This question regards the basic statistics of a normal distribution, but I can't figure it out. I have been given the mean and 95% confidence intervals for a distribution, but would like to know the standard deviation. In my example:
$$ \mu=53.4\quad 95\%\ c.i.=(52.3, 54.3) $$
I had thought that the solution for $\sigma$ would be something like:
$$54.3=53.4+(SE*1.96)$$ $$SE=(54.3-53.4)/1.96=0.46$$ and then, $$SE=\frac{\sigma}{\sqrt{n}}$$ $$\sigma=0.46*\sqrt{n}$$
So, if I don't know the $n$, is this possible?