Fit a normal distribution to a data set and estimate population percentages from areas under the curve

Learning goal: Use the mean and standard deviation of a data set to fit it to a normal distribution, estimate population percentages with the 68–95–99.7 rule and with areas under the normal curve found from tables, calculators and spreadsheets, and recognize the data sets for which this procedure is not appropriate.

What a normal distribution is. A normal distribution is a model for a measurement variable. Its graph is a smooth, symmetric, bell-shaped curve: highest at the center, falling away equally on both sides, never quite touching the horizontal axis. Two numbers fix a normal curve completely. The mean μ sits at the center and under the peak; the standard deviation σ sets how wide the bell is, because it is the horizontal distance from the center to the point where the curve changes from bending downward (concave down) to bending upward (concave up). The curve is scaled so that the total area under it is 1, which stands for 100% of the population. That gives the rule that makes the model useful: the area under the curve between two values is the proportion of the population between those values. Every percentage question about a normal population is an area question.

The 68–95–99.7 rule. For every normal distribution, whatever its mean and standard deviation, about 68% of the population lies within one standard deviation of the mean, about 95% within two, and about 99.7% within three. (To two decimal places the percentages are 68.27%, 95.45% and 99.73%.) Because the curve is symmetric, each of those bands splits evenly: about 34% lies between μ and μ + σ, about 13.5% between μ + σ and μ + 2σ, about 2.35% between μ + 2σ and μ + 3σ, and about 0.15% beyond μ + 3σ, with the same figures on the left side. Half of the population, 50%, lies below the mean.

Fitting a data set. The rule is only worth applying when the data look like the model, so the fit has to be checked. Here is the hypothetical data set used through this lesson: the masses, in grams, of 20 granola bars taken from a production line. The numbers are invented for teaching and chosen so that the arithmetic is clean.

35.3, 36.5, 37.5, 38.4, 38.9, 39.2, 39.6, 39.8, 39.9, 40.1, 40.4, 40.5, 40.8, 40.9, 40.9, 41.4, 41.5, 42.2, 42.9, 43.3

The sum is 800.0 g, so the mean is x̄ = 800.0 ÷ 20 = 40.0 g. The standard deviation adds the squared deviations from the mean, divides and takes the square root. The squared deviations total 76.24; this lesson uses the sample standard deviation, which divides by n − 1 = 19 and is what a calculator labels s_x: s = √(76.24 ÷ 19) = √4.013 = 2.003, so s = 2.00 g to two decimals. (Dividing by n instead gives σ_x = 1.95 g; either is acceptable if you say which you used. This lesson uses s throughout.) The fitted model is therefore a normal distribution with mean 40.0 g and standard deviation 2.00 g.

Now check whether the data behave like that model. The median is 40.25 g, close to the mean, and the values spread out about equally on both sides of 40, so the shape is roughly symmetric with a single peak. Count how many of the 20 values fall within one standard deviation, from 38.0 to 42.0 g: 38.4 through 41.5, which is 14 values, or 70%; the rule predicts about 68%. Within two standard deviations, 36.0 to 44.0 g, lie 19 of 20 values, or 95%, exactly the rule's figure; only 35.3 g lies outside. All 20, or 100%, lie within three, against a predicted 99.7%. A sample of 20 will never match the model exactly, and 70% against 68% is the kind of agreement that a normal model is expected to show. The fit is reasonable, and the model can now be used.

Estimating population percentages. The point of fitting a model is to say something about the whole population of bars the line produces, not only the 20 that were weighed. What percentage of bars have a mass below 38 g? Since 38 = 40 − 2 is exactly one standard deviation below the mean, the area to its left is the 50% below the mean minus the 34% between 38 and 40: 16%. What percentage lies between 36 and 42 g? That band runs from μ − 2σ to μ + σ: 13.5% + 34% + 34% = 81.5%. What percentage exceeds 44 g, which is μ + 2σ? About 2.5%, or 2.28% from a table. In a production run of 5,000 bars, that predicts roughly 0.0228 × 5,000 ≈ 114 bars over 44 g (about 125 with the rounded 2.5%). None of these percentages is a count from the sample; they are estimates for the population made from the model.

Cutoffs that are not whole standard deviations. The 68–95–99.7 rule only helps when the cutoff sits exactly one, two or three standard deviations from the mean. For any other cutoff, first convert the value to a z-score, the number of standard deviations it lies from the mean: z = (x − μ) ÷ σ. Then find the area with a standard normal table, a calculator or a spreadsheet. A standard normal table lists, for each positive z, the area under the curve to the left of z; the extract below is what such a table shows, and the values are also what a spreadsheet returns.

z = 0.5 → 0.6915; z = 1.0 → 0.8413; z = 1.5 → 0.9332; z = 2.0 → 0.9772; z = 2.5 → 0.9938; z = 3.0 → 0.9987.

For a negative z use symmetry: the area to the left of −z equals 1 minus the area to the left of z. Many graphing calculators have a function of the form normalcdf(lower, upper, mean, standard deviation) that returns the area between two values directly. A spreadsheet function of the form NORM.DIST(x, mean, standard deviation, TRUE) returns the area to the left of x. Whichever tool is used, sketch the curve, shade the region wanted and check that the answer is the area of the shaded part, not its complement.

Model calculation: what percentage of bars have a mass below 37 g? z = (37 − 40) ÷ 2 = −1.5. The table gives 0.9332 to the left of +1.5, so the area to the left of −1.5 is 1 − 0.9332 = 0.0668, about 6.7%. Between 38 g and 43 g? The z-scores are −1.0 and +1.5; the area is 0.9332 − (1 − 0.8413) = 0.9332 − 0.1587 = 0.7745, about 77.5%. Above 43.5 g? z = 1.75, which is between the table's rows, so a calculator or spreadsheet is the right tool: the area to the right is 0.0401, about 4.0%.

When the procedure is not appropriate. Every data set has a mean and a standard deviation, but not every data set is shaped like a bell, and fitting a normal curve to the wrong data produces confident nonsense. A second hypothetical data set records how many days overdue 20 library books were when returned:

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 5, 6, 9, 14, 21, 33

The mean is 101 ÷ 20 = 5.05 days and s = 8.54 days, but the median is 1.5 days, far below the mean, and seven of the 20 values sit at the minimum, 0. The distribution is strongly skewed to the right: a pile at zero and a long tail. A normal model with μ = 5.05 and σ = 8.54 would put 27.7% of the population below 0 days, since z = (0 − 5.05) ÷ 8.54 = −0.59 and the area to its left is 0.2772, yet a negative number of overdue days is impossible. The empirical-rule check fails too: 17 of 20 values, 85%, lie within one standard deviation, not 68%. The numbers x̄ and s exist, but the procedure of fitting a normal distribution to them is not appropriate. The same is true of data with two peaks, such as the heights of a group made of eight-year-olds and their parents, and of small counts pressed against a boundary. The deciding check is always the shape: plot the data first, look for one peak, rough symmetry and thin tails, then confirm that the proportions within one and two standard deviations are near 68% and 95%. If those checks fail, describe the distribution with a five-number summary instead and do not report normal-curve percentages.

Worked area between two values: What percentage of bars have masses between 39 g and 41 g under the model with mean 40 g and standard deviation 2 g? The z-scores are (39 − 40) ÷ 2 = −0.5 and (41 − 40) ÷ 2 = +0.5. The area to the left of +0.5 is 0.6915; by symmetry the area to the left of −0.5 is 1 − 0.6915 = 0.3085. Subtract: 0.6915 − 0.3085 = 0.3830, so about 38.3% of bars lie within one gram of 40 g, that is, within half a standard deviation. A calculator's normalcdf(39, 41, 40, 2) returns the same 0.3830 in one step.

Worked cutoff from a percentage: The line manager wants the mass exceeded by only the heaviest 2.5% of bars. Working backwards through the rule, 2.5% is the area beyond μ + 2σ, so the cutoff is 40 + 2 × 2 = 44 g. For the lightest 16%, the cutoff is μ − σ = 38 g. Reading the table backwards works the same way: find the area, read off z, then convert with x = μ + zσ.

Worked count from a percentage: Bars under 37 g are rejected. In a run of 5,000 bars, how many rejections does the model predict? The area to the left of z = −1.5 is 0.0668, and 0.0668 × 5,000 = 334, so about 334 rejections. The answer is an estimate for the population produced by the line, not a count of the 20 weighed bars, of which only two (35.3 and 36.5 g) were under 37 g.

Worked recognition of an inappropriate fit: A survey records how many pets each of 40 hypothetical households owns: 22 answer 0, 11 answer 1, 5 answer 2, 1 answers 3 and 1 answers 9. The mean is (0 + 11 + 10 + 3 + 9) ÷ 40 = 0.825 pets, but more than half the households sit at the minimum, values are whole numbers, and the single 9 pulls the mean and standard deviation up. The shape is skewed and piled at a boundary. Reporting the percentage of households with a negative number of pets, which a normal model would do, is meaningless; the procedure is not appropriate here.

Represent and check: Draw a histogram of the 20 granola-bar masses using 1-gram bins from 35 to 44. Mark the mean at 40.0 and the points 38.0, 42.0, 36.0 and 44.0 on the axis. Count the bars inside each band and write the three proportions (within 1, 2 and 3 standard deviations) beside the histogram with the rule's 68%, 95% and 99.7% next to them. Then sketch the fitted normal curve over the histogram and shade the region for masses below 37 g; label its area 6.7%.

Independent performance: Collect at least 30 measurements of one continuous quantity, for example the time to walk a fixed corridor, hand span, or the mass of a set of identical-looking objects. (1) Compute the mean and standard deviation and say which standard deviation, s or σ, you used. (2) Plot the data and decide, with reasons about shape, symmetry and the 68–95 check, whether a normal model is appropriate. (3) If it is, estimate two population percentages for cutoffs of your choosing, one using the 68–95–99.7 rule and one using a z-score with a table and then a calculator or spreadsheet; the two tools must agree. (4) If it is not, explain which feature of the data breaks the model and what the model would wrongly predict. Success means every percentage is tied to a shaded region of a sketched curve, the z-scores are shown, and the appropriateness decision is argued from the data rather than assumed.