your ability to find probabilities, compute z‑scores, and apply inverse normal techniques in both standard and real‑world contexts. This walkthrough will guide you through the essential concepts, step‑by‑step methods, and common pitfalls so that you can approach every question with confidence.

在 IB 数学中，正态分布的计算是统计分析的基石。练习15J 通常旨在强化你求概率、计算 z 分数以及在标准情境和实际情境中应用逆正态技巧的能力。本详解将带你梳理核心概念、步骤方法和常见误区，让你自信应对每一道题。

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1. Understanding the Normal Distribution | 理解正态分布

A normal distribution is a continuous probability distribution that is symmetric about the mean μ. The shape of the curve is bell‑shaped, and it is fully described by two parameters: the mean μ and the standard deviation σ.

正态分布是一种关于均值 μ 对称的连续概率分布。曲线的形状是钟形的，由两个参数完全描述：均值 μ 和标准差 σ。

The total area under the normal curve is exactly 1, representing the total probability. For any interval, the area under the curve equals the probability that a random variable X falls within that interval.

正态曲线下的总面积恰好为 1，代表总概率。对于任意区间，曲线下的面积等于随机变量 X 落入该区间的概率。

Many real‑world measurements—such as heights, test scores, and measurement errors—approximately follow a normal distribution, making it a powerful tool for modelling.

许多现实世界的测量值——如身高、考试成绩和测量误差——近似服从正态分布，这使其成为一种强大的建模工具。

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

The standard normal distribution is a special case where μ = 0 and σ = 1. We denote it by Z ∼ N(0, 1). Tables and calculators provide cumulative probabilities for the standard normal variable Z, which makes it the key to working with any normal distribution.

标准正态分布是 μ = 0 且 σ = 1 的特殊情况。我们将其记作 Z ∼ N(0, 1)。表格和计算器提供了标准正态变量 Z 的累积概率，这使其成为处理任何正态分布的关键。

The cumulative distribution function Φ(z) gives the probability that Z is less than or equal to a given value: Φ(z) = P(Z ≤ z). By looking up z in the standard normal table, you can quickly find left‑tail probabilities.

累积分布函数 Φ(z) 给出 Z 小于或等于给定值的概率：Φ(z) = P(Z ≤ z)。通过查标准正态表，你可以快速找到左尾概率。

• Example: Φ(1.00) ≈ 0.8413, meaning about 84.13% of the data lies below 1 standard deviation above the mean.
• 示例：Φ(1.00) ≈ 0.8413，这意味着大约 84.13% 的数据位于均值上方一个标准差以下。

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3. Using Z‑Scores | 使用 Z 分数

A z‑score measures how many standard deviations an observation x is from the mean μ. The transformation formula is:

z 分数衡量观测值 x 距离均值 μ 多少个标准差。转换公式为：

z = (x − μ) / σ

Once you convert a raw score to a z‑score, you can use standard normal tables to find probabilities, regardless of the original mean and standard deviation.

一旦将原始分数转换为 z 分数，你就可以使用标准正态表求概率，而无需考虑原始的均值和标准差。

Key tip: Always compute the z‑score before consulting the table, unless your calculator directly handles non‑standard normal curves.

关键提示：在查表之前一定要先计算 z 分数，除非你的计算器直接处理非标准正态曲线。

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4. Finding Probabilities for a Given Range | 求给定区间的概率

To find P(a < X < b) for X ∼ N(μ, σ²), follow these steps:

对于 X ∼ N(μ, σ²)，求 P(a < X < b) 的步骤如下：

1. Standardise both endpoints: zₐ = (a − μ)/σ, z_b = (b − μ)/σ.
2. Find Φ(z_b) and Φ(zₐ) from the table.
3. The required probability is Φ(z_b) − Φ(zₐ).

1. 对两个端点进行标准化：zₐ = (a − μ)/σ, z_b = (b − μ)/σ。
2. 从表中查出 Φ(z_b) 和 Φ(zₐ)。
3. 所求概率为 Φ(z_b) − Φ(zₐ)。

If you are using a calculator, you can often enter μ, σ, and the bounds directly to obtain the probability without manual lookups.

如果你使用计算器，通常可以直接输入 μ、σ 和界限来获得概率，而无需手动查表。

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5. Tail Probabilities and Symmetry | 尾部概率与对称性

Because the curve is symmetric, probabilities for intervals like P(X > a) or P(X < a) can be found using complements. For example, P(Z > z) = 1 − Φ(z).

由于曲线是对称的，对于像 P(X > a) 或 P(X < a) 这样的区间概率可以通过补数求得。例如，P(Z > z) = 1 − Φ(z)。

Symmetry also helps with negative z‑values: Φ(−z) = 1 − Φ(z). This means you only need a table for positive z, which was a common feature of older exam setups.

对称性也有助于处理负的 z 值：Φ(−z) = 1 − Φ(z)。这意味着你只需要正 z 值的表，这在过去的考试设置中很常见。

When a question asks for “more than” or “at least”, always draw a sketch and decide whether to use the left‑tail or right‑tail area.

当问题问及“多于”或“至少”时，一定要画图并判断是使用左尾面积还是右尾面积。

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6. Inverse Normal: Finding Values from Probabilities | 逆正态分布：从概率求取值

Sometimes you are given a probability and need to find the corresponding value of the variable. This is called the inverse normal problem. For a standard normal distribution, you find the z such that P(Z ≤ z) = p; this z is denoted z_p or invNorm(p).

有时给出概率，需要求对应的变量值。这称为逆正态问题。对于标准正态分布，求使得 P(Z ≤ z) = p 的 z；这个 z 记作 z_p 或 invNorm(p)。

For a non‑standard normal X ∼ N(μ, σ²), first find the z‑score that corresponds to the given left‑tail probability, then transform back using:

对于非标准正态 X ∼ N(μ, σ²)，先求出对应给定左尾概率的 z 分数，然后利用以下公式转换回原值：

x = μ + z × σ

Be careful: if the question gives a right‑tail probability (e.g., top 10%), subtract it from 1 to obtain the left‑tail probability before using the inverse function.

注意：如果题目给出的是右尾概率（例如前 10%），在使用逆函数之前要先用 1 减去它，以获得左尾概率。

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7. Working with Non‑Standard Normal Variables | 处理非标准正态变量

Most exam questions do not give you a standard normal variable directly. You must identify μ and σ from the problem statement, then use the z‑score formula to link real‑world units to the standard normal table.

大多数考题不会直接给出标准正态变量。你必须从题目叙述中识别 μ 和 σ，然后使用 z 分数公式将现实世界的单位与标准正态表联系起来。

Always check whether the variance or the standard deviation is provided. If you are given the variance σ², take the square root to obtain σ. Mixing up variance and standard deviation is one of the most frequent errors in Exercise 15J.

务必检查给出的是方差还是标准差。如果给出的是方差 σ²，要开平方求得 σ。混淆方差和标准差是练习15J中最常见的错误之一。

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8. Solving Real‑World Problems | 解决实际问题

Normal distributions are often used to model real‑life scenarios: the lifespan of light bulbs, the fill volume of cans, or the time taken to complete a task. In such problems, you are typically asked to find the proportion meeting a specification or the value that separates a certain percentage of the population.

正态分布常常用于模拟现实情境：灯泡的寿命、罐子的填充量或完成任务所需的时间。在这类问题中，通常会要求找出满足某个规格的比例，或者划分出人口中某一百分比的分界值。

For instance: “A machine fills bottles with a mean of 500 ml and a standard deviation of 5 ml. Find the probability that a randomly chosen bottle contains between 495 and 505 ml.” The z‑scores are −1 and 1, and the probability is Φ(1) − Φ(−1) = 0.6826 (or about 68.3%).

例如：“一台机器灌装瓶子的均值为 500 毫升，标准差为 5 毫升。求随机选取的瓶子容量在 495 到 505 毫升之间的概率。”z 分数为 −1 和 1，概率为 Φ(1) − Φ(−1) = 0.6826（约 68.3%）。

Scenario	Key Steps
Quality control	Define limits → z‑scores → probability of defect → inverse for tolerance
Examination scores	Percentile ranking → use invNorm to find cut‑off score
Medicine	Dosage based on patient weight distribution → ensure safe interval

情景	关键步骤
质量控制	定义界限 → z 分数 → 次品概率 → 逆求公差
考试成绩	百分位排名 → 用 invNorm 求截断分数
医学	基于患者体重分布的剂量 → 确保安全区间

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9. Common Mistakes to Avoid | 常见错误要避免

Even small mistakes in normal distribution calculations can lead to completely wrong conclusions. Here are the traps most students fall into in Exercise 15J:

正态分布计算中哪怕一个小错误都可能导致完全错误的结论。以下是学生们在练习15J中最常落入的陷阱：

• Using variance instead of standard deviation in the z‑score formula. Always confirm whether σ or σ² is given.
• 在 z 分数公式中使用方差而不是标准差。务必确认给出的是 σ 还是 σ²。
• Forgetting to convert a right‑tail probability to a left‑tail one before using invNorm. For example, if you need the top 5%, use invNorm(0.95).
• 在使用 invNorm 之前忘记将右尾概率转换为左尾概率。例如，如果需要前 5%，应使用 invNorm(0.95)。
• Using the wrong table – some tables give the probability between 0 and z, not Φ(z). Know your formula booklet’s table format.
• 用错表格——有些表格给出的是 0 到 z 之间的概率，而不是 Φ(z)。要熟悉你的公式手册中表格的格式。
• Rounding z‑scores too early, which accumulates error in multi‑step questions.
• 过早舍入 z 分数，在多步问题中会累积误差。

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10. Practice Questions from Exercise 15J | 练习15J 样题

The following examples reflect the style and difficulty of a typical IB Exercise 15J. Work through them step by step, checking your understanding of both forward and inverse normal calculations.

以下例题反映了典型 IB 练习15J 的风格和难度。逐步练习，检查你对正向和逆向正态计算的理解。

Question 1: X is normally distributed with μ = 100 and σ = 15. Find P(85 < X < 115).

问题 1：X 服从正态分布，μ = 100，σ = 15。求 P(85 < X < 115)。

Solution: z₁ = (85 − 100)/15 = −1, z₂ = (115 − 100)/15 = 1. P(−1 < Z < 1) = Φ(1) − Φ(−1) = 0.8413 − 0.1587 = 0.6826.

问题 1 解答：z₁ = (85 − 100)/15 = −1，z₂ = (115 − 100)/15 = 1。P(−1 < Z < 1) = Φ(1) − Φ(−1) = 0.8413 − 0.1587 = 0.6826。

Question 2: The weights of apples are normally distributed with mean 180 g and standard deviation 30 g. Only apples weighing more than 230 g are classified as “premium”. What proportion of apples is premium?

问题 2：苹果重量服从均值为 180 克、标准差为 30 克的正态分布。只有重量超过 230 克的苹果被归类为“特级”。特级苹果的比例是多少？

Solution: P(X > 230) = 1 − P(X ≤ 230). z = (230 − 180)/30 = 1.667. Φ(1.667) ≈ 0.9522, so probability ≈ 1 − 0.9522 = 0.0478 (about 4.78%).

问题 2 解答：P(X > 230) = 1 − P(X ≤ 230)。z = (230 − 180)/30 = 1.667。Φ(1.667) ≈ 0.9522，因此概率 ≈ 1 − 0.9522 = 0.0478（约 4.78%）。

Question 3: Given that the top 20% of students receive a distinction in a test with scores N(65, 10²), find the minimum mark needed for a distinction.

问题 3：某次考试分数服从 N(65, 10²)，前 20% 的学生获得优秀。求获得优秀所需的最低分数。

Solution: We need the 80th percentile: invNorm(0.80) ≈ 0.8416 (z‑score). x = μ + zσ = 65 + 0.8416 × 10 = 73.416, so a mark of about 73 or 74, depending on rounding.

问题 3 解答：我们需要第 80 百分位数：invNorm(0.80) ≈ 0.8416（z 分数）。x = μ + zσ = 65 + 0.8416 × 10 = 73.416，因此大约需要 73 或 74 分，视舍入而定。

Consistent practice with these patterns will make Exercise 15J feel routine and straightforward.

持续练习这些模式，练习15J 将会变得常规且直接。

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