# Thread: Division by sqrt(n) in CLT and linear transformations

1. ## Division by sqrt(n) in CLT and linear transformations

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:

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 $\displaystyle$Y=bX+a$$, \displaystyle \bar{Y}=b\bar{X}+a$$ and $\displaystyle$\sigma^2_Y=|b|\sigma^2_X$$. Therefore, I wrote that the new mean is 16 and the new standard deviation is 8. But the solution says this: \displaystyle SD[T]=\sqrt{n}*SD[X]=\sqrt{4}*n=4$$!!! 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:

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: $\displaystyle$y=1.2x,s_y=|1.2|x=7.2$$, which confirms my original understanding of linear transformations. Second problem: A user experience researcher wants to know if all users prefer one website layout (Layout A) over another (Layout B). He samples n = 64 customers and asks them which layout they prefer. He finds a sample proportion of p̂ = .68 prefer Layout A, and wants to know if this is evidence that there is a preference in the population of all users, using α = .01. And according to the solution (π=.5 is the probability according to the null hypothesis): the central limit theorem tells us that the distribution of sample proportions will: have a mean of μM = π = .50 have a standard deviation of σM = √(π*(1-π) / n) = √(.5*.5/64) = .0625 But on the very same document, it says: For any population with a mean μ and standard deviation σ, the distribution of sample means from that population for sample size n will: 1. have a mean of μ written E[M], E[X̅], μM, μX̅ 2. have a standard deviation of σ / √n. And since the users are binomially distributed, the standard deviation of the sample should be \displaystyle \sqrt{nπ(1-π)}=4$$, and dividing by $\displaystyle$\sqrt{n}$$we get .5, not .0625. If anyone can explain to me what I am missing here I would be extremely grateful! 2. ## Re: Division by sqrt(n) in CLT and linear transformations Hey omerkorat. Try using only one thread for a question and check the other thread for a response. 3. ## Re: Division by sqrt(n) in CLT and linear transformations Originally Posted by omerkorat I recently tried to solve two problems, and do not understand their solution. In both cases, the problem was that I did not divide some expression by sqrt(n). First problem: According to the lecture notes, when transforming a random variable X by a function \displaystyle Y=bX+a$$, $\displaystyle$\bar{Y}=b\bar{X}+a$$and \displaystyle \sigma^2_Y=|b|\sigma^2_X$$. Therefore, I wrote that the new mean is 16 and the new standard deviation is 8. But the solution says this: $\displaystyle$SD[T]=\sqrt{n}*SD[X]=\sqrt{4}*n=4$$!!! I do not understand how it is possible given the facts about linear transformations. Perhaps I am missing something. Using your first formula, with \displaystyle \sigma= 2 and b= 2, \displaystyle \sigma^2_Y= 4*4= 16 so that \displaystyle \sigma_Y= 4, exactly what the second formula gives. if this is not enough, there is also a problem in the same worksheet: And the solution says: \displaystyle y=1.2x,s_y=|1.2|x=7.2$$, which confirms my original understanding of linear transformations.

Second problem:

And according to the solution (π=.5 is the probability according to the null hypothesis):

But on the very same document, it says:

And since the users are binomially distributed, the standard deviation of the sample should be $\displaystyle$\sqrt{nπ(1-π)}=4$$, and dividing by \displaystyle \sqrt{n}$$ we get .5, not .0625.
Your formula is wrong here. Using this method, you should be dividing by n, not $\displaystyle \sqrt{n}$. Instead of cancelling the n in the numerator, that will leave a $\displaystyle \sqrt{n}$ in the denominator, giving exactly the same formula as you had before.

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