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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?

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    $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Commented Jun 29, 2012 at 20:48
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    $\begingroup$ 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. $\endgroup$
    – atiretoo
    Commented Jun 30, 2012 at 21:28
  • $\begingroup$ @atiretoo I have fixed this typo, but I am still confused about why pnorm(q=2) would work if q is a quantile. $\endgroup$
    – Abe
    Commented Jul 6, 2012 at 20:19
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    $\begingroup$ 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 $2$. $\endgroup$
    – whuber ♦
    Commented Jul 6, 2012 at 20:48
  • $\begingroup$ What is 50th quantile? Median? If so 0.5 is definitely not the median of $N(0,1)$ distribution. $\endgroup$
    – mpiktas
    Commented Jul 9, 2012 at 13:59

1 Answer 1

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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 $1.96$.

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    $\begingroup$ So, if I want to calculate the two-tailed probability of observing x from a standard normal, it would be ifelse(pnorm(x) > 0.5, 2*(1-pnorm(x)), 2*pnorm(x))? I was hoping for something a little more straight forward; perhaps that there is another base function for that. $\endgroup$
    – Abe
    Commented Jun 29, 2012 at 21:09
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    $\begingroup$ Michael: your 2*(1-pnorm(x)) only makes sense for non-negative $x$ $\endgroup$
    – Henry
    Commented Jun 29, 2012 at 22:47
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    $\begingroup$ as @Henry points out, it should be 2*(1-pnorm(abs(x))) or 2*pnorm(-abs(x)) $\endgroup$
    – Macro
    Commented Jun 29, 2012 at 22:50
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    $\begingroup$ Or 2*pnorm(abs(x),lower=FALSE) ... all of these come up with the same answer. The answer by @Abe is less efficient because it calls pnorm 3 times to get the same result. I wonder if the OP regards 'the two tailed probability of observing abs(x) or greater as the answer to "what does pnorm(2) say about my assumption ..."? A strict answer to the question as posed is nothing. $\endgroup$
    – atiretoo
    Commented Jun 30, 2012 at 21:35
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    $\begingroup$ @Macro: Your second version should be much, much more accurate if the absolute value of $x$ is of any appreciable size. $\endgroup$
    – cardinal
    Commented Jul 2, 2012 at 18:14

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