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Module 5 · L13 of 15 ~40 min ⚡ +95 XP available

Normal Distribution Applications

You know the bell curve exists — now you'll use it. What percentage of bottles are underfilled? What mark lands a student in the top 8%? In this lesson you master two fundamental operations: finding probabilities given values (normal CDF) and finding values given probabilities (inverse normal). These skills appear in every field that touches statistics.

Today's hook — A machine fills bottles with mean 500 mL and standard deviation 8 mL. The company needs to know what percentage are underfilled (below 490 mL). Using the empirical rule, would you expect this to be closer to 10%, 1%, or 0.1%? Commit to a number before you start.

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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

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A machine fills bottles with $\mu = 500$ mL and $\sigma = 8$ mL. The company wants to know what percentage of bottles are underfilled (less than 490 mL). Without calculating — using the empirical rule, do you expect this percentage to be closer to $10\%$, $1\%$, or $0.1\%$? Explain your reasoning.

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The two operations

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Every normal distribution question is one of two types — and only two. Lock this into muscle memory before touching any numbers.

Forward: value → probability

Given $x$, find $P(X < x)$. Standardise with $z = \frac{x-\mu}{\sigma}$, then use tables or calculator.

Inverse: probability → value

Given $P(X < x)$, find $x$. Look up $z$ first, then un-standardise: $x = \mu + z\sigma$.

Key z-values to memorise

$z_{90\%} \approx 1.28$ · $z_{95\%} \approx 1.645$ · $z_{97.5\%} \approx 1.96$ · $z_{99\%} \approx 2.33$

The fundamental identity that makes everything work: standardising converts any $X \sim N(\mu, \sigma^2)$ into $Z \sim N(0, 1)$. Every table and calculator in existence works with the standard normal.

$$z = \dfrac{x - \mu}{\sigma}$$

What you'll master

Know

Key facts

Standardise before using tables: $z = (x - \mu)/\sigma$
$P(X > x) = 1 - P(X < x)$ (complement rule)
Inverse normal: find $z$ first, then $x = \mu + z\sigma$

Understand

Concepts

Why we standardise — every normal curve has the same shape
The difference between "find the probability" and "find the cut-off value"
How quality control uses both forward and inverse operations

Can do

Skills

Calculate $P(X < x)$, $P(X > x)$, and $P(a < X < b)$
Find the value $x$ corresponding to a given probability
Solve contextual problems in manufacturing, assessment, and science

Key terms

Normal CDFCumulative distribution function — gives the area (probability) to the left of a value.

StandardisationConverting $x$ to $z = (x - \mu)/\sigma$ to use standard normal tables.

z-scoreThe number of standard deviations a value is from the mean.

Inverse normalFinding the value $x$ given a probability — working backwards from the table.

Left-tail$P(X < x)$ — cumulative probability; what tables give directly.

Right-tail$P(X > x) = 1 - P(X < x)$ — use the complement rule.

Using the Normal CDF — finding probabilities

core concept

To find a probability for any normal distribution, the strategy is always the same four steps:

Identify $\mu$ and $\sigma$ from the problem
Standardise the boundary value(s) using $z = \frac{x - \mu}{\sigma}$
Find the corresponding probability from a z-table or calculator
Interpret the result in context

Three probability types:

The three fundamental normal probability types — all rely on the standard CDF.

Example: Exam scores are $N(72, 12^2)$. Find $P(X > 84)$.

$z = \frac{84 - 72}{12} = 1$. So $P(X > 84) = P(Z > 1) = 1 - P(Z < 1) \approx 1 - 0.8413 = 0.1587$. Approximately $15.9\%$ of students scored above 84.

Quality control context. A bolt must be between $49.5$ mm and $50.5$ mm to pass. If lengths are $N(50, 0.3^2)$: $z_{49.5} = -1.67$, $z_{50.5} = 1.67$, $P(-1.67 < Z < 1.67) \approx 0.905$. About $90.5\%$ pass inspection.

Left tail: $P(X < x) = P(Z < z)$ where $z = \frac{x-\mu}{\sigma}$; Right tail: $P(X > x) = 1 - P(Z < z)$ — always use complement

Pause — copy the three normal CDF cases: left tail $P(X < x) = P(Z < z)$; right tail $P(X > x) = 1 - P(Z < z)$; middle $P(a < X < b) = P(Z < z_b) - P(Z < z_a)$ — where $z = (x-\mu)/\sigma$ — into your book.

Did you get this? True or false: to find $P(X > 84)$ when $P(Z < 1) = 0.8413$, the answer is $0.8413$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · CHOCOLATE BARS — FORWARD

Chocolate bar weights are $N(250, 5^2)$ g. Find $P(X < 242)$.

$z = \dfrac{242 - 250}{5} = -1.6$

Standardise the boundary value.

PROBLEM 2 · CHOCOLATE BARS — INVERSE

Same distribution $N(250, 5^2)$. Find the weight exceeded by $90\%$ of bars.

$P(X > x) = 0.90 \Rightarrow P(X < x) = 0.10$

Translate: "exceeded by 90%" means only 10% are above — so 10% are below.

PROBLEM 3 · COMMUTE TIMES — TWO-PART

Commute times are $N(35, 8^2)$ minutes. (a) What percentage take more than 45 min? (b) Find the 90th percentile.

$z = \dfrac{45 - 35}{8} = 1.25$; $P(X > 45) = 1 - P(Z < 1.25) \approx 1 - 0.8944 = 0.1056$

About $10.6\%$ take more than 45 minutes.

Quick check: IQ scores are $N(100, 15^2)$. What IQ score is at the 90th percentile?

Common errors · the 3 traps that cost marks

Trap 01

Reading the wrong tail

$P(Z < 1) = 0.8413$ is the LEFT tail. For $P(Z > 1)$ you need $1 - 0.8413 = 0.1587$. Always check whether your table gives cumulative (left-tail) or upper-tail probabilities.

Trap 02

"Top 10%" vs. "bottom 10%"

"Top 10%" means $P(X > x) = 0.10$, so the cumulative left area is $0.90$ — look up $z = 1.282$, NOT $z = -1.282$. Draw a sketch every time.

Trap 03

Forgetting to un-standardise

After finding $z$ in an inverse problem, students forget the final step: $x = \mu + z\sigma$. A bare z-score is not a weight, time, or mark — it has no units in context.

Fill in the blank: To find $P(X > x)$ using a left-tail table, you compute $1$ minus ___.

Quick-fire practice · 5 calculations

For $X \sim N(100, 12^2)$, find $P(X < 112)$

For $X \sim N(100, 12^2)$, find $P(X > 88)$

Fish lengths are $N(35, 2^2)$ cm. What $\%$ are longer than 38 cm?

Exam marks $N(68, 10^2)$. Top $15\%$ get Distinction. Find minimum mark.

Components $N(50, 0.4^2)$ mm. Rejected if outside 49.2 to 50.8 mm. What $\%$ rejected?

Odd one out: Three of these are inverse normal problems and one is a forward (CDF) problem. Which is the odd one out?

Revisit your thinking

$490 = 500 - 10 = 500 - 1.25 \times 8 = \mu - 1.25\sigma$. This is between $1\sigma$ and $2\sigma$ below the mean. The left tail beyond $1\sigma$ is about $16\%$, and beyond $2\sigma$ is about $2.5\%$. So the answer lies between $2.5\%$ and $16\%$ — closer to $10\%$ than to $1\%$ or $0.1\%$. The exact value (from tables) is approximately $\mathbf{10.6\%}$. The empirical rule gives a powerful first estimate before touching a calculator.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next.

Short answer

ApplyBand 43 marks

Q1. The diameters of ball bearings are $N(12.00, 0.05^2)$ mm. Bearings are acceptable if diameter is between $11.90$ mm and $12.10$ mm. (a) Find the z-scores for the lower and upper acceptance limits. (b) Find the percentage of acceptable bearings. (c) Find the percentage that are too large. (3 marks)

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ApplyBand 53 marks

Q2. Commute times are $N(35, 8^2)$ minutes. (a) What percentage take more than 45 minutes? (b) Find the 90th percentile commute time. (c) If 500 employees qualify for a work-from-home policy (commute exceeds the 90th percentile), how many employees does the company have in total? (3 marks)

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AnalyseBand 63 marks

Q3. A cereal manufacturer states boxes contain "at least 375 g." The filling machine is normally distributed. The company wants fewer than $0.5\%$ of boxes underweight. (a) If $\mu = 380$ g, find the maximum allowable $\sigma$ to 1 decimal place. (b) A proposal suggests $\mu = 385$ g, $\sigma = 4$ g. Calculate the percentage of underweight boxes. (c) Manufacturing cost increases $3\%$ per extra gram of mean fill. Evaluate which setting is more cost-effective while meeting the underweight requirement. (3 marks)

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Comprehensive answers (click to reveal)

Drill 1: $z=1$, $P \approx 0.8413$. 2: $z=-1$, $P(Z>-1)\approx 0.8413$. 3: $z=1.5$, $\approx 6.7\%$. 4: $z\approx1.036$, $x\approx78.4$. 5: $z=\pm2$, rejected $\approx4.6\%$.

Q1 (3 marks): (a) $z_{\text{lower}}=-2$, $z_{\text{upper}}=2$ [1]. (b) $P(-2<Z<2)\approx0.9545$, so $\approx95.5\%$ acceptable [1]. (c) $P(Z>2)\approx0.0228$, so $\approx2.3\%$ too large [1].

Q2 (3 marks): (a) $z=1.25$, $P(X>45)\approx10.6\%$ [1]. (b) $z\approx1.282$, $x=35+1.282(8)=45.26$ min [1]. (c) $10\%=500$ employees, total $=5000$ [1].

Q3 (3 marks): (a) $z=-2.576$, $\sigma=5/2.576\approx1.9$ g [1]. (b) $z=(375-385)/4=-2.5$, $P\approx0.62\%$ — exceeds 0.5% so FAILS the requirement [1]. (c) Setting (a) at $\mu=380$ is cheaper ($\$380$ base) and meets the requirement; setting (b) at $\mu=385$ costs more AND fails — setting (a) wins [1].

Boss battle · The Quality Controller

earn bronze · silver · gold

Five timed questions on normal CDF and inverse normal. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

Enter the arena

Science Jump · platform challenge

Climb platforms by answering normal distribution questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 12 · The Normal Distribution Lesson 14 · The Binomial Distribution →

Module overview · Maths Advanced · Checkpoint 3