From a Familiar Rule to Useful Probability Estimates
A standard normal table or calculator can provide probabilities for many cutoffs. But some questions can be answered quickly without either: if a value lies between one, two, or three standard deviations from the mean, the 68-95-99.7 rule gives a useful estimate. Symmetry then lets us split those central areas into left and right pieces and combine pieces for other ranges.
The aim here is not to replace table or software calculations. It is to recognize when a question lines up with the rule and make a transparent approximation. The result is an estimated area under a normal curve, and therefore an estimated probability under the stated model.
Here, \(\mu\) is the mean and \(\sigma\) is the standard deviation. The rule is approximate, even when the variable follows a normal model. It is most directly useful for cutoffs at \(\mu\pm\sigma\), \(\mu\pm2\sigma\), and \(\mu\pm3\sigma\), or for combinations of ranges bounded by those cutoffs.
Breaking the Curve into Bands
The central areas in the rule include several smaller bands. Subtracting one central area from another gives the combined area in the bands between their boundaries. Symmetry of a normal distribution divides each such combined area equally between its left and right sides.
The area beyond three standard deviations on both sides is \(1-0.997=0.003\), so each outer tail has approximate area \(0.003/2=0.0015\). The area between the mean and one standard deviation on either side is \(0.68/2=0.34\). These pieces give a convenient set of approximate areas to combine.
| Region relative to the mean | Approximate area |
|---|---|
| From the mean to 1 SD on one side | 0.34 |
| Between 1 and 2 SDs on one side | 0.135 |
| Between 2 and 3 SDs on one side | 0.0235 |
| Beyond 3 SDs in one tail | 0.0015 |
“On one side” means either the left or the right side of the mean. For example, the area between \(\mu+\sigma\) and \(\mu+2\sigma\) is approximately 0.135; the area between \(\mu-2\sigma\) and \(\mu-\sigma\) is also approximately 0.135. The same reasoning applies to the two bands between two and three standard deviations.
Worked Applications
Before using the rule, specify the model. Each example below assumes the quantity of interest follows the stated normal distribution. Thus, the rule applies to that model; if the distribution of real measurements is not reasonably normal, the estimates may not describe those measurements well.
Worked Example: A Range One Standard Deviation from the Mean
Suppose a hypothetical measurement \(X\) follows a normal model with mean 50 units and standard deviation 4 units. Estimate the probability that a measurement is between 46 and 54 units.
Check the model and cutoffs. The problem stipulates \(X\sim N(50,4^2)\). The requested endpoints are \(50-4=46\) and \(50+4=54\), exactly one standard deviation below and above the mean.
Apply the rule. This is the central interval within one standard deviation: $$ P(46<X<54)=P(\mu-\sigma<X<\mu+\sigma)\approx 0.68. $$
The estimated probability is 0.68, or about 68%. In this hypothetical normal model, about 68% of measurements are expected to fall between 46 and 54 units. This is an approximate model probability, not a claim that a particular sample must contain exactly 68% of its measurements in that range.
Worked Example: An Asymmetric Range
A hypothetical test score \(X\) is modeled as normal with mean 72 points and standard deviation 6 points. Estimate the probability that a score is between 60 and 78 points.
Check the model and convert the endpoints. The stipulated model is \(X\sim N(72,6^2)\). The lower endpoint is \(72-2(6)=60\), or two standard deviations below the mean. The upper endpoint is \(72+6=78\), or one standard deviation above it. The interval is therefore from \(\mu-2\sigma\) to \(\mu+\sigma\).
Split at the mean. From two standard deviations below the mean up to the mean is half of the central 95% area, approximately \(0.95/2=0.475\). From the mean to one standard deviation above it is half of the central 68% area, approximately \(0.68/2=0.34\). Adding the two non-overlapping areas gives $$ P(60<X<78)\approx 0.475+0.34=0.815. $$
The estimated probability is 0.815, or about 81.5%. In this model, about 81.5% of scores fall between 60 and 78 points. The interval is asymmetric around the mean, so it would be incorrect to use the central 68% or 95% as the answer without splitting and matching the requested bounds.
Worked Example: A Range Between Two Negative Cutoffs
A hypothetical package weight \(X\) is modeled as normal with mean 500 grams and standard deviation 10 grams. Estimate the probability that a package weighs between 470 and 490 grams.
Check the model and convert the endpoints. The stated model is \(X\sim N(500,10^2)\). Since \(470=500-3(10)\) and \(490=500-1(10)\), the requested range runs from three standard deviations below the mean to one standard deviation below it.
Add the two bands in that range. The band from three to two standard deviations below the mean has area approximately 0.0235. The band from two to one standard deviation below the mean has area approximately 0.135. Therefore, $$ P(470<X<490)\approx 0.0235+0.135=0.1585. $$
The estimated probability is 0.1585, or about 15.9%. In this hypothetical model, about 15.9% of packages weigh between 470 and 490 grams. Both cutoffs are on the same side of the mean, so the relevant area is the sum of the bands from \(-3\sigma\) to \(-2\sigma\) and from \(-2\sigma\) to \(-\sigma\), not a central area.
Worked Example: A One-Sided Tail Beyond Two Standard Deviations
Suppose a hypothetical laboratory result \(X\) follows a normal model with mean 100 units and standard deviation 8 units. Estimate the probability that a result exceeds 116 units.
Check the model and cutoff. The stipulated model is \(X\sim N(100,8^2)\), and \(116=100+2(8)\), so the cutoff is two standard deviations above the mean.
The central 95% lies between two standard deviations below and above the mean. The area outside that interval is approximately \(1-0.95=0.05\), split equally between the two tails by symmetry. Thus the right-tail probability is $$ P(X>116)\approx \frac{1-0.95}{2}=0.025. $$
The estimated probability is 0.025, or about 2.5%. In this model, about 2.5% of results exceed 116 units. The calculation uses the area in one tail; 5% is the combined area in both tails.
A Short Justification for the Band Areas
The arithmetic behind the band method follows directly from the nested central intervals and symmetry. The interval within one standard deviation is contained inside the interval within two standard deviations. Removing the smaller central interval from the larger leaves two equal bands, one on each side of the mean. Their combined probability is approximately \(0.95-0.68=0.27\); because the normal distribution is symmetric, either band has probability approximately \(0.27/2=0.135\).
Likewise, removing the central 95% interval from the central 99.7% interval leaves the two bands between two and three standard deviations. Their combined probability is approximately \(0.997-0.95=0.047\), so each has probability approximately \(0.047/2=0.0235\). Finally, the region beyond three standard deviations has total probability approximately \(1-0.997=0.003\), and symmetry gives approximately \(0.0015\) to each tail. This derivation justifies the pieces used in the examples; adding disjoint pieces gives the probability of their union.
When This Method Helps—and Its Limits
This method is useful when the question uses cutoffs exactly one, two, or three standard deviations from the mean, or when the requested interval can be assembled from the bands between those cutoffs. It also provides a quick reasonableness check: an interval from \(\mu-\sigma\) to \(\mu+\sigma\) should have an area near 0.68, while a single tail beyond \(\mu+2\sigma\) should have an area near 0.025.
The rule is not a table of exact probabilities for all possible cutoffs. If an endpoint is, for example, 1.4 standard deviations from the mean, the bands above do not determine the exact area up to that endpoint. A standard normal table or software is needed for a more precise calculation. Even when the endpoints align with the rule, report the result as approximate and avoid implying more precision than the rule supports.
The model matters as much as the arithmetic. The percentages describe areas under a normal curve. Applying them to actual measurements is appropriate only when a normal model is suitable for the situation. In examples where normality is stipulated, the calculation is conditional on that model; in an applied setting, the shape of the data and the context should support using a normal approximation.
Common Mistake
A frequent error is to treat 95% as the area in each tail beyond two standard deviations. In fact, 95% is the combined central area from \(\mu-2\sigma\) to \(\mu+2\sigma\). The area outside that central interval is approximately 5% altogether, and symmetry divides it into about 2.5% per tail. Another error is to use the full 27% between one and two standard deviations for just one side; that 27% combines the left and right bands, so one side has about 13.5%.
A reliable check is to ask whether the region is central, one-sided, or split across both sides of the mean. Identify its boundaries in standard-deviation units, use symmetry when appropriate, and add only the non-overlapping pieces inside the requested region.
Check Your Understanding
Use the 68-95-99.7 rule and symmetry to estimate each probability. Assume the stated variables follow normal models.
- For \(X\sim N(40,5^2)\), estimate \(P(35<X<45)\). State what interval this is relative to the mean and standard deviation.
- For a normal model with mean 80 and standard deviation 4, estimate the probability of a value between 76 and 84.
- For a normal model with mean 200 and standard deviation 10, estimate the probability of a value between 180 and 210. Show how you split the interval.
- For a normal model with mean 60 and standard deviation 3, estimate the probability of a value greater than 66. Explain why the answer is not 5%.
- Why is the area between one and two standard deviations above the mean approximately 0.135 rather than 0.27?