Math Core

Heights of adults, weights of cereal boxes, lifetimes of light bulbs, errors in careful measurements: when you graph data like these, the same bell shape shows up again and again. That shape is the normal distribution, and once you know its mean and standard deviation, you can estimate what fraction of the data falls in any range.

The shape of a normal curve

A histogram shows how data are spread out. If you collect more and more data and make the bars narrower and narrower, the tops of the bars start to trace a smooth curve. For many real quantities that curve is symmetric and bell-shaped.

Definition

Normal distribution

A normal distribution is a symmetric, bell-shaped distribution described by two numbers:

the mean $\mu$ (the center, where the peak is), and
the standard deviation $\sigma$ (how spread out the data are).

The total area under a normal curve is $1$ , or $100\%$ . The area over an interval equals the proportion of the data in that interval.

Some facts follow straight from the shape:

The curve is symmetric about the mean, so the mean, median and mode are all equal to $\mu$ .
Exactly half the area lies on each side of $\mu$ .
The curve never touches the horizontal axis, but it gets very close to it once you are more than about $3$ standard deviations from the mean.
A larger $\sigma$ gives a wider, flatter curve. A smaller $\sigma$ gives a taller, narrower one.

Below is the standard normal curve, the normal distribution with mean $0$ and standard deviation $1$ . The shaded region runs from one standard deviation below the mean to one standard deviation above it.

The standard normal curve. The shaded area between −1 and 1 is about 68% of the total.Open in grapher →

The 68–95–99.7 rule

Every normal distribution, no matter its mean or standard deviation, splits up the same way.

The 68–95–99.7 rule (empirical rule)

In a normal distribution:

about $68\%$ of the data 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 $\mu + 3\sigma$ .

Because the curve is symmetric, you can cut these regions in half. That gives the percentages in each band:

Band	$\mu - 3\sigma$ to $\mu - 2\sigma$	$\mu - 2\sigma$ to $\mu - \sigma$	$\mu - \sigma$ to $\mu$	$\mu$ to $\mu + \sigma$	$\mu + \sigma$ to $\mu + 2\sigma$	$\mu + 2\sigma$ to $\mu + 3\sigma$
Percent	$2.35\%$	$13.5\%$	$34\%$	$34\%$	$13.5\%$	$2.35\%$

The remaining $0.3\%$ is split between the two tails beyond $3\sigma$ , $0.15\%$ on each side. The value $13.5\%$ comes from $\dfrac{95 - 68}{2}$ , and $2.35\%$ comes from $\dfrac{99.7 - 95}{2}$ .

Worked example: Using the 68–95–99.7 rule

The heights of adult women in a large city are approximately normal with mean $64$ inches and standard deviation $2.5$ inches. Estimate the percent of women who are

between $61.5$ and $66.5$ inches tall;
taller than $69$ inches;
between $59$ and $66.5$ inches tall.

Solution. First mark the key values: $\mu - 2\sigma = 59$ , $\mu - \sigma = 61.5$ , $\mu = 64$ , $\mu + \sigma = 66.5$ , $\mu + 2\sigma = 69$ .

$61.5$ to $66.5$ is within $1$ standard deviation of the mean: about $68\%$ .
$69$ is $2$ standard deviations above the mean. About $95\%$ lie within $2\sigma$ , so $5\%$ lie outside, split evenly into two tails: $\dfrac{5\%}{2} = 2.5\%$ .
$59$ to $64$ is half of the middle $95\%$ , which is $47.5\%$ . $64$ to $66.5$ is half of the middle $68\%$ , which is $34\%$ . Total: $47.5\% + 34\% = 81.5\%$ .

Heights with mean 64 and standard deviation 2.5. The shaded region from 59 to 66.5 holds about 81.5% of the women.Open in grapher →

z-scores

The 68–95–99.7 rule only works at whole numbers of standard deviations. To handle any value, measure how far it is from the mean in units of standard deviations.

Definition

z-score

The z-score of a data value $x$ is

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

A positive $z$ means $x$ is above the mean; a negative $z$ means it is below. For example, $z = -1.5$ means one and a half standard deviations below the mean.

z-scores also let you compare values from different distributions: the value whose z-score is farther from $0$ is more unusual relative to its own group.

Worked example: Comparing with z-scores

Maya scored $82$ on a history test where the mean was $74$ and the standard deviation was $6$ . Jordan scored $88$ on a chemistry test where the mean was $80$ and the standard deviation was $8$ . Who did better compared with their class?

z_{\text{Maya}} = \frac{82 - 74}{6} = \frac{8}{6} \approx 1.33, \qquad z_{\text{Jordan}} = \frac{88 - 80}{8} = 1.

Maya's score is $1.33$ standard deviations above her class mean, while Jordan's is only $1$ standard deviation above. Relative to her class, Maya did better, even though Jordan's raw score is higher.

Areas from a z-table

A z-table lists the area under the standard normal curve to the left of a z-score, which is the proportion of data below that value. Here is a short table.

$z$	$0$	$0.25$	$0.5$	$0.75$	$1$	$1.25$	$1.5$	$1.75$	$2$	$2.5$	$3$
Area to the left	$0.5000$	$0.5987$	$0.6915$	$0.7734$	$0.8413$	$0.8944$	$0.9332$	$0.9599$	$0.9772$	$0.9938$	$0.9987$

For a negative z-score, use symmetry: the area to the left of $-z$ equals the area to the right of $z$ , which is $1$ minus the table value. For example, the area to the left of $-1.5$ is $1 - 0.9332 = 0.0668$ .

Three kinds of area questions

Below a value: $P(X < x) =$ the table value for its z-score.
Above a value: $P(X > x) = 1 -$ the table value.
Between two values: subtract the smaller table value from the larger one.

Worked example: Battery lifetimes

The lifetime of a certain battery is normally distributed with mean $40$ hours and standard deviation $4$ hours. Find the probability that a randomly chosen battery lasts

less than $45$ hours;
more than $34$ hours;
between $38$ and $46$ hours.

Solution.

$z = \dfrac{45 - 40}{4} = 1.25$ . The table gives $P(X < 45) = 0.8944$ .
$z = \dfrac{34 - 40}{4} = -1.5$ . The area to the left of $-1.5$ is $1 - 0.9332 = 0.0668$ , so $P(X > 34) = 1 - 0.0668 = 0.9332$ . (By symmetry, the area to the right of $-1.5$ equals the area to the left of $1.5$ .)
$z = \dfrac{38 - 40}{4} = -0.5$ and $z = \dfrac{46 - 40}{4} = 1.5$ . The area left of $-0.5$ is $1 - 0.6915 = 0.3085$ . So

P(38 < X < 46) = 0.9332 - 0.3085 = 0.6247.

Common mistake

The table gives the area to the left of $z$ . For "more than" questions, don't read the table value directly; subtract it from $1$ . A quick sketch of the curve with the region shaded will tell you whether your answer should be more or less than $0.5$ .

Tip

You can also work backward. Since $0.8413$ of the area is below $z = 1$ , a value one standard deviation above the mean is at about the $84$ th percentile. In the battery example, $40 + 4 = 44$ hours is the $84$ th percentile.

Practice

Use the 68–95–99.7 rule or the z-table in this lesson.

Practice 1

Which statement is not true of every normal distribution?

Practice 2

SAT section scores are approximately normal with mean $500$ and standard deviation $100$ . About what percent of scores fall between $400$ and $600$ ?