Normal distribution and sample distribution standard deviation The distribution of weights of 1000 students is normal with a mean of 55 kg and a variance of 25 kg. 100 random samples of size 16 are taken from this population. Determine the following:
i) The mean and standard deviation of the sampling distribution?
How do i find the standard deviation?
 A: You are probably studying, or about to study, the Central Limit Theorem. You should find some explanation of this in your text--adjacent to the discussion of the CLT. In particular, a sample mean of observations from a normal population is normally distributed.
If $X_1, X_2, \dots, X_{16}$ are a random sample from $Norm(\mu = 55, \sigma = 5),$
then the sample mean $\bar X \sim Norm(\mu = 55, \sigma = 5/\sqrt{16}) = Norm(5, 5/4).$
Key steps in the derivation of $E(\bar X) = \mu$ are as follows:
$$E(\bar X) = E\left(\frac{1}{n}\sum_{i=1}^n X_i\right)
= \frac{1}{n}E\left(\sum_{i=1}^n X_i\right) = \frac{1}{n}\sum_{i=1}^n E(X_i)
= \frac{1}{n}(n\mu) = \mu.$$
Notice that each of the $n$ terms in the last summation is $\mu.$
Key steps in the derivation of $Var(\bar X) = \sigma^2/n$ are as follows,
where the first step uses the identity $Var(aY) = a^2Var(Y):$
$$Var(\bar X) = Var\left(\frac{1}{n}\sum_{i=1}^n X_i\right)
= \left(\frac{1}{n}\right)^2 Var\left(\sum_{i=1}^n X_i\right) \\= \frac{1}{n^2}\sum_{i=1}^n Var(X_i)
= \frac{1}{n^2}(n\sigma^2) = \sigma^2/n.$$
Then $SD(\bar X) = \sqrt{Var(\bar X)} = \sqrt{\sigma^2/n} = \sigma/\sqrt{n}.$
Example: As a consequence, the probability that any one individual in the population
weighs between 50 and 60kg is $P(50 <X_i < 60) \approx .68$ However,
the probability that the average weight of 16 individuals lies between 50 and 50kg is $P(50 < \bar X < 60) \approx 1.$ 
 diff(pnorm(c(50,60), 55, 5))
 ## 0.6826895
 diff(pnorm(c(50,60), 55, 5/4))
 ## 0.9999367

The plot below shows the normal PDF of the population and the normal PDF of $\bar X.$

