The Normal Distribution and the 68-95-99.7 Rule

A shape that turns up everywhere

Measure the heights of a thousand adult women, the weights of cereal boxes off a production line, or the scores on a large standardized test. Draw a histogram of each, and the same shape keeps appearing: a single peak in the middle, falling away evenly on both sides.

That bell-shaped curve is the normal distribution.

• normal curve

Feature	What it means
symmetric	as many values fall a given distance above the mean as below it
single peak	the mean, median and mode are the same value
thin tails	values far from the mean are rare

A normal distribution is completely described by two numbers: its mean, which places the peak, and its standard deviation, which sets how wide the bell is.

The 68-95-99.7 rule

For any normal distribution, the share of values within a given number of standard deviations of the mean is always the same:

Within	Share of the values
$1$ standard deviation	about $68\%$
$2$ standard deviations	about $95\%$
$3$ standard deviations	about $99.7\%$

This is also called the empirical rule.

Adult women’s heights in the US are roughly normal, with mean $64$ inches and standard deviation $3$ inches.

$$64 \pm 3 = 61 \text{ to } 67 \text{ in} \quad\Rightarrow\quad \text{about } 68\%$$ $$64 \pm 6 = 58 \text{ to } 70 \text{ in} \quad\Rightarrow\quad \text{about } 95\%$$ $$64 \pm 9 = 55 \text{ to } 73 \text{ in} \quad\Rightarrow\quad \text{about } 99.7\%$$

Why symmetry gives the rest

The rule only states the middle bands. Symmetry supplies everything else.

If $95\%$ lies within two standard deviations, the other $5\%$ is split evenly between the two tails — $2.5\%$ on each side.

Region	Share
more than $2$ SD above the mean	$2.5\%$
between the mean and $1$ SD above	$34\%$
between $1$ and $2$ SD above	$13.5\%$
below the mean	$50\%$

The $34\%$ is half of $68\%$. The $13.5\%$ is half of the difference between $95\%$ and $68\%$.

What share of women are taller than $70$ inches?

$70$ is two standard deviations above the mean, so about $2.5\%$.

Every one of these answers is half of something, because the curve is symmetric. That is the whole method.

z-scores

A z-score says how many standard deviations a value sits from the mean.

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

$z$	Means
$0$	exactly the mean
$2$	two standard deviations above
$-1.5$	one and a half standard deviations below

z-scores put values from different scales on one common scale.

Ana scores $85$ on a test with mean $75$ and standard deviation $5$. Ben scores $90$ on a different test with mean $80$ and standard deviation $10$. Who did better relative to their class?

$$z_{\text{Ana}} = \frac{85 - 75}{5} = 2 \qquad z_{\text{Ben}} = \frac{90 - 80}{10} = 1$$

Ana did better. Ben’s raw score is higher, but Ana sits twice as far above her class average, measured in each test’s own spread.

For values between the whole-number z-scores, a table or a calculator’s normal distribution function gives the exact percentage. A spreadsheet’s NORM.DIST does the same.

When the normal model does not fit

The rule is a property of the bell shape. Data without that shape does not obey it, and applying it anyway produces confident, wrong percentages.

Data	Why it is not normal
household income	skewed right — a long tail of high earners
waiting time for a bus	cannot go below zero, and piles up near it
test with a hard ceiling	scores bunch against the maximum
a mix of two groups	two peaks rather than one

Household incomes in a county have mean $70,000 and standard deviation $50,000.

Two standard deviations below the mean is $70,000 − $100,000 = −$30,000. A normal model would claim $2.5\%$ of households earn less than a negative amount. The impossible answer is the evidence that the model is wrong, not a curiosity to report.

Check the shape before using the rule. A histogram, a dot plot or a box plot settles it in seconds, and skipping that check is how real reports end up quoting nonsense.

Worked examples

Common mistakes

Practice problems