math.wiki
Statistics › Distributions › z-scores and the empirical rule
Open on the map
Algebra 2 · S.ID.A.4

z-scores and the empirical rule

A moth with a wingspan of 81 mm and a cabbage weighing 9 kg are both big for their kind. Which is more out of the ordinary, when one is measured in millimeters and the other in kilograms?

You need
The normal distributionAlgebra 2
Unlocks
Linear transformations of a random variableStatisticsThe standard normal distributionAlgebra 2Correlation from standardized valuesStatistics
Where you are 8 in Distributions
Statistics7
Distributions8
▴ 2 aboveDensity curves: area, mean and medianSampling distributions7
z-scores and the empirical rule
The central limit theoremThe standard normal distribution▾ Show all 8 (1 more)

Why this exists

A moth of one species has a wingspan of 81 mm. Is that large? The number alone cannot say. If wingspans of this species are usually close to 80 mm, then 81 is ordinary; if they are usually close to 50 mm, it is a giant. Even knowing the mean is not enough: 9 mm above the mean is a lot when wingspans barely vary and nothing special when they vary widely.

What answers the question is distance from the mean measured in the right unit, and the right unit is the Standard deviation, the typical distance of a value from the mean (interpreting a standard deviation). Counting in standard deviations does more than that. Millimeters and kilograms cannot be compared, but “one and a half standard deviations above the mean” means the same thing for a moth, a cabbage or a phone call. That count is the value’s z-score. And for a variable with a normal model, the z-score alone tells you roughly what share of values lie beyond it, through three numbers that are worth knowing by heart: the empirical rule.

Definition

The z-score of a value x is the number of standard deviations it lies above or below the mean. For a distribution with mean \mu and standard deviation \sigma,

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

For a sample described by its mean \bar{x} and standard deviation s, the same idea reads \displaystyle z = \frac{x - \bar{x}}{s}. Either way, the top is a distance from the mean and the bottom is the size of one standard deviation.

  • A value above the mean has a positive z-score, a value below it a negative one, and the mean itself has z = 0.
  • The z-score has no units. The top and the bottom are both in the units of x, and the units cancel. For this reason a z-score is sometimes called a value in standard units.
  • Read the other way, a value with z-score z is x = \mu + z\sigma.

The empirical rule, also called the 68-95-99.7 rule, says that for a variable with a normal model,

  • about 68\% of values lie within one standard deviation of the mean, between \mu - \sigma and \mu + \sigma;
  • about 95\% lie within two standard deviations, between \mu - 2\sigma and \mu + 2\sigma;
  • about 99.7\% lie within three standard deviations, between \mu - 3\sigma and \mu + 3\sigma.

In z-scores: about 68\% of values have z between -1 and 1, about 95\% between -2 and 2, and about 99.7\% between -3 and 3.

Intuition

A second ruler under the first. Wingspans of one species of moth are modeled by N(72, 6), in millimeters. Under the usual axis, marked 54, 60, \dots, 90, lay a second ruler whose marks are one standard deviation apart and whose zero sits at the mean: 72 becomes 0, 78 becomes 1, 66 becomes -1, and so on. The z-score of a value is simply its reading on this second ruler. A wingspan of 81 mm is 9 mm above the mean, and 9 mm is one and a half steps of 6, so its reading is 1.5.

54 −3 60 −2 66 −1 72 0 78 1 84 2 90 3 mm
z
 81 mm has z = 1.5 Wingspan (mm), with its z-score below
Figure 1Two scales on one axis. The top row is wingspan in millimeters; the bottom row counts standard deviations from the mean, so 81 mm sits at z = 1.5, halfway between the 1 and the 2.

Every normal curve on one ruler. A normal curve is fixed by its mean and standard deviation, and every normal curve is the same bell slid to its mean and stretched by its standard deviation. Read on the second ruler, the slide and the stretch are undone, and every normal curve looks exactly alike. So the share of the area between, say, z = -1 and z = 1 is the same for every normal curve. That is why the empirical rule can state its three numbers once, for all of them.

Slices of the bell. The three shares nest inside each other, and symmetry splits each difference evenly between the two sides. The middle 68\% is two slices of 34\%. The step from 68\% to 95\% adds 27\%, which is 13.5\% on each side. The step from 95\% to 99.7\% adds 4.7\%, which is 2.35\% on each side. What is left beyond three standard deviations is 0.3\%, or 0.15\% in each tail.

\mu - 3\sigma
\mu - 2\sigma
\mu - \sigma
\mu
\mu + \sigma
\mu + 2\sigma
\mu + 3\sigma
 34% 13.5% 2.35% 0.15% 34% 13.5% 2.35% 0.15% 68% 95% 99.7%
Figure 2The empirical rule. Every normal curve, whatever its mean and standard deviation, splits this way at whole numbers of standard deviations from the mean.

Results

Position from the z-score. The sign of z says which side of the mean a value is on, and |z| says how far, in standard deviations. Two values from different distributions, in different units, can be compared by their z-scores: the one with the larger |z| is the further from typical for its own distribution.

Changing units leaves z-scores alone. If every value is changed to a + bx with b > 0, as in a change from Celsius to Fahrenheit or from minutes to seconds, every z-score stays exactly the same. If b < 0, every z-score changes sign and keeps its size.

z-scores of a whole data set. Converting every value in a data set to its z-score, using that data set’s own mean and standard deviation, gives a new set with mean 0 and standard deviation 1. This is Standardizing a data set.

The empirical rule, slice by slice. For a normal model, the shares of values in the whole-number bands of z are about:

Band of zShare
below -30.15\%
-3 to -22.35\%
-2 to -113.5\%
-1 to 034\%
0 to 134\%
1 to 213.5\%
2 to 32.35\%
above 30.15\%

Adding neighboring slices gives the share of any range whose ends are whole numbers of standard deviations from the mean. In particular, about 16\% of values lie more than one standard deviation above the mean, and about 2.5\% more than two above.

The exact shares. Rounded to four places, the exact shares of a normal curve within one, two and three standard deviations of the mean are 0.6827, 0.9545 and 0.9973. The rule’s 68\%, 95\% and 99.7\% are these, rounded.

Why it is true

No units. In \displaystyle \frac{x - \mu}{\sigma} the top is a difference of two values, so it carries the units of x, and \sigma, a distance, carries the same units. Dividing one by the other cancels them, the way \displaystyle \frac{12 \text{ cm}}{4 \text{ cm}} = 3.

Changing units. Suppose each value x becomes x' = a + bx. The mean becomes a + b\mu and the standard deviation becomes |b|\sigma (adding a moves every value, and the mean, by a and leaves the spread alone; multiplying by b multiplies every distance by |b|, as in shifting and rescaling a data set). So the new z-score is

\begin{aligned} z' &= \frac{(a + bx) - (a + b\mu)}{|b|\sigma} \\ &= \frac{b}{|b|} \cdot \frac{x - \mu}{\sigma}. \end{aligned}

The fraction \displaystyle \frac{b}{|b|} is 1 when b > 0 and -1 when b < 0, so z' = z or z' = -z.

Mean 0 and standard deviation 1. The z-scores of a data set are its values with \bar{x} taken away and then divided by s. Taking away \bar{x} moves the mean to \bar{x} - \bar{x} = 0 and leaves the spread alone; dividing by s keeps the mean at 0 and divides the standard deviation s by s, giving 1.

One rule for every normal curve. A value x lies within k standard deviations of the mean exactly when \mu - k\sigma \le x \le \mu + k\sigma. Take away \mu and divide by \sigma (a positive number, so the inequalities keep their direction): this is the same as -k \le z \le k. As the The normal distribution lesson showed, every normal curve, read in z-scores, is one and the same curve, the standard normal curve. So the share within k standard deviations is the area of that one curve between -k and k, whatever \mu and \sigma are. Those areas, found by calculus or technology, are 0.6827, 0.9545 and 0.9973 for k = 1, 2, 3.

The slices. The curve is symmetric about its mean, so each band on the right has the same area as its mirror image on the left. Hence the share between 1 and 2 is half of 95\% - 68\%, the share between 2 and 3 is half of 99.7\% - 95\%, and each tail beyond 3 is half of 100\% - 99.7\%.

Worked examples

Example 1

Finding z-scores

Wingspans of a species of moth are modeled by N(72, 6), in millimeters. Find the z-scores of wingspans of 81 mm, 63 mm, 72 mm and 90 mm, and say what each means.

Solution

Each z-score is the distance from the mean divided by the standard deviation:

\begin{aligned} \frac{81 - 72}{6} &= \frac{9}{6} = 1.5, \\ \frac{63 - 72}{6} &= \frac{-9}{6} = -1.5, \\ \frac{72 - 72}{6} &= 0, \\ \frac{90 - 72}{6} &= \frac{18}{6} = 3. \end{aligned}

A wingspan of 81 mm is one and a half standard deviations above the mean, and 63 mm is the same distance below it: the same size of z-score, opposite signs, as two values equally far from the mean on opposite sides should have. A wingspan of 72 mm is exactly the mean. A wingspan of 90 mm is three standard deviations above the mean, out where the curve has nearly reached the axis: by the empirical rule only about 0.15\% of these moths are that wide or wider.

Example 2

From a z-score back to a value

For the same moths, N(72, 6), which wingspan has a z-score of -2.25? Which has a z-score of 0.5?

Solution

A z-score of z means z standard deviations from the mean, so the value is x = \mu + z\sigma. This comes from solving \displaystyle z = \frac{x - \mu}{\sigma} for x: multiply both sides by \sigma, then add \mu.

For z = -2.25,

x = 72 + (-2.25)(6) = 72 - 13.5 = 58.5.

For z = 0.5,

x = 72 + (0.5)(6) = 72 + 3 = 75.

A wingspan of 58.5 mm has z-score -2.25 and one of 75 mm has z-score 0.5. As a check, \displaystyle \frac{58.5 - 72}{6} = \frac{-13.5}{6} = -2.25.

Example 3

Comparing values from different distributions

At a vegetable show, Zofia’s cabbage weighs 9 kg and Thaddeus’s runner bean is 48 cm long. The cabbages at the show have mean 6.5 kg and standard deviation 1.25 kg; the runner beans have mean 42 cm and standard deviation 4 cm. Which entry is more exceptional for its kind?

Solution

The raw figures cannot be compared: the cabbage is 2.5 kg above its mean and the bean 6 cm above its mean, and kilograms and centimeters are different things. The z-scores have no units, so they can be compared. For the cabbage,

z = \frac{9 - 6.5}{1.25} = \frac{2.5}{1.25} = 2,

and for the bean,

z = \frac{48 - 42}{4} = \frac{6}{4} = 1.5.

The cabbage is two standard deviations above the mean cabbage; the bean is one and a half above the mean bean. The cabbage is further from typical for its kind, so it is the more exceptional entry, even though 2.5 looks smaller than 6.

Example 4

Using the empirical rule

Lengths of carrots from one field are modeled by N(18, 2), in centimeters. Use the empirical rule to estimate the share of carrots (a) between 14 and 22 cm, (b) longer than 20 cm, (c) between 16 and 24 cm, (d) shorter than 12 cm.

Solution

First turn each boundary into a z-score, since the rule speaks in standard deviations: 14 and 22 are z = \mp 2, 20 is z = 1, 16 is z = -1, 24 is z = 3 and 12 is z = -3.

(a) Between z = -2 and z = 2: about 95\%. The exact share for a normal curve is 0.9545.

(b) Above z = 1: the middle 68\% leaves 32\% outside, split equally between the two tails by symmetry, so about 16\% of carrots are longer than 20 cm.

(c) From z = -1 to z = 3. The part from -1 to 0 is half of 68\%, which is 34\%, and the part from 0 to 3 is half of 99.7\%, which is 49.85\%. Together that is about 83.85\%.

(d) Below z = -3: the middle 99.7\% leaves 0.3\%, half in each tail, so about 0.15\%.

The rule’s numbers are rounded, so the answers are estimates; (c) given as “about 84\%” says all the rule can honestly claim.

Example 5

Reading a model from the empirical rule

The times a batch of two-part resin takes to set hard are modeled by a normal curve, and about 95\% of batches take between 22 and 46 minutes. Find the mean and standard deviation of the model. What share of batches take longer than 40 minutes, and what share take between 22 and 40 minutes?

Solution

The middle 95\% of a normal curve runs from \mu - 2\sigma to \mu + 2\sigma. Its center is the mean, so

\mu = \frac{22 + 46}{2} = 34 \text{ minutes},

and its width is 4\sigma, so 4\sigma = 46 - 22 = 24 and \sigma = 6 minutes.

A setting time of 40 minutes has \displaystyle z = \frac{40 - 34}{6} = 1, so about 16\% of batches take longer.

From 22 minutes (z = -2) to 40 minutes (z = 1) is half of 95\% plus half of 68\%: 47.5\% + 34\% = 81.5\%.

Example 6

Finding the mean and standard deviation from two z-scores

Heights of a kind of shrub, in centimeters, have a distribution in which a shrub of 50 cm has z-score -1 and a shrub of 86 cm has z-score 2. Find the mean and the standard deviation.

Solution

Each z-score gives an equation through x = \mu + z\sigma:

\begin{aligned} 50 &= \mu - \sigma, \\ 86 &= \mu + 2\sigma. \end{aligned}

Taking the first equation from the second removes \mu: 36 = 3\sigma, so \sigma = 12 cm. Then \mu = 50 + 12 = 62 cm. In words: the two heights are 2 - (-1) = 3 standard deviations apart and 86 - 50 = 36 cm apart, so one standard deviation is 12 cm.

As a check, \displaystyle \frac{50 - 62}{12} = -1 and \displaystyle \frac{86 - 62}{12} = 2.

Example 7

When the empirical rule does not apply

In a survey of households, the number of pets per household has mean 1.2 and standard deviation 1.5. Find the z-score of a household with 6 pets. Can the empirical rule say what share of households have fewer than \mu - \sigma pets?

Solution

The z-score needs only the mean and the standard deviation, so it is fine for any distribution:

z = \frac{6 - 1.2}{1.5} = \frac{4.8}{1.5} = 3.2.

This household is 3.2 standard deviations above the mean, a long way out.

The empirical rule is another matter. One standard deviation below the mean is 1.2 - 1.5 = -0.3 pets. If the rule applied, about 16\% of households would have fewer than -0.3 pets, which is impossible: no household has a negative number of pets. The distribution is piled against 0 with a long tail to the right (skewed right), not normal, and the rule’s shares belong to normal curves only. For the same reason, the z-score of 3.2 says the household is far from the mean but cannot be turned into “about 0.07\% have more” without a model that fits.

Boundaries

The empirical rule needs a normal model. Its three numbers are areas under a normal curve. For a skewed, two-peaked or flat distribution they can be badly wrong, so check the shape first (checking normality with a plot). The z-score itself has no such condition: it is defined for any distribution with a standard deviation.

A guarantee for any shape is much weaker. For any distribution at all, at least \displaystyle 1 - \frac{1}{k^2} of the values lie within k standard deviations of the mean (Chebyshev’s inequality): at least 75\% within two and at least about 89\% within three. The normal shape is what raises these to 95\% and 99.7\%.

Whole numbers of standard deviations only. The rule gives shares between z = -3, -2, \dots, 3. The share below z = 1.5, or the value at the 90th percentile, needs the full curve: normal calculations read them from a table or technology.

The standard deviation must be positive. If \sigma = 0 every value equals the mean and \displaystyle \frac{x - \mu}{\sigma} is undefined.

Use the matching mean and standard deviation. A z-score is relative to one distribution. The same wingspan has a different z-score among a different species of moth, and z-scores computed against different distributions say how each value sits in its own group, not how the values compare in size.

Where this goes next

The z-score turns any normal curve into one curve, the standard normal distribution N(0, 1), and normal calculations use it to find the share beyond any z-score, not just the whole-number ones the empirical rule covers. Going the other way, from a share back to a value, is finding the value at a percentile. Applying the conversion to a whole data set is Standardizing a data set, and correlation can be built from the products of two variables’ z-scores (correlation from standardized values). How z-scores move when every value is shifted or scaled is part of linear transformations of a random variable.

Further on, inference keeps asking the same question this lesson asks of a wingspan: how many standard deviations from what was expected is this? A Test statistic is a z-score of a sample’s result, and the normal approximation to a binomial puts counts on the same ruler.

Exercises

Finding z-scores

  1. A variable is modeled by N(150, 20). Find the z-scores of 190, 120 and 150.
  2. A data set has mean 3.2 and standard deviation 0.4. Find the z-score of the value 2.5.
  3. A sample has mean 48 and standard deviation 8. Find the z-score of 66 and say what it means.

From a z-score back to a value

  1. In a distribution with mean 40 and standard deviation 10, which value has z-score -0.5?
  2. A model is N(200, 15). Which value has z-score 2.4?
  3. A model is N(9.5, 1.2). Which value has z-score -1.5?
  4. In a distribution with standard deviation 5, the value 64 has z-score -2. Find the mean.
  5. In one distribution the value 31 has z-score 1.25 and the value 19 has z-score -0.25. Find the mean and the standard deviation.

Comparing

  1. At an athletics club, the javelin throws have mean 46 m and standard deviation 3 m, and the shot put throws have mean 12.5 m and standard deviation 0.8 m. One athlete throws the javelin 52 m and the shot 14.1 m. Which throw is better, compared with the club?
  2. Wilhelmina builds shelf A in 40 minutes and shelf B in 25 minutes. Build times for shelf A have mean 52 minutes and standard deviation 8 minutes; for shelf B, mean 30 minutes and standard deviation 4 minutes. Shorter is better. Which build was the more impressive?
  3. Calls to a help line last a mean of 12 minutes, with standard deviation 3 minutes. Find the z-score of an 18-minute call. Each call is billed at 2 dollars plus 0.50 dollars a minute, so a call of m minutes costs 2 + 0.5m dollars. Convert the mean, the standard deviation and the 18-minute call to dollars, and find the z-score again.

The empirical rule

  1. Lifetimes of a kind of battery are modeled by N(500, 40), in hours. About what share last between 460 and 540 hours? Between 420 and 580 hours?
  2. For the same batteries, N(500, 40), about what share last longer than 580 hours? Less than 460 hours?
  3. For the same batteries, about what share last between 420 and 460 hours?
  4. For the same batteries, about what share last between 540 and 620 hours?
  5. A model is N(12, 0.5). Between which two values do the middle 99.7\% of values lie?
  6. About 95\% of the values of a normal model lie between 86 and 110. Find the mean and the standard deviation, and the share of values above 104.
  7. A machine fills cans of soup with amounts modeled by N(250, 3), in milliliters. In a batch of 600 cans, about how many hold more than 256 ml?

Explaining

  1. Explain why a z-score has no units.
  2. Explain where the 2.35\% for the band from z = 2 to z = 3 comes from, using only the rule’s three numbers and symmetry.
  3. A value in a data set that looks normal has z-score 4. What does the empirical rule suggest about how often such a value occurs, and what might its appearance mean?
  4. Explain why the z-scores of all the values in a data set have mean 0.
  5. Explain why billing every call at 2 + 0.5m dollars, as in exercise 11, leaves every z-score unchanged.
Answers
  1. 2, -1.5 and 0: \displaystyle \frac{190 - 150}{20} = 2, \displaystyle \frac{120 - 150}{20} = -1.5, and 150 is the mean.
  2. z = -1.75: \displaystyle \frac{2.5 - 3.2}{0.4} = \frac{-0.7}{0.4} = -1.75.
  3. z = 2.25: \displaystyle \frac{66 - 48}{8} = \frac{18}{8} = 2.25. The value 66 is two and a quarter standard deviations above the sample mean.
  4. 35: x = 40 + (-0.5)(10) = 35.
  5. 236: x = 200 + 2.4 \times 15 = 200 + 36 = 236.
  6. 7.7: x = 9.5 + (-1.5)(1.2) = 9.5 - 1.8 = 7.7.
  7. \mu = 74: 64 = \mu - 2 \times 5, so \mu = 64 + 10 = 74.
  8. \mu = 21, \sigma = 8. The values are 1.25 - (-0.25) = 1.5 standard deviations apart and 31 - 19 = 12 apart, so \displaystyle \sigma = \frac{12}{1.5} = 8; then \mu = 19 + 0.25 \times 8 = 21.
  9. Equally good: the javelin has \displaystyle z = \frac{52 - 46}{3} = 2 and the shot \displaystyle z = \frac{14.1 - 12.5}{0.8} = 2. Both throws are two standard deviations above the club mean.
  10. Shelf A. Its z-score is \displaystyle \frac{40 - 52}{8} = -1.5 and shelf B’s is \displaystyle \frac{25 - 30}{4} = -1.25. Shorter is better, so the more negative z-score, further below the mean, is the more impressive.
  11. z = 2 both times. In minutes, \displaystyle \frac{18 - 12}{3} = 2. In dollars the mean is 2 + 0.5 \times 12 = 8, the standard deviation 0.5 \times 3 = 1.5 (adding 2 does not change a spread) and the call costs 2 + 0.5 \times 18 = 11, so \displaystyle z = \frac{11 - 8}{1.5} = \frac{3}{1.5} = 2.
  12. About 68\% and about 95\%: 460 and 540 are z = \mp 1, and 420 and 580 are z = \mp 2.
  13. About 2.5\% and about 16\%. 580 is z = 2, and the 5\% outside \pm 2 splits into two tails of 2.5\%. 460 is z = -1, and the 32\% outside \pm 1 splits into two tails of 16\%.
  14. About 13.5\%: from z = -2 to z = -1, half of 95\% - 68\%.
  15. About 15.85\%: from z = 1 to z = 3, 13.5\% + 2.35\%.
  16. From 10.5 to 13.5: 12 \mp 3 \times 0.5.
  17. \mu = 98, \sigma = 6, and about 16\% above 104. The middle 95\% is centered at \displaystyle \frac{86 + 110}{2} = 98 and is 4\sigma = 24 wide; 104 is \displaystyle z = \frac{104 - 98}{6} = 1.
  18. About 15 cans: 256 is \displaystyle z = \frac{256 - 250}{3} = 2, about 2.5\% lie above it, and 2.5\% of 600 is 15.
  19. The top of \displaystyle \frac{x - \mu}{\sigma} is a difference of two values, in the units of the variable, and the bottom, a standard deviation, is a distance in the same units. Dividing two quantities in the same units cancels the units, leaving a pure number: a count of standard deviations.
  20. Within two standard deviations is about 95\% and within three about 99.7\%, so the two bands from 2 to 3 standard deviations, one on each side, hold 99.7\% - 95\% = 4.7\% together. A normal curve is symmetric, so the two bands are equal: \displaystyle \frac{4.7\%}{2} = 2.35\% each.
  21. By the rule only about 0.15\% of values lie more than three standard deviations above the mean, so a value four above is rarer still (the exact normal share is about 0.003\%, around one in thirty thousand). Such a value is either a genuinely unusual case, a mistake in recording, or a sign that the normal model does not fit the tails of these data. It is worth checking rather than accepting or deleting without thought.
  22. The z-scores are \displaystyle \frac{x - \bar{x}}{s}. Their mean is \displaystyle \frac{1}{s} times the mean of the differences x - \bar{x}, and those differences add to 0, because the values above the mean exactly balance those below it (that is what makes \bar{x} the mean). So the mean of the z-scores is \displaystyle \frac{0}{s} = 0.
  23. The change is m \mapsto 2 + 0.5m. Adding 2 moves every cost and the mean by the same amount, so each distance from the mean is unchanged; multiplying by 0.5 multiplies each distance from the mean and the standard deviation by 0.5. The z-score is the ratio of those two, and multiplying the top and bottom of a fraction by the same positive number leaves it unchanged.
kg1128 · updated 2026-10-19 · CC-BY-SA-4.0