Standard deviation and the normal distribution

σ inside the formula

The height of the normal curve at any point x is given by its probability density function:

f(x) = 1 / (σ√(2π)) · e^(−(x − μ)² / (2σ²))

σ appears twice, and each appearance has a job. In the exponent, the distance from the mean, x − μ, is divided by σ before it is squared. That makes σ a scale parameter: the curve only "sees" distances measured in standard deviations. A point 14 cm from the mean of a curve with σ = 7 cm has exactly the same relative height as a point 2 cm from the mean of a curve with σ = 1 cm, because both are 2σ away.

The factor in front, 1 / (σ√(2π)), is the normalising constant. It is whatever it takes to make the total area under the curve equal 1, and it contains σ because stretching the curve sideways by σ would otherwise multiply the area by σ. Substituting z = (x − μ) / σ turns dx into σ dz, and that σ cancels the one in the constant. The area is 1 for every σ.

The bell curve generator draws this function for any μ and σ.

Where you can see σ on the curve

The bell has a dome in the middle that curves downward and two tails that curve upward. The points where one kind of bend changes into the other, the inflection points, sit at exactly μ − σ and μ + σ. Differentiating the density twice gives

f″(x) = f(x) · [(x − μ)² − σ²] / σ⁴

f(x) is always positive, so the sign of f″ is the sign of (x − μ)² − σ². That is negative when x is within σ of the mean and positive beyond it, and zero precisely at μ ± σ. At those points the curve's height is e^(−1/2) ≈ 0.6065 of its peak height. So if you are handed a plotted bell curve with no labels, you can estimate σ by finding where the curve stops bending inward, or where it has dropped to about 61% of its maximum.

Changing σ: wider, flatter, same area

The peak of the curve is at x = μ, where the exponent is zero and the density is 1 / (σ√(2π)). The peak height is therefore inversely proportional to σ: double the standard deviation and the peak halves, while the curve spreads twice as wide.

The height is a density, in "share per unit", not a probability. With heights in centimetres and σ = 7, a peak of 0.0570 means roughly 5.7% of people fall in the one-centimetre slice around the mean (the exact figure for 174.5 to 175.5 cm is 5.69%). With σ = 14 the same slice holds only 2.85%, because the same total of 100% is spread over twice the range.

Standardising: every normal curve is one curve

Because the formula depends on x only through (x − μ) / σ, every normal distribution is the standard normal (mean 0, standard deviation 1) shifted and stretched. The conversion is the z-score:

z = (x − μ) / σ x = μ + z·σ

The first form turns a value into a count of standard deviations, which is what the z-score calculator does. The second form runs it backwards, turning a z from a table into a value in your own units. Any area question about any normal distribution reduces to an area under the standard normal curve, written Φ(z).

Exact coverage and tail areas

The rounded 68–95–99.7 figures are covered on the empirical rule page and in what is standard deviation. The table below gives the exact values, adds the fourth standard deviation, and splits the outside share into the two-tailed and one-tailed probabilities that tests and control charts actually use.

Two things stand out. The tails shrink much faster than the distance grows: going from 3σ to 4σ cuts the one-sided tail by a factor of more than 40. And the familiar multipliers 1.645, 1.96 and 2.576 are simply the z-values that make the coverage a round 90%, 95% and 99%. A one-sided question ("how many exceed the upper limit?") uses half the two-tailed share, because the curve is symmetric.

A worked example: adult heights

Suppose adult heights in a population are normal with μ = 175 cm and σ = 7 cm. What share of people are between 170 cm and 185 cm, and how tall is someone at the 90th percentile?

First, convert both limits to z-scores:

z₁ = (170 − 175) / 7 = −0.7143 z₂ = (185 − 175) / 7 = +1.4286

Then take the area to the left of each and subtract:

Φ(1.4286) = 0.9234 Φ(−0.7143) = 0.2375 Share between = 0.9234 − 0.2375 = 0.6859

About 68.6% of people are between 170 and 185 cm. The range is not symmetric about the mean, so the empirical rule cannot give this directly; the z-scores can.

For the 90th percentile, run the conversion the other way. The z-value with 90% of the area below it is 1.2816, so

x = μ + z·σ = 175 + 1.2816 × 7 = 183.97 cm

One in ten people is taller than about 184 cm. By the same method, the 25th percentile is 175 − 0.6745 × 7 = 170.28 cm, and the share taller than 190 cm (z = 2.1429) is 1.61%. The normal distribution calculator does the forward direction and the inverse normal calculator the percentile direction.

Why so much data is close to normal

The normal curve turns up so often because of the central limit theorem: a sum or average of many independent contributions tends toward a normal distribution, whatever the shape of each contribution, provided none of them dominates. Height is the combined result of many genes and environmental effects; measurement error is the sum of many small disturbances. Both come out close to normal.

The theorem also says what happens to σ. The mean of n independent values with standard deviation σ has a standard deviation of σ / √n, so averages are both more normal and narrower than the raw data. The central limit theorem calculator applies this to sample means.

Checking whether your data is normal enough

Every percentage above assumes a normal distribution. Real data is never exactly normal, so the practical question is whether it is close enough that the percentages are useful. Four checks, from quickest to most informative:

1. Draw a histogram. Look for a single peak, rough symmetry, and no hard wall on one side. Incomes, waiting times and anything bounded at zero with a long right tail usually fail here.
2. Compute the skewness. A normal distribution has skewness 0. As a rough guide, values between −0.5 and 0.5 are close to symmetric, and values beyond ±1 mean the normal percentages will be noticeably off in one tail.
3. Count what is inside ±1s and ±2s. With 200 observations, a normal distribution would put about 137 within one standard deviation of the mean and about 191 within two. If you find 160 and 185, the data has a sharper peak and heavier tails than a normal curve, and extreme values will be more common than the table suggests.
4. Make a normal probability plot. Sort the data and plot each value against the z-score its rank would have under a normal distribution. Normal data lies on a straight line whose slope is σ and whose intercept is μ. Curvature at the ends shows heavy or light tails; a bow across the whole plot shows skew.

What to do when it is not normal

The standard deviation is still a valid measure of spread. What you cannot do is turn it into precise percentages with the normal table. You have three options.

Use a bound that holds for any distribution. Chebyshev's theorem guarantees that at least 1 − 1/k² of the values lie within k standard deviations: at least 75% within ±2s and at least 88.9% within ±3s. These are minimums, not estimates, and they are much weaker than the normal figures. The Chebyshev's theorem calculator works them out for any k.

Describe the data with the median and the interquartile range instead. They are not pulled around by a long tail. For normal data the IQR is about 1.349σ, so the IQR divided by 1.349 gives a σ estimate that outliers cannot inflate; if that estimate is much smaller than the ordinary standard deviation, the tails are heavier than normal.

Transform the data. Right-skewed, positive data such as incomes or reaction times often becomes close to normal after taking logarithms, and the normal methods can then be used on the log scale. The lognormal distribution calculator handles this case directly.