I recently tried to solve two problems, and do not understand their solution. In both cases, the problem was that I did not divide some expression by sqrt(n).

First problem:

According to the lecture notes, when transforming a random variable X by a function $Y=bX+a$, $\bar{Y}=b\bar{X}+a$ and $\sigma^2_Y=|b|\sigma^2_X$. Therefore, I wrote that the new mean is 16 and the new standard deviation is 8. But the solution says this: $SD[T]=\sqrt{n}*SD[X]=\sqrt{4}*n=4$!!! 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: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?

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

Second problem:

And according to the solution: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 p̂ = .68 prefer Layout A, and wants to know if this is evidence that there is a preference in the population of all users, using α = .01.

But on the very same document, it says: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

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.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.

If anyone can explain to me what I am missing here I would be extremely grateful!