The Normal Distribution Advanced. This lesson goes beyond core Algebra II. You can skip it.

Learning goals

Read probability as area under a density curve
Note that a single value has probability zero
Describe the standard curve, symmetric and fast-decaying
Apply the 68-95-99.7 rule as rounded percentages
Convert with $z = \tfrac{x - \mu}{\sigma}$ to use one table
Walk back to $x = \mu + z\sigma$ from a percentage

A shape that keeps turning up

Start with something you already built. In the Binomial Theorem lesson you learned that flipping a coin $n$ times produces $\binom{n}{k}$ different ways to get exactly $k$ heads. You also learned that those counts are the rows of Pascal’s triangle. Take $n = 10$ . There are $2^{10} = 1024$ equally likely sequences of heads and tails, so the probability of exactly $k$ heads is

P(k \text{ heads}) = \frac{\binom{10}{k}}{1024}.

Now draw those eleven probabilities as bars. The binomial coefficients $1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1$ are large in the middle and tiny at the ends. The plain reason is that there are many ways to get five heads and only one way to get ten. The bars pile into a hump.

The eleven probabilities for the number of heads in 10 coin flips, drawn as bars, with a bell curve laid over them. The bars are just row 10 of Pascal's triangle divided by 1024. Nobody put the curve there; it is the shape the binomial coefficients already have.

Coin flips are not special. What they are is a sum of many small independent effects: ten little pushes, each one either up or down, added together. Human height is the same kind of thing, a total of many genetic and nutritional nudges. A measurement error is a total of many tiny disturbances in the instrument and the air. A Galton board, a peg-covered board down which a ball bounces left or right at each row, is that idea made physical. The balls really do pile up in a bell at the bottom, because the number of paths to each slot is a binomial coefficient.

That is the pattern: add up many independent small effects and the total tends toward this one shape, provided no single one of them dominates the total and a few other technical conditions hold that this course will not state precisely. That proviso is not a technicality, and we will come back to it. When one ingredient is allowed to run away with the sum, the bell may not form, which is part of why some famous quantities are not normal. The theorem that makes all of this precise is the Central Limit Theorem. It is one of the deepest results in mathematics, it is the reason the bell turns up in so many unrelated places, and proving it is far beyond this course. What this lesson can do honestly is describe the curve and work with it.

Probability as area, and one thing we cannot do

Coin flips are discrete: there are finitely many outcomes, each with its own probability, and the probabilities add to $1$ . Height is not like that. A person can be $170$ cm tall, or $170.4$ cm, or $170.4013$ cm. There are infinitely many possible values and no way to hand each one a probability that still totals $1$ .

So the whole set-up has to change. For a continuous quantity we do not attach probability to points. We draw a curve, called a density curve, and we agree on one rule. Let $X$ name the numerical result of the experiment, the height of a randomly chosen adult, say. Then a statement like $X < 5$ is itself an event in the sense of the previous lesson: the set of every outcome whose measured value is below $5$ , so $P(X < 5)$ is just that event’s probability, written in this new notation:

Probability is area under the curve. The curve never dips below the horizontal axis, the area between $a$ and $b$ is $P(a < X < b)$ , and the total area under the whole curve is $1$ .

Everything follows from that rule, including a genuine surprise. Ask for the probability that an adult is exactly $170$ cm tall, to infinite precision. The region above the single point $170$ has width zero, so it has no area at all:

P(X = c) = 0 \quad \text{for every single value } c.

That is not a rounding convention. For a continuous quantity, every individual value has probability zero, and probability lives only in intervals. A first reaction is that this makes the model absurd (surely somebody is exactly $170$ cm). It does not. Probability zero does not mean impossible; it means that one single point gets no share of the curve’s area, because area needs width and a point has none. In practice this rarely matters anyway, since “exactly $170$ cm” would mean matching infinitely many decimal places, which no measurement ever pins down. Real questions are always intervals: between $169.5$ and $170.5$ , above $180$ , below $150$ .

This has a consequence you should test yourself with. In the discrete probability you met earlier in this chapter, $P(X < 5)$ and $P(X \le 5)$ differ whenever $5$ is a possible outcome, because the second one also counts it. For a continuous quantity, the outcome $X = 5$ contributes zero, so

P(X < c) = P(X \le c).

The endpoint simply does not matter. Include it or exclude it, the area is the same.

Check your understanding

$X$ is a continuous quantity with a density curve. Which statement is true?

P(X < 5) = P(X \le 5)

, because

P(X = 5) = 0

P(X \le 5) > P(X < 5)

, because it also counts the outcome

X = 5

P(X = 5)

equals the height of the density curve at

5

P(X < 5) > P(X \le 5)

, because a strict inequality is easier to satisfy

The curve, and the debt

The standard normal curve is the density curve given by

\varphi(z) = \frac{1}{\sqrt{2\pi}}\, e^{-z^2/2}.

You can read every part of this. The exponent $-z^2/2$ is never positive, and since $(-z)^2 = z^2$ it gives the same value at $z$ and at $-z$ , so the curve is symmetric about $0$ . It is largest when the exponent is $0$ , that is when $z = 0$ , so the peak is at the center. And $z^2$ grows fast, so $e^{-z^2/2}$ collapses fast: at $z = 1$ the height is $e^{-0.5} \approx 0.6065$ of the peak, and at $z = 3$ it is $e^{-4.5} \approx 0.0111$ of the peak, about a ninetieth. That is why the tails look empty. The constant $\frac{1}{\sqrt{2\pi}} \approx 0.3989$ out front is not decoration: it is exactly the number that makes the total area come out to $1$ . Where that number comes from is not something this course can show you.

We cannot compute these areas by hand. Finding the area under a curve is what integral calculus is for, and you have not met calculus yet. Even worse: mathematicians have proven that this curve’s area cannot be reached by the ordinary algebra formulas that calculus normally produces, no matter how much calculus you eventually learn. That is not a gap in anybody’s cleverness; it is a theorem.

What people do instead is compute the areas numerically, to as many decimal places as they like, by adding up huge numbers of very thin strips. They then write the answers down in a table. The table is not a definition and it is not a cheat. It is the record of a computation nobody can shortcut. Every number you are about to look up was earned that way, and knowing that is the difference between using the table and understanding it.

Check your understanding

Put these three heights of the standard normal curve in order, tallest to shortest: $\varphi(-1)$ , $\varphi(2)$ , $\varphi(0)$ .

\varphi(0) > \varphi(-1) > \varphi(2)

\varphi(0) > \varphi(2) > \varphi(-1)

\varphi(-1) > \varphi(0) > \varphi(2)

\varphi(2) > \varphi(-1) > \varphi(0)

The 68-95-99.7 rule

Because the areas have been computed once and for all, we can state the headline results. For the normal curve, measuring distance from the center in standard deviations:

The normal curve, split into bands one standard deviation wide, using the 68-95-99.7 rule's own rounded bookkeeping. The exactly computed bands are a little different, as the text below explains.

For any normal distribution with mean $\mu$ and standard deviation $\sigma$ :

about $68\%$ of the values lie within $1$ standard deviation of the mean, between $\mu - \sigma$ and $\mu + \sigma$ ,
about $95\%$ lie within $2$ standard deviations, between $\mu - 2\sigma$ and $\mu + 2\sigma$ ,
about $99.7\%$ lie within $3$ standard deviations, between $\mu - 3\sigma$ and $\mu + 3\sigma$ .

This is the 68-95-99.7 rule, sometimes called the empirical rule. Be clear about its status: it is not something we derived, it is the numerical answer rounded off. The computed areas are $68.27\%$ , $95.45\%$ and $99.73\%$ . The rule is a memory aid, accurate to within half a percentage point, and when a question wants more precision you go to the table instead.

Two useful facts fall straight out of symmetry. The two halves of the central band are equal, so each holds about $34\%$ . And using the rule’s own rounded bookkeeping, what is left outside the $3$ -standard-deviation band is about $0.3\%$ , split evenly between two tails, so about $0.15\%$ of values sit more than $3$ standard deviations above the mean. Rare, but not impossible.

Keep the rule’s numbers and the computed numbers apart, because they are not the same and a careful student will notice. The band figures $34$ , $13.5$ and $2.35$ are the rule’s own bookkeeping, chosen so that they add up to $68$ , $95$ and $99.7$ . The computed band areas are $34.13\%$ , then $13.59\%$ , then $2.14\%$ , with $0.13\%$ left in each tail beyond $3$ standard deviations. The two agree to within a fraction of a percentage point, which is all the rule ever promised. So when a question says to use the 68-95-99.7 rule, use the rule’s numbers. When it says to use the table, use the table, and do not expect the last decimal place to match.

Worked example 1 Resting heart rates with the 68-95-99.7 rule

Adult resting heart rates are roughly normal with mean $\mu = 72$ beats per minute and standard deviation $\sigma = 8$ . What fraction of adults have a resting heart rate between $64$ and $88$ beats per minute?

First locate the two endpoints in standard deviations from the mean. Since $64 = 72 - 8$ , the value $64$ sits exactly $1$ standard deviation below the mean. Since $88 = 72 + 16 = 72 + 2(8)$ , the value $88$ sits exactly $2$ standard deviations above it. So the question asks for the area from $-1$ to $+2$ in standard-deviation units.

Read that off the band picture. From $-1$ to $0$ is $34\%$ , from $0$ to $+1$ is another $34\%$ , and from $+1$ to $+2$ is $13.5\%$ :

34\% + 34\% + 13.5\% = 81.5\%.

So about $81.5\%$ of adults fall in that range. The table, which we build next, gives $81.85\%$ for the same question. The gap of about a third of a percentage point is exactly the price of using a rounded rule. That gap is usually a price worth paying for an answer you can get in your head.

Check your understanding

For a normal distribution, about what percent of values lie between $\mu$ and $\mu + 2\sigma$ ?

47.5\%

68\%

95\%

34\%

Try it yourself. In the figure below, drag the two marks to one standard deviation each side of the mean and read the percentage: about 68. Now slide the mean without touching anything else. The marks stay where you put them, so they are no longer one spread each side, and the percentage falls away. Walk them back out to one spread each side of the new mean and it reads 68 again. That is the point: the answer depends only on how many standard deviations out the marks are, never on where the mean sits or how wide the curve is, which is why a single table can serve every normal distribution. Then slide the spread and watch the curve flatten as it widens, because the area underneath is always the whole of it.

How far out, in standard deviations?

Mean Spread Lower mark Upper mark

Mean 100, standard deviation 15. Between 85 and 115 lies about 68% of the area. That is one standard deviation each side of the mean, wherever the mean is.

A bell curve over a scale running from 40 to 160, with a dashed line at its mean and two upright marks enclosing a shaded band beneath the curve. The mean and the spread are set by sliders and the two marks can be dragged along the scale or stepped one unit at a time; the sentence below reports the share of the area the band encloses and how far the marks sit from the mean in standard deviations.

The z-score, and why one table is enough

There are infinitely many normal distributions, one for every choice of $\mu$ and $\sigma$ . Heights in centimeters, test scores out of $800$ , loaf weights in grams: all different curves. It would be hopeless to tabulate them all. We do not have to, and the reason is a piece of chapter 2.

Define the z-score of a value $x$ :

z = \frac{x - \mu}{\sigma}.

Read it as a unit conversion. The numerator $x - \mu$ measures how far $x$ is from the mean, in the original units (centimeters, points, grams). Dividing by $\sigma$ re-expresses that distance in standard deviations. A z-score of $1.5$ means “one and a half standard deviations above the mean”, whatever the original quantity was, and that sentence carries no units at all. That is the whole trick: z-scores are the common currency in which every normal distribution can be compared. Applying that same conversion to the whole quantity, not just one value, gives it a name: $Z = \frac{X - \mu}{\sigma}$ is the standardized version of $X$ , the same random result relabeled in standard deviations, so a statement like $P(a < Z < b)$ is just $P(\mu + a\sigma < X < \mu + b\sigma)$ written in the new units.

One curve, two rulers. A test score of 600 and a z-score of 1 mark the same place on the same curve, named in two different units. Standardizing does not change the picture; it changes the labels on the axis.

Now here is why the trick is allowed, which is a stronger claim than saying it is convenient. A normal distribution with mean $\mu$ and standard deviation $\sigma$ has the density curve

f(x) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-(x-\mu)^2/(2\sigma^2)},

and the claim is that this is not a new curve at all. It is $\varphi$ , shifted and rescaled, exactly the graph transformations you studied in chapter 2: shifted right by $\mu$ , and stretched horizontally, with its height correspondingly shrunk, by a factor of $\sigma$ .

Here is why that is enough to justify one shared table. Stretch a shape sideways by a factor of $\sigma$ while shrinking its height by that same factor, and its area does not change: a strip that gets $\sigma$ times wider and $\sigma$ times shorter has the same width times height as before. So shifting and rescaling $\varphi$ into $f$ leaves matching pieces of area exactly equal,

P(a < Z < b) = P(\mu + a\sigma < X < \mu + b\sigma),

which is the whole reason a single table can serve every normal distribution. Up to a shift and a rescale, there is only one normal curve.

Check your understanding

Jars of peanut butter are filled to a mean of $\mu = 340$ grams with standard deviation $\sigma = 4$ grams. One jar weighs $349$ grams. What is its z-score?

2.25

0.44

-2.25

9

Reading the standard normal table

The table records one thing: the area to the left of a given z-score. Write it

\Phi(z) = P(Z < z) = \text{the area under the standard curve to the left of } z.

Because a single point has no area, $P(Z < z)$ and $P(Z \le z)$ are the same number, so you never have to worry about which inequality the table means.

The table entry for z = 1 is the shaded area, 0.8413. It is the probability of landing anywhere to the left of 1 standard deviation above the mean, so about 84% of the distribution sits below that line.

Here is the table for $z$ from $0$ to $3$ . Two columns of pairs, to keep it compact.

$z$	$\Phi(z)$	$z$	$\Phi(z)$
$0.0$	$0.5000$	$1.6$	$0.9452$
$0.1$	$0.5398$	$1.7$	$0.9554$
$0.2$	$0.5793$	$1.8$	$0.9641$
$0.3$	$0.6179$	$1.9$	$0.9713$
$0.4$	$0.6554$	$2.0$	$0.9772$
$0.5$	$0.6915$	$2.1$	$0.9821$
$0.6$	$0.7257$	$2.2$	$0.9861$
$0.7$	$0.7580$	$2.3$	$0.9893$
$0.8$	$0.7881$	$2.4$	$0.9918$
$0.9$	$0.8159$	$2.5$	$0.9938$
$1.0$	$0.8413$	$2.6$	$0.9953$
$1.1$	$0.8643$	$2.7$	$0.9965$
$1.2$	$0.8849$	$2.8$	$0.9974$
$1.3$	$0.9032$	$2.9$	$0.9981$
$1.4$	$0.9192$	$3.0$	$0.9987$
$1.5$	$0.9332$

The table starts at $z = 0$ and never goes negative, which looks like half a table. It is a full one, because symmetry supplies the rest.

Negative z-scores: $\Phi(-z) = 1 - \Phi(z)$ #

The standard curve is symmetric about $0$ , since $\varphi(-z) = \varphi(z)$ . Reflecting the picture in the vertical axis therefore leaves the curve unchanged. That reflection also carries the region to the left of $-z$ exactly onto the region to the right of $+z$ . Two regions that are reflections of one another have the same area, so

P(Z < -z) = P(Z > z).

The total area under the curve is $1$ , and the region to the right of $z$ is everything that is not to its left, so $P(Z > z) = 1 - \Phi(z)$ . Putting those two sentences together,

\Phi(-z) = P(Z < -z) = P(Z > z) = 1 - \Phi(z).

So a table of positive z-scores is a table of all of them. For example $\Phi(-1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$ .

Three question types cover almost everything, and each is a subtraction away from the table:

Left of $z$ : read $\Phi(z)$ directly.
Right of $z$ : the area to the right is everything not to the left, so $P(Z > z) = 1 - \Phi(z)$ .
Between $a$ and $b$ : take the area left of $b$ and remove the area left of $a$ , giving $P(a < Z < b) = \Phi(b) - \Phi(a)$ .

Check your understanding

Using the table, what is $P(Z > 1.4)$ ?

0.0808

0.9192

0.5808

0.4192

Worked example 2 Comparing two scores from different tests

A student scores $82$ on a biology test where the class mean was $\mu = 74$ with $\sigma = 5$ , and $91$ on a chemistry test where the class mean was $\mu = 85$ with $\sigma = 6$ . Both classes’ scores are roughly normal, which is what makes a table lookup valid below. The raw score $91$ is higher. Which performance was stronger relative to the class?

The raw scores cannot be compared, because they are measured on different scales. Convert each to a z-score, which is the same scale for both.

For biology,

z_{\text{bio}} = \frac{82 - 74}{5} = \frac{8}{5} = 1.6.

For chemistry,

z_{\text{chem}} = \frac{91 - 85}{6} = \frac{6}{6} = 1.0.

The biology score is $1.6$ standard deviations above its class mean, while the chemistry score is only $1.0$ above its own. So the biology performance was the stronger one, even though its raw number was nine points lower.

The table turns those into percentiles: $\Phi(1.6) = 0.9452$ and $\Phi(1.0) = 0.8413$ , so the student beat about $94.5\%$ of the biology class and about $84.1\%$ of the chemistry class. This is what z-scores buy you: a way to compare performances that were never measured in the same units.

Worked example 3 A filling machine, with the table

A machine fills bags of flour. The weights are normal with mean $\mu = 1000$ grams and standard deviation $\sigma = 10$ grams.

Part (a): what fraction of bags weigh less than $985$ grams? Standardize the cutoff first:

z = \frac{985 - 1000}{10} = \frac{-15}{10} = -1.5.

The table has no negative entries, so use symmetry: $\Phi(-1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$ . About $6.68\%$ of bags are under $985$ grams.

Part (b): what fraction weigh between $1005$ and $1030$ grams? Standardize both endpoints:

z_1 = \frac{1005 - 1000}{10} = 0.5, \qquad z_2 = \frac{1030 - 1000}{10} = 3.0.

For an area between two values, take the area left of the upper one and remove the area left of the lower one:

P(0.5 < Z < 3.0) = \Phi(3.0) - \Phi(0.5) = 0.9987 - 0.6915 = 0.3072.

So about $30.72\%$ of bags land in that range. Notice that both parts were solved on the same table, even though the flour weights are nowhere near $0$ and nowhere near a spread of $1$ . The z-score did all the translating.

Going backwards, from a percentage to a value

Every question so far handed you a value and asked for an area. The reverse question is just as common, and just as easy once you read the table in the other direction. You are given an area and you want the value that produces it.

The tool is the z-score formula, solved for $x$ . From $z = \dfrac{x - \mu}{\sigma}$ , multiply both sides by $\sigma$ and add $\mu$ :

x = \mu + z\sigma.

In words: start at the mean and walk $z$ standard deviations. The procedure is to convert the percentage into an area to the left, find the z-score whose table entry matches that area, and then walk.

Worked example 4 Cutoffs for the top of a distribution

Test scores are normal with $\mu = 500$ and $\sigma = 100$ .

Part (a): a scholarship goes to the top $6.68\%$ . What is the cutoff score? If $6.68\%$ of the area lies above the cutoff, then the area below it is

1 - 0.0668 = 0.9332.

Now search the $\Phi$ column for $0.9332$ . It appears exactly, at $z = 1.5$ . So the cutoff sits $1.5$ standard deviations above the mean, and

x = \mu + z\sigma = 500 + 1.5(100) = 650.

A score of $650$ or better makes the top $6.68\%$ .

Part (b): an interview goes to the top $10\%$ . The area below the cutoff is $1 - 0.10 = 0.90$ , and now the table does not cooperate: $\Phi(1.2) = 0.8849$ and $\Phi(1.3) = 0.9032$ , and $0.9000$ falls between them. Our table is only printed in steps of $0.1$ , so the honest answer from this table is “a little below $z = 1.3$ ”, giving a cutoff of about

x \approx 500 + 1.3(100) = 630.

A finer table, printed in steps of $0.01$ , gives $\Phi(1.28) = 0.8997$ , which is as close to $0.90$ as makes no difference, and the standard cutoff is

x = 500 + 1.28(100) = 628.

Three z-scores come up constantly and are worth recognizing: $z = 1.28$ cuts off the top $10\%$ , $z = 1.645$ cuts off the top $5\%$ , and $z = 1.96$ cuts off the top $2.5\%$ . That last one is why $95\%$ of a normal distribution lies within $1.96$ standard deviations of the mean. It is worth noticing that this $1.96$ is not the same as the $2$ in the 68-95-99.7 rule. The rule’s $95$ is the rounded area out to $2$ standard deviations, which is really $95.45\%$ . Exactly $95\%$ needs $1.96$ .

Check your understanding

Delivery times are normal with $\mu = 30$ minutes and $\sigma = 5$ minutes. The table gives $\Phi(0.5) = 0.6915$ . What delivery time is the $69.15$ th percentile?

32.5

27.5

0.5

35

Common mistakes

Treating the height of the curve as a probability. The curve’s height at $z = 0$ is $\varphi(0) \approx 0.3989$ , and that is not the probability of anything. Only an area is a probability. This is the one idea that makes continuous distributions different, and it is why $P(X = c) = 0$ for every $c$ .
Fussing over $<$ versus $\le$ . For a continuous distribution they give the same answer, since the endpoint contributes zero area. $P(Z < 1.2)$ and $P(Z \le 1.2)$ are both $0.8849$ . (For the discrete probabilities earlier in this chapter the two differ whenever the endpoint is a possible outcome, so do not carry this habit backwards.)
Reading the table as the area to the right. $\Phi(z)$ is the area to the left. For a right-hand tail you must subtract: $P(Z > 1.4) = 1 - 0.9192 = 0.0808$ , not $0.9192$ .
Sign errors in the z-score. A value below the mean has a negative z-score. If $x = 985$ with $\mu = 1000$ and $\sigma = 10$ , then $z = -1.5$ , not $1.5$ . Then handle it with $\Phi(-z) = 1 - \Phi(z)$ rather than by looking up $1.5$ and hoping.
Dividing by the variance instead of the standard deviation. The denominator of the z-score is $\sigma$ , in the same units as $x$ . Dividing by $\sigma^2$ leaves you with something that is not a count of standard deviations.
Assuming everything is normal. It is not. Incomes, city populations, insurance claims and stock returns all have far heavier tails than the normal curve, so a “once in ten thousand years” event computed from a bell curve can turn up twice in a decade. Quantities like these are often driven by one huge term rather than many small balanced ones (a single blockbuster year, a single market crash), which is exactly the situation the Central Limit Theorem’s no single term dominating condition rules out. The bell is a model, and using it requires a reason to believe it fits.

Key takeaways

For a continuous quantity, probability is area under a density curve, the total area is $1$ , and a single value has probability $0$ . Because of that, $P(X < c) = P(X \le c)$ .
The standard normal curve is $\varphi(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}$ : symmetric about $0$ , peaking there, and dropping off fast because of the $-z^2$ in the exponent.
Its areas cannot be written as an algebra formula, a proven fact, so the values are computed numerically and recorded in a table. The table is a record of a computation, not a definition.
The 68-95-99.7 rule: about $68\%$ , $95\%$ and $99.7\%$ of the area lies within $1$ , $2$ and $3$ standard deviations of the mean. These are rounded from $68.27\%$ , $95.45\%$ and $99.73\%$ .
The z-score $z = \frac{x - \mu}{\sigma}$ converts a value into “standard deviations from the mean”. It is a shift and a rescale, so every normal curve is the standard curve in different units, and one table serves them all.
To read the table: $\Phi(z) = P(Z < z)$ is the area to the left; $P(Z > z) = 1 - \Phi(z)$ ; $P(a < Z < b) = \Phi(b) - \Phi(a)$ ; and $\Phi(-z) = 1 - \Phi(z)$ .
To go backwards from a percentage to a value, find the z-score with the required area to its left, then walk $x = \mu + z\sigma$ .
Sums of many small independent effects tend toward this shape provided no single effect dominates the total and a few other technical conditions this course does not state precisely also hold, which is the Central Limit Theorem. That proviso is also why the curve is not universal: some real quantities are driven by one huge term rather than many small ones, and assuming a bell without a reason is a mistake.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

The standard normal distribution (free textbook) (OpenStax)
Using the normal distribution (free textbook) (OpenStax)
Normal distribution (Math is Fun)
Standard normal distribution table (careful, it gives the area from 0 to z, so add 0.5 to get the area-to-the-left values this lesson uses) (Math is Fun)
The quincunx (a working Galton board) (Math is Fun)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why no elementary formula gives this area

The lesson states the result: the area under $\varphi(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}$ cannot be written down with an algebra formula, no matter how much calculus you know. Here is that claim, made precise.

The area function of $\varphi$ would be a formula $F(z)$ that hands you the area from $0$ out to any $z$ , in one step, with no numerical add-up required. Calculus calls such a formula an antiderivative. For most functions built from powers, roots, exponentials, logarithms and trigonometric functions, an antiderivative built from those same tools, an elementary antiderivative, exists. For $e^{-z^2/2}$ it provably does not.

Get the direction of that claim right, because it is easy to state backwards. The difficulty is not in writing $e^{-z^2/2}$ down, which takes one line. It is in writing down its area function using only powers, roots, exponentials, logarithms and trigonometric functions, in any combination. A theorem from the 1830s, due to Joseph Liouville, gives a test that any candidate area function built from those tools would have to pass, and the area function of $e^{-z^2/2}$ fails it. This is not a gap in anybody’s cleverness; it is a proven fact about which functions have an elementary antiderivative and which do not. (Mathematicians do still write this particular area function down, using a special function built for exactly this purpose, the error function. That sidesteps the theorem rather than contradicting it: the error function is not built from the ordinary toolbox the theorem rules out.)

That is exactly why the table exists. The area is computed numerically instead, by adding up huge numbers of very thin strips, a method that works whether or not an elementary antiderivative exists, and the results are recorded once so nobody has to repeat the computation.

Every normal curve is the standard curve in different units

The lesson states the result: $f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2/(2\sigma^2)}$ is $\varphi$ shifted and rescaled, and shifting and rescaling preserve area, which is why one table can serve every normal distribution. Here is that argument in full.

Every normal curve is the standard curve in different units#

Start from the general density and substitute the z-score. If $z = \dfrac{x-\mu}{\sigma}$ , then squaring gives $z^2 = \dfrac{(x-\mu)^2}{\sigma^2}$ , so the exponent in $f(x)$ is

-\frac{(x-\mu)^2}{2\sigma^2} = -\frac{z^2}{2}.

Substituting that back, and pulling the $\frac{1}{\sigma}$ out to the front,

f(x) = \frac{1}{\sigma} \cdot \frac{1}{\sqrt{2\pi}}\, e^{-z^2/2} = \frac{1}{\sigma}\, \varphi(z) = \frac{1}{\sigma}\, \varphi\!\left(\frac{x-\mu}{\sigma}\right).

So $f$ is built from $\varphi$ by exactly two moves from the transformations lesson. Replacing the input by $\frac{x-\mu}{\sigma}$ shifts the graph right by $\mu$ and multiplies every horizontal distance from the center by $\sigma$ . Those two effects are what put the peak at $\mu$ and set the width. Chapter 2 names a horizontal scaling by the input multiplier, never by its reciprocal, and the input multiplier here is $\frac{1}{\sigma}$ . So in that lesson’s vocabulary this move is a horizontal compression by $\frac{1}{\sigma}$ . It is the same move under both names, since dividing the inputs by $\frac{1}{\sigma}$ is multiplying them by $\sigma$ . The geometric phrasing is the one that matters below, and it has the advantage of staying correct whether $\sigma$ is bigger or smaller than $1$ . Multiplying the output by $\frac{1}{\sigma}$ then divides every height by $\sigma$ , which is what keeps a wide curve from also being a tall one.

Those two moves cancel each other as far as area is concerned, and that is the heart of it. Cut the region under $\varphi$ into thin vertical strips, each one close to a rectangle of width $w$ and height $h$ . The horizontal move multiplies every width by $\sigma$ ; the vertical move divides every height by $\sigma$ . A strip that had area $wh$ now has area

(\sigma w)\left(\frac{h}{\sigma}\right) = wh,

the very same area. Since every strip keeps its area, so does the whole region assembled from them.

Two honest remarks about that step. The strips only approximate the region, and the claim that the approximation becomes exact is one we are taking on trust. The table already asked the very same trust of you a moment ago, when its numbers turned out to come from adding up huge numbers of very thin strips. And notice what the argument does and does not do. It never computes an area. It shows only that two areas are equal, because a map that multiplies widths by $\sigma$ and divides heights by $\sigma$ leaves every area alone. That is why the argument is allowed even though computing either area is not: equality is cheap here, and evaluation is what costs calculus.

The total area under $f$ is therefore still $1$ , which is what makes $\frac{1}{\sigma\sqrt{2\pi}}$ the correct constant, and, more usefully, corresponding pieces of area match up:

P(a < Z < b) = P(\mu + a\sigma < X < \mu + b\sigma).

Any probability question about $X$ is therefore the same question about the standard curve, once each value is rewritten as a z-score. One table can serve every normal distribution because, up to a shift and a rescale, there is only one normal curve.

A bit of history (optional)

For a long time this curve belonged mostly to astronomers. It described the errors in a hundred telescope readings of a single star. Measure a thousand people and you would not expect that same shape. A person, after all, is not a mistake.

Adolphe Quetelet, a Belgian astronomer, looked anyway. In the 1830s he gathered the chest measurements of thousands of Scottish soldiers. They scattered into the same bell he knew from the sky. From that he drew an extraordinary conclusion. Nature was aiming at an ideal he called l’homme moyen, the average man, and every real person was an error around that target.

The measurements were sound; the interpretation was not. There is no target, and nobody is a miss. What Quetelet had stumbled on was the pattern this lesson describes. A human chest is a total of many small influences, and such sums do tend to a bell. The result carries conditions, though: the influences must be roughly independent, none may swamp the rest, and a few further technical conditions this course does not state precisely must hold too. Where those hold the bell is earned, and where they fail it has merely been assumed. His average man was a mean, not a design.

The habit outlived the mistake, and ranking people against a curve later did real harm. Keep that warning tied to the word normal. The curve is a model of the world, never a verdict on it.