Algebra 2 · S.ID.A.4

The standard normal distribution

Every normal model has its own mean and standard deviation, so it seems every one needs its own table of areas. Is there one curve whose areas answer the questions for all of them?

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Standardizing a data setAlgebra 2 z-scores and the empirical ruleAlgebra 2

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Normal calculations: areas and percentilesAlgebra 2 The t distribution and degrees of freedomStatistics

Where you are 8 in Distributions

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▴ 3 above Sampling distributions7 z-scores and the empirical rule The central limit theorem

The standard normal distribution

Normal calculations: areas and percentiles

Why this exists

A normal model answers questions with areas: the share of values below some cut is the area under the curve to the left of it. But the curve has no simple area formula. Its areas have to be worked out numerically, and there is a different curve for every mean and every standard deviation. A printed table for each one would fill a library, and every new model would need a new table.

Standardizing removes the problem. Measure every value by its z-score, the number of standard deviations it lies from the mean, and every normal curve turns into the same curve: the one with mean 0 and standard deviation 1. So a single table of areas, worked out once, serves every normal model there is. That one curve is the standard normal distribution, and this lesson is about reading it: areas to the left, to the right and between two cuts, and the cut that has a given area to its left.

Definition

The standard normal distribution is the normal distribution with mean 0 and standard deviation 1, written N(0, 1). A variable with this distribution is usually called Z, and its values are z-scores. Its density curve is the graph of

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

The share of the distribution below a value z is the area under this curve to the left of z. That area is written \Phi(z), so

\Phi(z) = P(Z < z).

Here P(\dots) is read “the probability that”, or equally “the share of the distribution for which”. \Phi is the capital Greek letter phi, and \varphi is the small one.

Intuition

The one bell. The curve N(0, 1) peaks above z = 0 at height \displaystyle \frac{1}{\sqrt{2\pi}} \approx 0.399, changes its bending at z = -1 and z = 1, and by z = \pm 3 it is almost on the axis. Its numbers along the bottom are already in standard deviations, so the axis is the z-score scale itself.

Every normal curve is this one, relabeled. Take any normal curve N(\mu, \sigma). Writing each value x as its z-score \displaystyle \frac{x - \mu}{\sigma} slides the center to 0 and squeezes or stretches the spread to 1. The The normal distribution lesson showed that sliding and rescaling a normal variable leaves it normal, so what comes out is a normal curve with mean 0 and standard deviation 1: exactly N(0, 1). The picture does not change; only the numbers written under it change, from the variable’s own units to z-scores. A region under the original curve and the matching region under N(0, 1) have the same area.

The table. Because there is only one curve, its areas can be worked out once and printed. A standard normal table gives \Phi(z), the area to the left of z. A full table lists every z to two decimal places; the excerpt below is a selection, and it is all this lesson needs. Many tables list only z \geq 0, because symmetry, below, gives the rest.

z	\Phi(z)	z	\Phi(z)
0.00	0.5000	1.28	0.8997
0.25	0.5987	1.29	0.9015
0.50	0.6915	1.50	0.9332
0.67	0.7486	1.64	0.9495
0.68	0.7517	1.65	0.9505
0.75	0.7734	1.96	0.9750
0.84	0.7995	2.00	0.9772
0.85	0.8023	2.33	0.9901
1.00	0.8413	2.50	0.9938
1.04	0.8508	3.00	0.9987
1.25	0.8944

Every entry is rounded to four decimal places. A calculator or spreadsheet gives the same areas for any z at all; the table is the same function written out at a few points.

Figure 1The area to the left of z = 1.25 is \Phi(1.25) \approx 0.8944, read straight from the table.

Results

The whole and the half. The total area under the curve is 1, and the curve is symmetric about z = 0, so

\Phi(0) = 0.5.

Left, right and between. For any numbers z, a and b with a < b:

\begin{aligned} P(Z < z) &= \Phi(z), \\ P(Z > z) &= 1 - \Phi(z), \\ P(a < Z < b) &= \Phi(b) - \Phi(a). \end{aligned}

Symmetry. For any z,

\begin{aligned} P(Z > z) &= P(Z < -z), \\ \Phi(-z) &= 1 - \Phi(z). \end{aligned}

A symmetric middle. For any c > 0,

P(-c < Z < c) = 2\Phi(c) - 1.

Strict or not makes no difference. A single point has no area under a density curve, so P(Z \leq z) = P(Z < z) and P(Z \geq z) = P(Z > z).

Percentiles. For a share p between 0 and 1, the p cut, or 100pth percentile, of Z is the value z with \Phi(z) = p. There is exactly one such z. The percentiles for p and 1 - p are negatives of each other. A few that turn up constantly:

percentile	z	percentile	z
50th	0	95th	1.645
75th	0.67	97.5th	1.96
90th	1.28	99th	2.33

Standardizing. If X has distribution N(\mu, \sigma), then \displaystyle Z = \frac{X - \mu}{\sigma} has distribution N(0, 1), and for any a < b

\begin{aligned} &P(a < X < b) \\ &\quad = P\!\left(\frac{a - \mu}{\sigma} < Z < \frac{b - \mu}{\sigma}\right). \end{aligned}

Why it is true

Symmetry. The density depends on z only through z^2, so \varphi(-z) = \varphi(z): the curve is the same height at z and at -z, and the left half is the mirror image of the right. The region to the right of z is the mirror image of the region to the left of -z, so they have the same area. That gives P(Z > z) = P(Z < -z), and since the area right of z is 1 - \Phi(z), also \Phi(-z) = 1 - \Phi(z). With z = 0 the two halves are mirror images, so each is \displaystyle \frac{1}{2}.

Left, right and between. The region left of z and the region right of z together make the whole, with area 1, which gives P(Z > z) = 1 - \Phi(z). For a < b, the region left of b is the region left of a together with the strip from a to b, so the strip has area \Phi(b) - \Phi(a). The symmetric middle is the strip from -c to c, so its area is

\begin{aligned} \Phi(c) - \Phi(-c) &= \Phi(c) - (1 - \Phi(c)) \\ &= 2\Phi(c) - 1. \end{aligned}

Points have no area. The region above a single point is a line segment, and a segment has no width, so its area is 0. Adding or removing the point z does not change an area.

Exactly one percentile. As z moves to the right, \Phi(z) gains area all the time, since the curve is above the axis everywhere, so \Phi is strictly increasing. It runs from near 0 far to the left to near 1 far to the right without jumping, so each share p strictly between 0 and 1 is reached once and only once. If \Phi(z) = p, symmetry gives \Phi(-z) = 1 - p, so the 1 - p cut is -z.

Standardizing. By the rules for shifting and rescaling, \displaystyle \frac{X - \mu}{\sigma} has mean \displaystyle \frac{\mu - \mu}{\sigma} = 0 and standard deviation \displaystyle \frac{\sigma}{\sigma} = 1, as in Standardizing a data set; and a shifted, rescaled normal variable is still normal. So it is N(0, 1). Since \sigma > 0, subtracting \mu and dividing by \sigma keeps inequalities the same way round, so a < X < b happens exactly when \displaystyle \frac{a - \mu}{\sigma} < Z < \frac{b - \mu}{\sigma}. The same event has the same probability.

Where the table comes from. \Phi(z) is the area under \varphi from far left to z. That area cannot be written using the usual functions (powers, roots, exponentials and logarithms), so it is computed numerically, to as many decimal places as needed, and printed.

Worked examples

Example 1

Reading an area to the left

Find P(Z < 1.25), and say what it means for the place of z = 1.25 in the distribution.

Solution

An area to the left is exactly what the table lists. The row for 1.25 gives

P(Z < 1.25) = \Phi(1.25) \approx 0.8944.

So about 0.8944, or 89.44\%, of the standard normal distribution lies below 1.25, and z = 1.25 sits between the 89th and 90th percentiles. This is the shaded area in the first figure. Because a single point has no area, P(Z \leq 1.25) is the same number.

Example 2

An area to the right, and its mirror image

Find P(Z > 1.25), P(Z < -1.25) and P(Z > -1.25).

Solution

The area to the right of 1.25 is the whole, 1, take away the area to the left:

P(Z > 1.25) = 1 - 0.8944 = 0.1056.

The curve is symmetric about 0, so the tail to the left of -1.25 is the mirror image of the tail to the right of 1.25:

P(Z < -1.25) = P(Z > 1.25) \approx 0.1056.

That is how a table listing only positive z answers a question about a negative one. Finally, the area to the right of -1.25 is the whole without that left tail, 1 - 0.1056 = 0.8944, the same as the area left of 1.25; again the mirror image.

Figure 2The curve is its own mirror image in z = 0, so the tail to the right of 1.25 and the tail to the left of -1.25 have the same area, 1 - 0.8944 = 0.1056.

Example 3

An area between two cuts

Find P(-0.5 < Z < 1.5).

Solution

The region left of 1.5 is the region left of -0.5 together with the strip between them, so the strip’s area is the difference of two left areas:

P(-0.5 < Z < 1.5) = \Phi(1.5) - \Phi(-0.5).

The table gives \Phi(1.5) \approx 0.9332. For the negative cut, use symmetry: \Phi(-0.5) = 1 - \Phi(0.5), which is about 1 - 0.6915 = 0.3085. So

P(-0.5 < Z < 1.5) \approx 0.9332 - 0.3085 = 0.6247.

As a check on size: the strip contains the middle, where the curve is tallest, and runs further on the right than on the left, so it should hold well over half the area, and it does.

Figure 3The area between -0.5 and 1.5 is the area left of 1.5 take away the area left of -0.5.

Example 4

The empirical rule, sharpened

The empirical rule says that about 68\%, 95\% and 99.7\% of a normal distribution lie within one, two and three standard deviations of the mean. Find P(-1 < Z < 1), P(-2 < Z < 2) and P(-3 < Z < 3) from the table.

Solution

Each is a symmetric middle, so each is 2\Phi(c) - 1:

\begin{aligned} P(-1 < Z < 1) &\approx 2(0.8413) - 1 = 0.6826, \\ P(-2 < Z < 2) &\approx 2(0.9772) - 1 = 0.9544, \\ P(-3 < Z < 3) &\approx 2(0.9987) - 1 = 0.9974. \end{aligned}

These are the rule’s 68\%, 95\% and 99.7\% to more places. (Worked from unrounded areas, the three are 0.6827, 0.9545 and 0.9973; doubling a four-place entry can move the last digit.) Because every normal curve standardizes to this one, the same three shares hold for every normal model, which is why the empirical rule never mentions \mu or \sigma.

Example 5

Finding the cut for a given area

Find the 90th percentile of Z, and the 10th.

Solution

Now the area is known and the cut is not, so the table is read backwards: look among the areas for 0.9000 and read off its z. It falls between two entries:

\Phi(1.28) \approx 0.8997, \qquad \Phi(1.29) \approx 0.9015.

0.9000 is 0.0003 above the first and 0.0015 below the second, so the nearer is 1.28; the 90th percentile is about z = 1.28. (A calculator gives 1.2816.) The last figure shows the cut, with 0.90 of the area to its left and 0.10 to its right.

The 10th percentile has area 0.10 to its left. By symmetry it is the mirror image of the cut with 0.10 to its right, which is the one just found, so the 10th percentile is about z = -1.28.

Figure 4The 90th percentile of Z is the cut with area 0.90 to its left, about z = 1.28.

Example 6

The middle 95%

Find the number c for which the middle 95\% of the standard normal distribution lies between -c and c.

Solution

The middle holds 0.95, so the two tails share the remaining 0.05. They are mirror images, so each tail is 0.025. The cut c then has the left tail and the middle to its left:

\Phi(c) = 0.025 + 0.95 = 0.975.

The table has \Phi(1.96) \approx 0.9750, so c \approx 1.96. Check with the symmetric middle rule: 2(0.9750) - 1 = 0.95. The empirical rule’s “two standard deviations” for 95\% is this 1.96, rounded.

Example 7

One table for every normal curve

Two variables are modeled by N(3, 0.4) and N(600, 75). For each, find the value 1.25 standard deviations above the mean, and the share of the distribution above that value.

Solution

The values are 3 + 1.25(0.4) = 3.5 and 600 + 1.25(75) = 693.75.

Each of these has z-score 1.25 in its own distribution. Standardizing turns each model into N(0, 1) and carries the region above the value onto the region of N(0, 1) above z = 1.25, with the same area. So for both models

P(Z > 1.25) \approx 1 - 0.8944 = 0.1056.

About 0.1056 of each distribution lies above its value, although the two curves have nothing else in common. The share depends only on how many standard deviations out the cut is. Doing this for a real model, turning a measurement into a z-score and the share back into a statement about the measurement, is the work of normal calculations.

Boundaries

Normal models only. Standardizing any data set gives mean 0 and standard deviation 1, but it keeps the shape. A skewed variable standardizes to a skewed set of z-scores, not to N(0, 1), and the table’s areas say nothing about it. The table is for variables a normal model fits.

The table is rounded. Entries are given to four decimal places and z to two, so answers built from them carry small rounding errors, as in the fourth example. A cut read backwards from the table is the nearest listed z, not the exact one.

Far tails are small, never zero. A table that stops at z = 3.49 or so prints values like 0.9998, and beyond it the area left over is tiny but positive, because the curve never meets the axis.

The height is not an area. \varphi(1.25) \approx 0.183 is the height of the curve at 1.25, a density, not a share. Only areas are shares, and the share at a single point is 0.

Notation. Some tables list the area between 0 and z instead of the area to the left of z. Read a table’s heading before using it: for z = 1.25 such a table prints 0.3944, which is 0.8944 - 0.5.

Where this goes next

With the one curve read, normal calculations apply it to real normal models: turn a measurement into a z-score, read the share, and turn a percentile of Z back into a measurement, the step finding the value at a percentile uses. A normal random variable is the same curve read as chance, and the The central limit theorem explains why so many averages standardize to N(0, 1), which is what makes the table the working tool of Sampling distributions and inference. The cuts 1.645, 1.96 and 2.33 return there constantly. When the standard deviation must be estimated from the sample, the standardized value follows not N(0, 1) but the The t distribution and degrees of freedom, a bell with heavier tails that approaches N(0, 1) as samples grow.

Exercises

Use the table in the lesson for every exercise.

Areas to the left

Find P(Z < 0.75).
Find P(Z < 2.33).
Find P(Z < -1.5).
Find P(Z \leq 0).

Areas to the right and symmetry

Find P(Z > 0.5).
Find P(Z > -2).
Find P(Z > 1.96).
Given \Phi(1.04) \approx 0.8508, find P(Z < -1.04) and P(Z > -1.04).

Areas between

Find P(0 < Z < 1.5).
Find P(-1.25 < Z < 0.75).
Find P(1 < Z < 2).
Find P(-0.25 < Z < 2.5).
Find the share of the standard normal distribution that lies more than one standard deviation from the mean, that is, P(Z < -1 \text{ or } Z > 1).

Finding the cut

Find the value of z with area 0.75 to its left, to two decimal places.
Find the value of z with area 0.05 to its right.
Find the 10th percentile of Z.
Find c so that P(-c < Z < c) = 0.98.
Find the 20th percentile of Z, to two decimal places.

Every normal curve

For any normal model, what share of the distribution lies more than two standard deviations above the mean? For the model N(64, 8), what value is that cut?
For any normal model, what share of the distribution lies within 1.5 standard deviations of the mean?

Explaining

Explain, with a sketch of the curve in mind, why \Phi(-z) = 1 - \Phi(z) for every z.
Explain why P(Z < 1.25) and P(Z \leq 1.25) are the same number.
Explain why one table of areas for N(0, 1) is enough for every normal model, whatever its mean and standard deviation.
Explain why the 90th and 10th percentiles of Z are negatives of each other, and say which percentile of Z is 0.

Answers

About 0.7734, read from the table: \Phi(0.75) \approx 0.7734.
About 0.9901, read from the table: \Phi(2.33) \approx 0.9901.
About 0.0668. By symmetry \Phi(-1.5) = 1 - \Phi(1.5), which is about 1 - 0.9332 = 0.0668.
Exactly 0.5. Half the area lies on each side of 0, and the point 0 itself adds no area.
About 0.3085. The area right of 0.5 is 1 - \Phi(0.5), about 1 - 0.6915 = 0.3085.
About 0.9772. By symmetry, P(Z > -2) = P(Z < 2), and that is \Phi(2) \approx 0.9772.
About 0.0250. The area right of 1.96 is 1 - \Phi(1.96), about 1 - 0.9750 = 0.0250.
About 0.1492 and 0.8508. P(Z < -1.04) = 1 - 0.8508 = 0.1492, and P(Z > -1.04) is its complement, 1 - 0.1492 = 0.8508, which by symmetry is also \Phi(1.04).
About 0.4332. The area is \Phi(1.5) - \Phi(0), about 0.9332 - 0.5 = 0.4332.
About 0.6678. The area is \Phi(0.75) - \Phi(-1.25), and \Phi(-1.25) = 1 - 0.8944, so it is about 0.7734 - 0.1056 = 0.6678.
About 0.1359. The area is \Phi(2) - \Phi(1), about 0.9772 - 0.8413 = 0.1359.
About 0.5925. The area is \Phi(2.5) - \Phi(-0.25), and \Phi(-0.25) = 1 - 0.5987 = 0.4013, so it is about 0.9938 - 0.4013 = 0.5925.
About 0.3174. Each tail is 1 - \Phi(1) \approx 0.1587, and the two tails are equal, so the share is 2(0.1587) = 0.3174; equally, 1 - 0.6826.
z \approx 0.67. 0.75 lies between \Phi(0.67) \approx 0.7486 and \Phi(0.68) \approx 0.7517, and is nearer the first (0.0014 against 0.0017).
z \approx 1.645. Area 0.05 to the right means 0.95 to the left, and 0.95 lies exactly halfway between \Phi(1.64) \approx 0.9495 and \Phi(1.65) \approx 0.9505.
z \approx -1.28. The 90th percentile is 1.28, and the 10th is its mirror image.
c \approx 2.33. The tails share 0.02, so each is 0.01, and \Phi(c) = 0.99; the table has \Phi(2.33) \approx 0.9901.
z \approx -0.84. The 80th percentile has \Phi(z) = 0.80; 0.80 lies between 0.7995 (at 0.84) and 0.8023 (at 0.85) and is nearer the first, so it is 0.84. The 20th is its mirror image.
About 0.0228, and the cut is 80. Every normal model standardizes to N(0, 1), so the share is P(Z > 2) \approx 1 - 0.9772 = 0.0228. For N(64, 8) the cut is 64 + 2(8) = 80.
About 0.8664. Standardized, the share is P(-1.5 < Z < 1.5) = 2\Phi(1.5) - 1, about 2(0.9332) - 1 = 0.8664, the same for every normal model.
\Phi(-z) is the area of the tail to the left of -z. The curve is symmetric about 0, so that tail is the mirror image of the tail to the right of z, and has the same area. The tail to the right of z is the whole, 1, without the area to the left of z, which is \Phi(z). So \Phi(-z) = 1 - \Phi(z).
The two differ only by the single value 1.25. Above a single point the region under the curve is a line segment, which has no width and so no area. Including it or not leaves the area the same.
Standardizing turns N(\mu, \sigma) into N(0, 1): the z-score \displaystyle \frac{x - \mu}{\sigma} has mean 0, standard deviation 1, and is still normal. A cut x becomes the cut \displaystyle \frac{x - \mu}{\sigma}, and the region beyond it has the same area before and after, since the same values are counted. So every area of every normal model is an area of N(0, 1), which one table gives.
The 90th percentile z has 0.10 to its right. By symmetry the cut -z has 0.10 to its left, so -z is the 10th percentile. The value 0 has half the area on each side, so it is the 50th percentile, the median.