Mean and SD for a normal distribution given value of a percentile Is it possible to calculate the mean and standard deviation of a normal distribution given a value x is the 25th percentile?
It feels like this is enough information but I can't seem to figure this out.
 A: No, this is not possible: as whuber notes, you are trying to determine two parameters (the mean and the standard deviation), having only one data point.
For instance, a 25th percentile at $1$ can come from an uncountably infinite one-parameter family of normal distributions, because for any mean $\mu>1$ (do you see why $\mu$ must be larger than $1$?), you can find a standard deviation $\sigma$ such that $q_{0.25}(\mu,\sigma^2)=1$. A few examples, with the 25th percentile at $1$ marked in red:

R code:
perc <- 1
means <- c(2,4,6,8)
sds <- sapply(means,function(mm)
  optimize(function(ss)(qnorm(0.25,mm,ss)-perc)^2,interval=c(0.01,20)))

xx <- seq(-12,12,by=.01)+perc
opar <- par(mai=c(.5,.8,.8,0),mfrow=c(2,2),las=1)
    for ( ii in seq_along(means) ) {
        plot(xx,dnorm(xx,means[ii],sds[1,ii]$minimum),type="l",xlab="",ylab="",
   main=paste0("M = ",means[ii],", SD = ",round(sds[1,ii]$minimum,2)))
        abline(v=perc,col="red")
    }
par(opar)

Of course, if you have two (different) percentiles, you can determine a normal distribution (as long as the percentiles "make sense", i.e., they are not "the wrong way around").
A: If you know the mean, you can determine the variance.
Let's assume that $q_{0.25} = -0.67$ and $\mu = \mu_0$ with unknown $\sigma^2$. Then $$f_X(x) = \dfrac{1}{\sigma\sqrt{2\pi}}\exp\bigg(-\dfrac{1}{2}\dfrac{(x - \mu_0)^2}{\sigma}\bigg)$$
$$
F_X(x) = \int_{-\infty}^xf_X(u)du \implies F_X(0.25) = \int_{-\infty}^{0.25}f_X(u)du = 0.25
$$
This solution will be hard (I think impossible) to determine analytically, but you can use some kind of numerical approximation. As an example where the solution is possible, consider the same for $q_{0.25} = 0.5$ for an exponential distribution:
$$
F_X(0.25) = 1-\exp(-\lambda \times  0.25) = 0.5
\\
1-0.5 = \exp(-\lambda \times  0.25)\\
\log(0.5) = -0.25\lambda\\
\lambda = -4log(0.5) \approx 2.77$$
If you know the variance, you can determine the mean.
You would go through a similar calculation and numerical approximation as above but with known variance and unknown mean.
If you do not know the mean or the variance, there are infinitely-many possibilities, and you can use the calculations in the former two to come up with them. Pick a mean of $0$...what is the variance? Pick a variance of $1$...what is the mean? Pick a variance of $9$...what is the mean? Pick a mean of $-0.5$...what is the variance? Every one of these results in $q_{0.25} = -0.67$.
