Mean and standard deviation of a sample distribution

The question goes like this:

The distribution of the weights of 1000 students is normal with a mean of 55kg and a variance of 25kg^2. 100 random sample sizes of size 16 are taken from this population.

Find the mean and the standard deviation of the sampling distribution.

The answer is 55kg for the mean and 1.25kg for the standard deviation.

My idea is the sample mean and the population mean are the same, thus giving the answer 55kg but for standard deviation i have no idea how to calculate it.

Thanks for the help in advance!

• @Cm7F7Bb edited the question – Ong Nov 30 '15 at 10:33

The standard deviation of the sampling distribution of $\bar{X}$ is $\sigma_\bar{X}=\frac{\sigma_X}{\sqrt{n}}$, where $\sigma_X$ is the standard deviation of $X$ and $n$ is the sample size.
Since the variance of $X$ is $25kg^2$, the standard deviation of $X$ is $\sigma_X=5kg$. Then the standard deviation of the sampling distribution of $\bar{X}$ is $\sigma_{\bar{X}}=\frac{\sigma_X}{\sqrt{n}}=\frac{5}{\sqrt{16}}=1.25kg$.