# sample size for normal distribution

i have a small set of data. I'm sure that their distribution is normal, with standard deviation $$\sigma=10$$. I'm interested in the mean value, $$\mu$$ and its uncertainty. Which is the minimum sample size i must have ? And which will be the uncertainty on the mean value ? I think the uncertainty on the mean value is $$\sigma_\mu=\sigma/\sqrt{N}=10/\sqrt{N}$$, where $$N$$ is the sample size. Which is the min value for $$N$$ i should have ? Thank you Andrea

One approach is to use a 95% confidence interval of the form $$\bar X \pm 1.96\sigma/\sqrt{n},$$ where $$\pm 1.96$$ cut probability 2.5% from the upper and lower tails, respectively, of the standard normal distribution. (This leaves 95% in the central part of the distribution.)
Then. as you say the 'standard error of the mean' is $$SD(\bar X) = \sigma/\sqrt{n}.$$ Also, $$M = 1.96\sigma/\sqrt{n}$$ is called the 'margin of error' for the sample mean $$\bar X$$ as an estimate of the population mean $$\mu.$$
So, if you know $$\sigma$$ and you know what size $$M$$ you want, then you can solve for $$n$$ to get $$n = (1.96\sigma/M)^2.$$ It is customary to round up to the next larger integer.