Math Core

Many distributions (heights, measurement errors, test scores, weights of manufactured parts) have the same bell-shaped pattern: unimodal, symmetric, and trailing off evenly in both directions. The normal distribution is the mathematical model for that shape. It lets you answer questions like "what percent of values are above $80$ ?" or "how high must a score be to reach the top $10\%$ ?" with just the mean and standard deviation. It will be central to everything in the second half of AP Statistics.

Density curves

When a histogram has a very large number of values and very narrow bins, its outline becomes a smooth curve. A density curve is a curve that models the overall shape of a distribution.

Definition

Density curve

A density curve is a curve that is always on or above the horizontal axis and has an area of exactly $1$ underneath it. The area under the curve above any interval equals the proportion of values in that interval.

The mean of a density curve is its balance point, and the median is the point that divides the area in half. For a symmetric curve they are equal; for a right-skewed curve, the mean is pulled out toward the tail, to the right of the median. Because a density curve is an idealized model, its mean and standard deviation are written $\mu$ and $\sigma$ .

Normal distributions

A normal distribution is described by a symmetric, bell-shaped density curve. It is completely determined by two numbers: its mean $\mu$ , which sits at the center of the bell, and its standard deviation $\sigma$ , which controls the width. We write $N(\mu, \sigma)$ . The points where the curve changes from curving downward to curving upward (the inflection points) are exactly $1$ standard deviation on either side of the mean.

The empirical rule (68–95–99.7 rule)

In any normal distribution, approximately

$68\%$ of the values fall within $1\sigma$ of the mean,
$95\%$ fall within $2\sigma$ of the mean, and
$99.7\%$ fall within $3\sigma$ of the mean.

The standard normal curve N(0, 1). The shaded area between −1 and 1 is about 68% of the total.Open in grapher →

Symmetry fills in the rest. For example, since $68\%$ is within $1\sigma$ , the other $32\%$ is split equally between the two tails, so $16\%$ of values are more than $1\sigma$ above the mean.

Worked example: Using the empirical rule

The weights of eggs from a farm are approximately normal with mean $60$ grams and standard deviation $4$ grams, $N(60, 4)$ .

What percent of eggs weigh between $52$ and $68$ grams?
What percent weigh more than $64$ grams?
What percent weigh between $56$ and $72$ grams?

Solutions.

$52$ and $68$ are $2\sigma$ below and above the mean, so about $95\%$ .
$64$ is $1\sigma$ above the mean. Half of the $32\%$ outside $\mu \pm \sigma$ is above it: $16\%$ .
$56$ is $1\sigma$ below the mean and $72$ is $3\sigma$ above. The area from $56$ to $60$ is half of $68\%$ , or $34\%$ ; from $60$ to $72$ it is half of $99.7\%$ , or $49.85\%$ . The total is about $83.85\%$ .

Egg weights, N(60, 4). The shaded region from 56 to 72 grams holds about 83.85% of the eggs.Open in grapher →

z-scores: standardizing

To compare values from different distributions, or to use a normal table, convert each value to a z-score.

Definition

z-score

The z-score (standardized score) of a value $x$ is

z = \frac{x - \text{mean}}{\text{standard deviation}}.

It tells how many standard deviations $x$ is above (positive $z$ ) or below (negative $z$ ) the mean.

If $X$ has a normal distribution, its z-scores follow the standard normal distribution $N(0, 1)$ . That's why a single table of areas works for every normal distribution.

Worked example: Comparing with z-scores

Priya scored $680$ on a test where scores had mean $560$ and standard deviation $80$ . Marcus scored $31$ on a different test where scores had mean $25$ and standard deviation $5$ . Relative to the other test takers, who did better?

z_{\text{Priya}} = \frac{680 - 560}{80} = 1.5, \qquad z_{\text{Marcus}} = \frac{31 - 25}{5} = 1.2.

Priya's score is $1.5$ standard deviations above her test's mean, while Marcus's is $1.2$ standard deviations above his. Priya did better relative to her group.

Finding areas and percentiles

For intervals that aren't whole numbers of standard deviations, use technology (such as normalcdf on a calculator) or a table of standard normal areas. Here is an excerpt. Each entry is the area to the left of $z$ .

$z$	Area left of $z$	Area left of $-z$
$0.00$	$0.5000$	$0.5000$
$0.50$	$0.6915$	$0.3085$
$0.84$	$0.7995$	$0.2005$
$1.00$	$0.8413$	$0.1587$
$1.28$	$0.8997$	$0.1003$
$1.33$	$0.9082$	$0.0918$
$1.50$	$0.9332$	$0.0668$
$1.60$	$0.9452$	$0.0548$
$1.645$	$0.9500$	$0.0500$
$1.75$	$0.9599$	$0.0401$
$2.00$	$0.9772$	$0.0228$
$2.50$	$0.9938$	$0.0062$

Normal calculations in four steps

State the distribution and the question, and draw a curve with the region shaded.
Standardize: find the z-score of each boundary.
Find the area: left of $z$ from the table; right of $z$ is $1$ minus the table value; between two z-scores, subtract the smaller table value from the larger.
Answer in context.

Worked example: Battery life

The battery life of a certain phone is approximately normal with mean $18$ hours and standard deviation $1.5$ hours. What proportion of phones last between $16$ and $20.25$ hours?

Battery life, N(18, 1.5), with the region from 16 to 20.25 hours shadedOpen in grapher →

Standardize. $z = \dfrac{16 - 18}{1.5} \approx -1.33$ and $z = \dfrac{20.25 - 18}{1.5} = 1.5$ .

Area. The area left of $1.5$ is $0.9332$ and the area left of $-1.33$ is $0.0918$ . The area between is $0.9332 - 0.0918 = 0.8414$ .

Context. About $84\%$ of these phones have a battery life between $16$ and $20.25$ hours. (Technology, which doesn't round $z$ , gives $0.8420$ .)

Sometimes you know the area and need the value. That's an inverse normal problem: find the z-score with the right area to its left (from the table's middle column, or invNorm), then solve $x = \mu + z\sigma$ .

Worked example: Finding a percentile

For the phones above, how long must a battery last to be in the top $10\%$ ?

The top $10\%$ means $90\%$ of the area is to the left. The table shows that $z = 1.28$ has area $0.8997 \approx 0.90$ to its left. So

x = 18 + 1.28(1.5) = 19.92 \text{ hours}.

A battery must last about $19.92$ hours to be in the top $10\%$ , that is, at or above the $90$ th percentile.

Is a normal model appropriate?

Only use normal calculations when the data are approximately normal. Check that a dot plot or histogram is unimodal and roughly symmetric, and that about $68\%$ and $95\%$ of the values fall within $1$ and $2$ standard deviations of the mean. Technology can also draw a normal probability plot: if the points lie close to a straight line, a normal model is reasonable. Strongly skewed data should not be modeled with a normal curve.

Common mistake

The table gives the area to the left of $z$ . For "more than" questions, subtract the table value from $1$ . Before you finish, glance at your sketch: if the shaded region is less than half the curve, your answer must be less than $0.5$ .

Practice

Use the empirical rule or the table in this lesson. Technology answers are accepted too.

Practice 1

IQ scores are approximately normal with mean $100$ and standard deviation $15$ . Use the empirical rule to find the percent of people with IQ scores between $85$ and $130$ .