Normal Distribution

The normal distribution is a continuous probability model with a symmetric, bell-shaped density. Its two parameters are the mean $\mu$ and the positive standard deviation $\sigma$ . The mean locates the centre; the standard deviation sets the horizontal scale.

A normal curve is a model, not a default label for all numerical data. It can describe some measurements or sampling distributions well, but its suitability must be checked against the context and observations. NIST’s normal-distribution reference gives its density, cumulative probabilities and main properties.

On this page

Read the parameters and notation

This page writes

X\sim N(\mu,\sigma^2),\qquad \sigma>0,

to identify a normal distribution with mean $\mu$ and variance $\sigma^2$ . Some software requests the standard deviation rather than the variance, so always check what its second input means.

For the synthetic model $X\sim N(50,5^2)$ , the mean is 50, the standard deviation is 5, and the variance is 25. The model’s mean, median and mode coincide at 50.

The density is

f(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).

It is positive at every real input, approaches zero in either tail, and has total area 1. That unbounded support matters: a normal model can assign small probabilities to impossible physical values, so it may be unsuitable near an important boundary.

Probability is area, not curve height

For a continuous normal variable, the probability of falling between two numbers is the area under the curve between them. The curve’s height at a point is a density, not the probability of that exact value.

P(a<X<b)=\int_a^b f(x)\,dx.

You do not need to evaluate this integral by hand to use a normal model. A cumulative probability table or statistical tool supplies the relevant areas.

A normal probability is an area

Specified mathematical model N(50, 5²), not a fit to observed learner data.

Diagram explanation and exact data

The highlighted area from 45 to 55 equals P(−1 < Z < 1), approximately 0.682689. The total area over the full real line is 1; the plotted window does not truncate the mathematical distribution.

A normal probability is an area: exact reference data
Quantity	Value
Mean	50
Standard deviation	5
Variance	25
P(45 < X < 55)	0.6826894921
P(X > 60)	0.02275013195
90th percentile	56.40775783

In a continuous model, $P(X=50)=0$ . This does not mean a recorded measurement can never read 50: a rounded recording usually represents a small interval of underlying values. For interval probabilities, including or excluding an exact endpoint makes no difference under the continuous model.

A density height can exceed 1 when the curve is narrow. The restriction is that total area, not maximum height, equals 1.

Standardise to the standard normal

The standard normal variable has mean 0 and standard deviation 1. If $X$ follows the stated normal model, then

Z=\frac{X-\mu}{\sigma}\sim N(0,1).

For $X\sim N(50,5^2)$ , a value of 60 corresponds to

z=\frac{60-50}{5}=2.

It is two standard deviations above the mean. A value of 45 corresponds to $z=-1$ . Standardising changes the units, not the underlying probability assigned to corresponding intervals.

The symbol $\Phi(z)$ denotes the cumulative probability $P(Z\leq z)$ . Some printed tables instead give the area from 0 to $z$ . Read the table’s heading before substituting its entry into a calculation.

Calculate a central interval

For the interval from 45 to 55,

P(45<X<55) =P(-1<Z<1) =\Phi(1)-\Phi(-1) \approx0.682689.

The model therefore assigns about 68.27% probability to this interval. It does not require exactly 68 or 69 of the next 100 observations to fall inside it.

For any normal distribution, the familiar empirical rule is approximately:

Calculate a central interval: reference table
Distance from the mean	Model probability
Within one standard deviation	68.27%
Within two standard deviations	95.45%
Within three standard deviations	99.73%

These percentages are consequences of the normal model. They are not automatic guarantees for arbitrary symmetric or numerical datasets.

Calculate a tail probability

For a value above 60 in the same model,

P(X>60)=P(Z>2)=1-\Phi(2)\approx0.022750.

That is about 2.28%. The cumulative probability below 60 is about 97.72%, not 2.28%.

Symmetry gives $P(X<40)\approx0.022750$ . Combining both outer tails gives $P(|X-50|>10)\approx0.045500$ . Distinguish a one-tail question from a two-tail question before reading the probability.

A shaded interval and a written probability statement should refer to the same event. A common graphing error is to shade the middle while reporting the tail, or to report one tail for a two-sided question.

Find a value from a percentile

The 90th percentile is the value below which the model places 90% probability. First find the corresponding standard-normal quantile:

z_{0.90}=\Phi^{-1}(0.90)\approx1.281552.

Convert back to the original scale:

x_{0.90}=50+5(1.281552)\approx56.408.

About 10% of the model’s outcomes exceed this value. The 90th percentile is not 90% of the maximum: a normal distribution has no finite maximum.

For a central 95% interval, the two remaining tails each contain 2.5%, so use approximately $z=\pm1.959964$ . For this model the interval is approximately $[40.200,59.800]$ . It is an interval for individual modelled values, not a confidence interval for an unknown mean.

Check whether a normal model is reasonable

Inspect the data for asymmetry, multiple groups, extreme tails and boundaries. Consider how the measurements were produced. A mixture of two different populations may have two peaks even when each group individually has a simpler distribution.

A histogram and a normal probability plot can help assess shape, but small samples are limited evidence. Rounding and discrete outcomes also affect how a continuous model fits.

Keep the distribution of individuals separate from the distribution of a sample mean. Raw data need not be normal for a large-sample mean approximation to be useful under suitable conditions. Conversely, a bell-shaped histogram cannot establish independent observations or unbiased sampling.

Practice questions

For $X\sim N(50,5^2)$ , what are the mean, standard deviation and variance?
Find the z-score for 60 and the approximate probability of a value above 60.
Is the interval containing about 95% of individual modelled values automatically a 95% confidence interval for the population mean?

Check your answers

The mean is 50, the standard deviation is 5, and the variance is 25. This page uses variance in the second position of $N(\mu,\sigma^2)$ .
$z=(60-50)/5=2$ . The upper-tail probability is $1-\Phi(2)\approx0.02275$ , or about 2.28%.
No. An individual-value interval describes population spread. A confidence interval estimates an unknown parameter using sampling variability and a specified procedure.

Sources and further reading

NIST/SEMATECH handbook: Normal distribution.