Division by $\sqrt{n}$ in CLT and linear transformations I recently tried to solve two problems$^1$, and do not understand their solution. In both cases, the problem was that I did not divide some expression by $\sqrt n$.
First problem:

Imagine that students bring a mean of 4 framed photographs to
  campus, with a standard deviation of 2 photographs.If students are
  randomly assigned to live in quad-style dorm rooms (four students per
  room), what is the mean and standard deviation of the distribution of
  total number of framed photographs per dorm room?

According to the lecture notes, when transforming a random variable $X$ by a function $Y=bX+a,$ $\bar{Y}=b\bar{X}+a$ and $\sigma_Y=|b|\sigma_X.$ Therefore, I wrote that the new mean is 16 and the new standard deviation is 8. But the solution says this: $SD[T]=\sqrt{n} \cdot \operatorname{SD}[X]=\sqrt{4}\cdot n=4$!!!   I do not understand how it is possible given the facts about linear transformations. And if this is not enough, there is also a problem in the same worksheet:

Suppose that a set of exam scores has a mean of 80 with a standard
  deviation of 6. Calculate the mean and standard deviation of the new,
  transformed distributions of exam scores if each score is increased by
  20%.

And the solution says: $y=1.2x$, $s_y=|1.2|s_x=7.2$, which confirms my original understanding of linear transformations.
Second problem:

A user experience researcher wants to know if all users prefer one
  website layout (Layout A) over another (Layout B). He samples n = 64
  customers and asks them which layout they prefer. He finds a sample
  proportion of $\widehat p = 0.68$ prefer Layout A, and wants to know if this is
  evidence that there is a preference in the population of all users,
  using $\alpha = .01$.

And according to the solution:

the central limit theorem tells us that the distribution of sample
  proportions will: 
    have a mean of μM = π = .50 have a standard
  deviation of σM = √(π*(1-π) / n) = √(.5*.5/64) = .0625

But on the very same document, it says:

For any population with a mean μ and standard deviation σ, the
  distribution of sample means from that population for sample size n
  will: 1. have a mean of μ written E[M], E[X̅], μM, μX̅ 2. have a
  standard deviation of σ / √n.

And since the users are binomially distributed, the standard deviation of the sample should be $\sqrt{nπ(1-π)}=4$, and dividing by $\sqrt{n}$ we get .5, not .0625.
If anyone can explain to me what I am missing here I would be extremely grateful!
$^1$ From the contents of Psych 10 at Stanford. Specifically, the first problem is number 17 from practice problem set 3, and the second problem is from slide 34 in the 1/30/17 slides.
 A: In the first two cases, you are dealing with an average $\bar X$ if $n$ normal
observations, $X_i \sim \mathsf{Norm}(\mu, \sigma).$
Thus, $E(X_i) = \mu$ and $Var(X_i) = \sigma^2.$ Then to find $E(\bar X),$
we proceed as follows.
$$E(\bar X) = E\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = 
\frac{1}{n}E\left(\sum_{i=1}^n X_i\right)
=\frac{1}{n}\left(E(X_1) + E(X_2) + \cdots + E(X_n)\right)\\ = \frac{1}{n}n\mu = \mu,$$
where we have used  $E(aY) = aE(Y)$, where $a = 1/n$, in the second step and an extended
version of $E(X + Y) = E(X) + E(Y)$ to get the last term in the first line. 
The proof that $Var(\bar X) = \sigma^2/n$ uses $Var(aY) = a^2Var(Y)$, again for $a = 1/n$, which is how we get $n$ in the denominator. 
[It seems you have a mistake in your formula for variances above, where you are summarizing lecture notes: For standard deviations, $SD(bY) = \sigma_{bY} =|b|\sigma_Y =|b|SD(Y)$, but not for variances.] 
Also, because the $X_i$ are independent, the variance of their sum is the sum of their variances. Putting it all together, we have
$$Var(\bar X) = Var\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = 
\frac{1}{n^2}Var\left(\sum_{i=1}^n X_i\right)  
=\frac{1}{n^2}\left(Var(X_1) + \cdots + Var(X_n)\right) \\= \frac{1}{n^2}n\sigma^2 = \sigma^2/n,$$
Then finally, taking square roots, we have $SD(\bar X) = \sigma/\sqrt{n},$
which is the equation that puzzled you. If you look around in your textbook
you can probably find a similar discussion somewhere. 
