# How to interpret the results of the R pnorm and analogous functions?

The help for the pnorm function states:

It says that pnorm gives the "distribution function", but it seems that it gives the quantile,

for example, pnorm(q = 0, 0, 1) returns 0.5, which suggests that q=0 refers to the 50th quantile of a N(0,1). I understand what the "normal probability density function" is, but not why pnorm is called a 'distribution function'.

The R help says that the q argument is a "vector of quantiles", but it appears in practice that q represents an observed value.

What I want to know is: if I observe '2', what does pnorm(2) say about my assumption that it came from a N(0,1) distribution?

• Not all distributions have a density function. All of them have a cumulative distribution function (CDF), which thereby deserves to be called "distribution function." Everything else can be derived from it. In particular, its inverse is the quantile function.
– whuber
Commented Jun 29, 2012 at 20:48
• The statement "R help says that the q argument is a vector of probabilities ..." is incorrect. The documentation under ?pnorm clearly says that the argument q is a vector of quantiles. Commented Jun 30, 2012 at 21:28
• @atiretoo I have fixed this typo, but I am still confused about why pnorm(q=2) would work if q is a quantile.
– Abe
Commented Jul 6, 2012 at 20:19
• A percentile is a value of a variable corresponding to a certain percent (which is a value of the cumulative distribution function). Thus the q=2 in an expression like pnorm(q=2) refers to a value of a standard normal distribution. The result, 0.977 = 97.7%, says that 97.7% of a normal distribution lies at or below $2$.
– whuber
Commented Jul 6, 2012 at 20:48
• What is 50th quantile? Median? If so 0.5 is definitely not the median of $N(0,1)$ distribution. Commented Jul 9, 2012 at 13:59

pnorm gives you the cumulative probability distribution at a specified value of $$x$$. This is the cumulative probability for a standard normal distribution. So in your example, the quantity I call $$x$$ is specified as $$0$$ and $$0.5$$ as the answer is $$P[X\leq 0]$$ for a random variable $$X$$ with a $$N(0,1)$$ distribution. If you took $$x=1.96$$ you would get $$0.975$$ because a standard normal distribution has a probability of $$0.025$$ in the upper right tail above $$1.96$$.
• Michael: your 2*(1-pnorm(x)) only makes sense for non-negative $x$ Commented Jun 29, 2012 at 22:47
• @Macro: Your second version should be much, much more accurate if the absolute value of $x$ is of any appreciable size. Commented Jul 2, 2012 at 18:14