Normal Distribution — SL 4.7 Normal distribution; z-scores and probability calculations (technology-supported) - IB Mathematics Applications and Interpretation

SECTION 1

Historical Context & Motivation

Imagine measuring the heights of every student in your school and plotting the results on a graph. You would likely see a familiar shape: most students cluster near the average, with fewer and fewer students as you move toward very short or very tall extremes. This symmetrical, bell-shaped pattern appears so frequently in nature, science, and everyday life that mathematicians spent centuries developing a precise way to describe it. The normal distribution is that mathematical description, and it is one of the most powerful tools in all of statistics.

The story of the normal distribution begins with a quest to understand errors in scientific measurement. Early astronomers noticed that when they repeatedly measured the position of a star, their measurements scattered around the true value in a predictable way. This pattern of random variation needed a mathematical model, and several brilliant thinkers contributed pieces of the puzzle over more than a century.

1733 • De Moivre's Discovery

Abraham de Moivre, a French mathematician living in London, first described the bell-shaped curve as an approximation to the binomial distribution. He published his findings in The Doctrine of Chances, laying the groundwork for probability theory.

1809 • Gauss and the 'Error Curve'

Carl Friedrich Gauss used the bell curve to model errors in astronomical observations, which is why it is sometimes called the Gaussian distribution. He showed that measurement errors naturally follow this shape.

1835 • Quetelet Applies It to People

Adolphe Quetelet demonstrated that human characteristics like height and chest measurements follow the normal distribution. This was the first application of the curve beyond physical science.

1893 • Pearson Coins 'Normal'

Karl Pearson popularised the term 'normal distribution,' establishing it as the standard name used in statistics textbooks worldwide. He also developed many statistical tests that rely on normality.

The central question that drove these mathematicians remains the same one you will explore in this lesson: if data naturally clusters around a central value, how can we calculate the probability of observing any particular range of values? The normal distribution, combined with z-scores and modern calculator technology, gives us a precise and efficient way to answer that question.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need to understand the key building blocks of the normal distribution. Every normal distribution is completely determined by just two numbers, and every calculation you perform in this topic comes back to the same fundamental ideas.

Mean (μ)

The mean (μ, the Greek letter 'mu') is the centre of the distribution. It marks the peak of the bell curve and is the value around which all data clusters. Moving the mean shifts the entire curve left or right on the number line.

Standard Deviation (σ)

The standard deviation (σ, 'sigma') measures the spread of data around the mean. A small σ produces a tall, narrow bell; a large σ creates a wider, flatter bell. It controls how 'spread out' the data is.

Z-Score

A z-score tells you how many standard deviations a data point is from the mean. It standardises any normal distribution so that you can compare values from different contexts (e.g., exam scores in different subjects).

Symmetry & the 68-95-99.7 Rule

The normal curve is perfectly symmetrical about μ. Approximately 68% of data falls within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ. This empirical rule provides quick mental estimates.

Notation: X ~ N(μ, σ²)

In IB notation, we write X ~ N(μ, σ²) to mean 'the random variable X follows a normal distribution with mean μ and variance σ².' Your GDC uses μ and σ (not σ²), so keep this distinction in mind.

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KEY TAKEAWAY

Think of the normal distribution like a target at an archery range. Most arrows land near the bullseye (the mean), and fewer arrows land as you move further from centre. The standard deviation is like the skill level of the archer — a smaller σ means arrows are tightly grouped (skilled archer), while a larger σ means they are scattered widely (beginner). The z-score simply counts how many 'rings' away from the bullseye a particular arrow landed.

SECTION 3

The Bell Curve Visualised

The diagram below shows a standard normal distribution (μ = 0, σ = 1) with the empirical rule regions shaded. Study how the percentages relate to the number of standard deviations from the mean.

The standard normal distribution with μ = 0 and σ = 1. The cyan band covers ±1σ (68.2%), the violet band covers ±2σ (95.4%), and the pink band covers ±3σ (99.7%). Notice the perfect symmetry about the mean.

Notice several important features. First, the curve never actually touches the horizontal axis — it extends infinitely in both directions, getting ever closer to zero but never reaching it. In practice, almost all data (99.7%) falls within three standard deviations. Second, the peak of the curve occurs exactly at the mean. Third, the curve is perfectly symmetrical, so the probability of being a certain distance above the mean is identical to the probability of being that same distance below the mean.

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GDC Tip

On a TI-84, the normal distribution functions are found under 2nd → DISTR. On a TI-Nspire or Casio, look in the Statistics/Distributions menu. You will primarily use normalcdf (to find probabilities for a range) and invNorm (to find a value given a probability).

SECTION 4

Mathematical Framework

The IB syllabus does not require you to memorise the full probability density function, but understanding the z-score formula is essential. Every probability calculation you perform begins with converting raw data values into z-scores, or using your GDC to handle that conversion automatically.

Z-SCORE FORMULA

z = (x − μ) / σ

Where x = the data value, μ = the population mean, and σ = the population standard deviation. A positive z-score means the value is above the mean; a negative z-score means it is below.

PROBABILITY — LESS THAN A VALUE

P(X < x) = normalcdf(−∞, x, μ, σ)

On your GDC, use a very large negative number (like −1 × 10⁹) for −∞. This gives the probability that X takes a value less than x. Equivalently, P(Z < z) on the standard normal.

PROBABILITY — BETWEEN TWO VALUES

P(a < X < b) = normalcdf(a, b, μ, σ)

This calculates the area under the curve between a and b. Remember: probability equals area under the normal curve.

INVERSE NORMAL

x = invNorm(p, μ, σ)

Given a probability p (area to the left), this returns the value x such that P(X < x) = p. Use this when a problem gives you a percentage and asks for the boundary value.

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IB Exam Note

On IB exams, you are expected to use your GDC for normal distribution calculations. Always write down the distribution, the probability statement (e.g., P(X > 75)), the GDC command you used, and the answer. Showing this working is how you earn method marks even if your final answer has a rounding error.

SECTION 5

Understanding Z-Scores & Probability Regions

The z-score is the bridge between any normal distribution and the standard normal distribution. When you calculate a z-score, you are effectively asking: 'If I rescale this distribution so that the mean is 0 and the standard deviation is 1, where does my data point land?' This standardisation process lets you compare values from completely different contexts, like comparing your maths score to your science score even when the exams have different means and spreads.

Left: A normal distribution with μ = 70, σ = 10 and a raw score of 80 highlighted in red. Right: The same score mapped onto the standard normal distribution as z = 1. The reference table below shows common z-scores and their cumulative probabilities.

The diagram above illustrates the standardisation process. A score of 80 on an exam with mean 70 and standard deviation 10 corresponds to a z-score of exactly 1.0. This means the score is one standard deviation above the mean. Using our GDC or the reference table, we can immediately state that approximately 84.13% of students scored below 80 on this exam, because P(Z < 1) ≈ 0.8413.

Z-scores also allow powerful comparisons. Suppose you scored 80 on Maths (μ = 70, σ = 10) and 85 on English (μ = 78, σ = 5). Your Maths z-score is (80 − 70)/10 = 1.0, while your English z-score is (85 − 78)/5 = 1.4. Even though the raw scores look similar, your English performance was relatively stronger because you were 1.4 standard deviations above the class average versus only 1.0 in Maths.

SECTION 6

Worked Example — Exam Score Probability

Let's work through a complete IB-style problem from start to finish. Pay attention to how we state the distribution, write the probability statement, and use the GDC.

Finding Probabilities and Inverse Normal Values

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Step 1 — Read the Problem

The heights of adult women in a certain country are normally distributed with a mean of 162 cm and a standard deviation of 7.5 cm. Let X represent the height of a randomly selected woman.
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Step 2 — State the Distribution

Write the distribution using correct notation:

X ~ N(162, 7.5²)
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Step 3 — Part (a): Find P(X > 170)

We want the probability that a woman is taller than 170 cm. First, note that P(X > 170) = 1 − P(X < 170). On your GDC, enter normalcdf(170, 1 × 10⁹, 162, 7.5), or equivalently 1 − normalcdf(−1 × 10⁹, 170, 162, 7.5). The z-score is z = (170 − 162) / 7.5 = 1.067.

P(X > 170) ≈ 0.143 (14.3%)
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Step 4 — Part (b): Find P(155 < X < 175)

We want the probability that a woman's height is between 155 cm and 175 cm. On the GDC, enter normalcdf(155, 175, 162, 7.5). This calculates the area under the curve between these two bounds.

P(155 < X < 175) ≈ 0.783 (78.3%)
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Step 5 — Part (c): Find the height exceeded by only 10% of women

If 10% of women are taller, then 90% are shorter. We need the value x such that P(X < x) = 0.90. Use invNorm(0.90, 162, 7.5) on your GDC. This is an inverse normal problem — given a probability, find the data value.

x ≈ 171.6 cm
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Step 6 — Verify and Communicate

Always check your answer makes sense. The value 171.6 cm is above the mean of 162 cm, which is logical since only the top 10% exceed it. On the IB exam, write your final answer with appropriate units and rounding (3 significant figures unless stated otherwise).

SECTION 7

Strengths & Limitations of the Normal Model

The normal distribution is incredibly useful, but it is not a universal model for all data. Understanding when it works well and when it fails is an important part of statistical literacy. The table below summarises the key strengths and limitations you should keep in mind.

Aspect	Strengths	Limitations
Applicability	Models many natural phenomena (heights, test scores, measurement errors) remarkably well.	Fails for skewed data (income, house prices), discrete counts, or bounded data (percentages, ages).
Simplicity	Only two parameters (μ and σ) completely define the distribution. Easy to communicate and compute.	Real data may have multiple peaks (bimodal), heavy tails, or asymmetry that two parameters cannot capture.
Technology	GDCs and software compute exact probabilities instantly, eliminating the need for printed z-tables.	Over-reliance on technology without understanding can lead to misapplication (e.g., using normalcdf on non-normal data).
Tail behaviour	The 68-95-99.7 rule gives quick estimates without any calculation.	Tails extend to ±∞, which is physically impossible for many variables (e.g., negative heights). Extreme values may be unrealistic.

Key strengths and limitations of the normal distribution model

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KEY TAKEAWAY

The normal distribution is like a map — incredibly useful for navigating, but not the territory itself. A map of London is great for finding streets in London, but useless if you are actually in Tokyo. Similarly, the normal model works beautifully when data is roughly symmetric and bell-shaped, but applying it blindly to skewed or bounded data is like using the wrong map. Always check whether the normal model is appropriate before using it.

SECTION 8

Connection to Advanced Topics

The normal distribution you are learning in SL 4.7 is a foundation for many more advanced ideas in statistics. Even within the IB Applications and Interpretation course, you will see it appear again in hypothesis testing, confidence intervals, and regression. Understanding where this topic leads can help you appreciate why it is worth mastering now.

SL 4.7 (What You Learn Now)	Advanced Extension
Calculate P(X < x) using normalcdf	In hypothesis testing, you compare a test statistic to a normal distribution to decide whether to reject a null hypothesis.
Use invNorm to find boundary values	Confidence intervals use invNorm to find the critical value (e.g., z = 1.96 for 95% confidence).
Z-scores standardise individual values	The Central Limit Theorem (HL) shows that sample means follow a normal distribution even when the population is not normal.
Single normal distribution with given μ and σ	In regression, the residuals (errors) are assumed to be normally distributed, which validates the model's predictions.

How SL 4.7 concepts extend into more advanced statistics

One of the most remarkable results in all of mathematics is the Central Limit Theorem. It states that if you take the average of a sufficiently large random sample from any distribution, that average will be approximately normally distributed — regardless of the shape of the original data. This explains why the normal distribution appears so often in practice: many real-world measurements are themselves averages or sums of many small random effects. While you will not be tested on the formal theorem at SL level, knowing it exists helps explain why the bell curve is everywhere.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A dataset has mean μ = 50 and standard deviation σ = 8. A data point has a z-score of −1.5. Explain in words what this z-score tells you about where this data point sits relative to the rest of the data.

Show answer

A z-score of −1.5 means the data point is 1.5 standard deviations below the mean. Since σ = 8, the data point is 1.5 × 8 = 12 units below the mean, so its value is 50 − 12 = 38. This tells us the data point is in the lower portion of the distribution — using the empirical rule, roughly 93.3% of data lies above this point (since P(Z > −1.5) ≈ 0.933). The negative sign indicates the value is below average, and the magnitude 1.5 tells us it is moderately far from the centre.

Question 2: BASIC CALCULATION

The masses of apples from a farm are normally distributed with mean 185 g and standard deviation 12 g. Find the probability that a randomly selected apple has a mass less than 200 g.

Show answer

Let X represent the mass of an apple, so X ~ N(185, 12²). We need P(X < 200). First, calculate the z-score: z = (200 − 185) / 12 = 15/12 = 1.25. Using your GDC: normalcdf(−1 × 10⁹, 200, 185, 12) ≈ 0.894. Therefore, there is approximately an 89.4% probability that a randomly selected apple has a mass less than 200 g. This makes sense since 200 g is 1.25 standard deviations above the mean, so we expect most apples to be lighter.

Question 3: INTERMEDIATE

The time students take to complete a test is normally distributed with mean 45 minutes and standard deviation 8 minutes. (a) Find the probability that a student takes between 35 and 55 minutes. (b) If 120 students take the test, how many would you expect to finish in less than 30 minutes?

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(a) Let X ~ N(45, 8²). P(35 < X < 55) = normalcdf(35, 55, 45, 8) ≈ 0.789. About 78.9% of students take between 35 and 55 minutes. (b) First find P(X < 30) = normalcdf(−1 × 10⁹, 30, 45, 8) ≈ 0.0304. The expected number of students finishing in less than 30 minutes is 120 × 0.0304 ≈ 3.65, so we would expect approximately 4 students. Note that the z-score for 30 minutes is (30 − 45)/8 = −1.875, confirming this is nearly 2 standard deviations below the mean, so very few students finish this quickly.

Question 4: APPLIED

A battery manufacturer claims that the lifetime of their batteries is normally distributed with mean 500 hours and standard deviation 40 hours. They want to offer a warranty: any battery that lasts fewer than k hours gets a free replacement. If they want to replace no more than 5% of batteries, find the value of k.

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Let X ~ N(500, 40²). We need to find k such that P(X < k) = 0.05. This is an inverse normal problem. Using the GDC: k = invNorm(0.05, 500, 40) ≈ 434.2 hours. The manufacturer should set the warranty threshold at approximately 434 hours. To verify: the z-score corresponding to the 5th percentile is invNorm(0.05, 0, 1) ≈ −1.645. Then k = 500 + (−1.645)(40) = 500 − 65.8 = 434.2 hours, confirming our answer. Batteries lasting fewer than 434 hours represent the bottom 5% of the distribution.

Question 5: CRITICAL THINKING

Two students are comparing their performance. Anika scored 78 on a chemistry test where the class mean was 65 and the standard deviation was 10. Ben scored 82 on a biology test where the class mean was 70 and the standard deviation was 8. (a) Who performed better relative to their class? Justify with z-scores. (b) If both tests are normally distributed, what percentage of students did each outperform?

Show answer

(a) Anika's z-score: z = (78 − 65)/10 = 1.30. Ben's z-score: z = (82 − 70)/8 = 1.50. Ben performed better relative to his class because his z-score is higher — he was 1.5 standard deviations above his class mean compared to Anika's 1.3 standard deviations. (b) For Anika: P(Z < 1.30) = normalcdf(−1 × 10⁹, 1.30, 0, 1) ≈ 0.903, so she outperformed about 90.3% of her class. For Ben: P(Z < 1.50) = normalcdf(−1 × 10⁹, 1.50, 0, 1) ≈ 0.933, so he outperformed about 93.3% of his class. This problem illustrates the power of z-scores: they allow fair comparison across different distributions by standardising both scores to the same scale.

SUMMARY

Lesson Summary

Summary

The normal distribution is a continuous, symmetric, bell-shaped probability distribution defined by two parameters: the mean (μ), which locates the centre, and the standard deviation (σ), which controls the spread. The 68-95-99.7 empirical rule tells us that approximately 68%, 95%, and 99.7% of data falls within 1, 2, and 3 standard deviations of the mean respectively. We write X ~ N(μ, σ²) to denote a normally distributed random variable.

A z-score standardises any value using z = (x − μ) / σ, telling us how many standard deviations the value is from the mean. For probability calculations, use normalcdf(lower, upper, μ, σ) on your GDC to find the probability of a range. For inverse normal problems — where you are given a probability and need to find the boundary value — use invNorm(p, μ, σ). Always state the distribution, write the probability statement, record the GDC command, and give your answer to 3 significant figures.