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Year 12 Maths Standard 2 (2027) The normal distribution

The Empirical Rule (68-95-99.7)

20 practice questions 0 video lessons Theory + worked examples

Master the empirical rule (68-95-99.7) for NSW Year 12 Mathematics Standard 2. For normally distributed data, about 68% of the values lie within one standard deviation of the mean, about 95% within two, and about 99.7% within three - the fast way to read percentages off a bell curve.

You will learn to find the interval that holds a given percentage, work out the percentage in a one-sided or across-the-mean region using the 34%, 13.5% and 2.35% segments, and estimate how many of a batch fall in a range with expected frequency - core Standard 2 skills for heights, masses, test scores and machine-fill volumes.

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Theory

The empirical rule (68-95-99.7) tells you how normally distributed data spreads around the mean. This Year 12 Standard 2 (NSW) guide shows how to find the percentage of data within 1, 2 or 3 standard deviations, handle more-than and less-than regions, and estimate how many of a batch fall in a range.

The empirical rule (also called the 68-95-99.7 rule) describes how normally distributed data spreads around the mean. For a bell-shaped data set, a fixed percentage of the values falls within each whole number of standard deviations (SD) of the mean.

About 68% of the data lies within 1 SD of the mean, about 95% within 2 SD, and about 99.7% within 3 SD. Because the curve is symmetric, each half breaks into segments of 34% (mean to 1 SD), 13.5% (1 to 2 SD) and 2.35% (2 to 3 SD).

This Year 12 Standard 2 (NSW) tool lets you read off the percentage of data in a region, work out a more-than or less-than percentage, or estimate how many of a batch fall in a range \((\text{percentage}\times n)\) - all without a \(z\)-table, as long as the boundaries sit at whole standard deviations.

The empirical rule percentagesA normal curve with the per-segment percentages 2.35, 13.5, 34, 34, 13.5, 2.35 marked between the standard-deviation lines; they add to 68, 95 and 99.7 percent. 2.35% 13.5% 34% 34% 13.5% 2.35% x 20 30 40 50 60 70 80
The six segments (2.35%, 13.5%, 34%) add to 68%, 95% and 99.7%.
68% within one standard deviationA normal curve, mean 50, SD 10; the region from 40 to 60 (one SD each side) is shaded and holds about 68 percent of the data. x 20 30 40 50 60 70 80
About \(68\%\) of the data lies within one standard deviation (here \(40\) to \(60\)).

Take the mean \(\mu\) and standard deviation \(\sigma\). The middle band that reaches \(k\) standard deviations each side of the mean is:

\[\mu \pm 1\sigma \approx 68\%,\quad \mu \pm 2\sigma \approx 95\%,\quad \mu \pm 3\sigma \approx 99.7\%\]
μ±1σ68%,μ±2σ95%,μ±3σ99.7%

The percentage in each segment (one side of the mean) is fixed:

Segmentmean to 1 SD1 to 2 SD2 to 3 SD
Percentage34%13.5%2.35%

Each half of the curve holds \(50\%\), so for a one-sided region start from the mean and add or subtract segments:

\[P(X > \mu + 1\sigma) = \tfrac{100\% - 68\%}{2} = 16\%\]
P(X>μ+1σ)=16%
Expected count. To find how many of \(n\) items fall in a range, multiply: \(\text{expected number} = \text{percentage} \times n\).

Using the empirical rule

  1. Write down the mean \(\mu\) and standard deviation \(\sigma\).
  2. Count the SDs. For each boundary work out \(k = \dfrac{\text{value}-\mu}{\sigma}\); it should be a whole number \(1\), \(2\) or \(3\).
  3. Sketch a bell curve and mark the region you want.
  4. Add the segments. Use \(34\%\), \(13.5\%\), \(2.35\%\) for each band, and \(50\%\) for a whole half; for a one-sided region start at the mean.
  5. For a count, multiply the percentage by the number of items \(n\).
Example 1 — 95% interval
The masses of newborn babies are normally distributed with a mean of \(3.4\) kg and a standard deviation of \(0.4\) kg. Between which two masses do about \(95\%\) of the babies lie?
Solution

The middle \(95\%\) reaches two SD each side of the mean.

Baby masses, mean 3.4 kgBaby mass m (kg) normally distributed, mean 3.4, SD 0.4; the region from 2.6 to 4.2 kg (two SD each side) is shaded and holds about 95 percent of the babies. m 2.2 2.6 3 3.4 3.8 4.2 4.6
\(2\,\text{SD}\)\(=\)\(2 \times 0.4 = 0.8\)
\(\text{lower}\)\(=\)\(3.4 - 0.8 = 2.6\)
\(\text{upper}\)\(=\)\(3.4 + 0.8 = 4.2\)
2.6 kg to 4.2 kg

About \(95\%\) have a mass between 2.6 kg and 4.2 kg.

Example 2 — One-sided percentage
The time to commute to work is normally distributed with a mean of \(30\) min and a standard deviation of \(5\) min. What percentage of commutes take more than 35 minutes?
Solution

\(35 = 30 + 1\times 5\), so \(35\) min is one SD above the mean.

Commute time, mean 30 minCommute time t (minutes) normally distributed, mean 30, SD 5; the tail above 35 min (one SD above the mean) is shaded and holds about 16 percent of the commutes. t 15 20 25 30 35 40 45
\(\text{within }1\,\text{SD}\)\(=\)\(68\%\)
\(\text{both tails}\)\(=\)\(100\% - 68\% = 32\%\)
\(\text{one tail}\)\(=\)\(\dfrac{32\%}{2} = 16\%\)
16%

About 16% of the commutes take more than \(35\) minutes.

Example 3 — Region across the mean
A machine fills cups with a mean of \(200\) mL and a standard deviation of \(10\) mL (normally distributed). What percentage of cups hold between 190 and 220 mL?
Solution

Find how many SDs each boundary is from the mean, then add the segments.

Cup fill, mean 200 mLCup volume V (mL) normally distributed, mean 200, SD 10; the region from 190 (one SD below) to 220 (two SD above) is shaded and holds about 81.5 percent of the cups. V 170 180 190 200 210 220 230
\(190\)\(=\)\(200 - 1\times 10\ \ (1\,\text{SD below})\)
\(220\)\(=\)\(200 + 2\times 10\ \ (2\,\text{SD above})\)
\(\text{total}\)\(=\)\(34\% + 34\% + 13.5\%\)
\(\)\(=\)\(81.5\%\)
81.5%

About 81.5% of the cups hold between \(190\) and \(220\) mL.

Example 4 — How many of \(n\)
The heights of \(600\) tomato plants are normally distributed with a mean of \(80\) cm and a standard deviation of \(6\) cm. How many plants are expected to be between 68 and 92 cm tall?
Solution

\(68\) and \(92\) are two SD each side of the mean, so use \(95\%\).

Plant heights, mean 80 cmPlant height h (cm) normally distributed, mean 80, SD 6; the region from 68 to 92 cm (two SD each side) is shaded and holds about 95 percent of the plants. h 62 68 74 80 86 92 98
\(68\)\(=\)\(80 - 2\times 6,\ \ 92 = 80 + 2\times 6\)
\(\text{within }2\,\text{SD}\)\(=\)\(95\%\)
\(\text{count}\)\(=\)\(0.95 \times 600 = 570\)
570 plants

About 570 of the \(600\) plants are between \(68\) and \(92\) cm tall.

Common pitfalls

Only whole SD boundaries. The empirical rule works when a value is exactly \(1\), \(2\) or \(3\) SD from the mean. For an in-between value you need a \(z\)-score, not this rule.
Within vs more/less than. "Within" is two-sided; "more than" or "less than" is one-sided and splits at the mean, so begin from \(50\%\).
Do not double count. \(68\%\) already covers both sides; a single segment from the mean to \(1\) SD is \(34\%\), not \(68\%\).

Frequently asked questions

What is the 68-95-99.7 rule?

It is the empirical rule for normally distributed data: about 68% of the values lie within one standard deviation of the mean, about 95% within two standard deviations, and about 99.7% within three standard deviations.

How do I find the range that holds 95% of the data?

Go two standard deviations each side of the mean. The 95% range is from mean minus 2 times SD up to mean plus 2 times SD. For 68% use one SD each side, and for 99.7% use three SD each side.

What percentage is between the mean and one standard deviation above it?

About 34%. The 68% within one SD is split evenly by the symmetric curve, so each side holds 34%. The next band (1 to 2 SD) holds 13.5% and the outer band (2 to 3 SD) holds 2.35%.

How do I work out a 'more than' or 'less than' percentage?

A one-sided region splits at the mean, where 50% lies on each side. Start from 50% and add or subtract the 34%, 13.5% and 2.35% segments up to your boundary. For example, more than one SD above the mean is (100% minus 68%) divided by 2, which is 16%.

When can I not use the empirical rule?

Only when the boundary is a whole number of standard deviations from the mean (1, 2 or 3 SD). If the value falls between whole SDs you must calculate a z-score and use a z-table or calculator instead.

How do I find how many items are expected in a range?

Work out the percentage of data in the range using the empirical rule, then multiply by the total number of items n. For example, 95% of 600 items is 0.95 times 600, which is 570 items.