How big should be a sample size? Question:
Assuming the population standard deviation σ = 3, how large should a sample be to estimate the population mean µ with a margin of error not exceeding 0.5?
My answer:
SE<= 0.5=3/√n ---> n=36
But probably the correct answer is:

So, can anyone explain more about the second solution?
 A: A 95% confidence interval for normal population $\mu,$ where the SD is $\sigma,$
based on a sample of size $n$ is of the form $\bar X \pm 1.96\sigma/\sqrt{n}.$
As in @statistician's Comment, the 'standard error' of $\bar X$ is $SD(\bar X) = \sigma/\sqrt{n}.$
And the 'margin of error' of the confidence interval is $\Delta = 1.96\sigma/\sqrt{n}.$ You know $\sigma = 3$ and you want to find $n$ so that $\Delta = 0.5$ (or perhaps a bit smaller).
So solve  $\Delta = 1.96\sigma/\sqrt{n}$ for $n$ in terms of $\Delta$ and $\sigma.$ That gives the equation $n = (1.96\sigma/\Delta)^2,$ to which you
refer in your question. So $n \approx 11.76^2 = 138.4.$ Rounding up to
the nearest integer, you get $n = 139$ as the sample size that gives $\Delta$ just a bit smaller than the required 0.5.
Note: Your solution $n = 36$ gives SE$ =3/\sqrt{36} = 0.5,$ equivalent to $\Delta = 1.96(0.5),$ but not to $\Delta = 0.5.$
Roughly speaking, to cut $\Delta$ in half requires a four-fold increase in $n,$ 
and dividing by 1.96 is almost cutting $\Delta$ in half, which explains why
$4(36) = 144$ is not far from the correct answer $n = 139.$
