S3 need help panicking

how do i do part b
since the variables are added would u just add the means and then add the standard deviations^2?

here is the question part b

$X_i$ are independent, so yes all you need to do is add together the mean and variance of $2X_1, 3X_{10}$ to yield the distribution of $(2X_1 + 3X_{10})$

i'm confused man here is the answer to part b
i understand how to get the variance (2/5)^2+(3/5)^2=13/25
but why is the mean still mew

Because the mean is multiplied by whatever you're multiplying the r.v. by. So, since both rv's have the mean as $\mu$ to start with, they will then have $2\mu$ and $3\mu$. Adding the rv's yields $5\mu$ and then dividing the sum of the rv's by 5 hence divides this mean by 5 to arrive back to $\mu$

one more question if you look at part c and f
can you please explain why E(Xi-mew)=0
and what happens when you need to find the E((Xi-mew)/standard deviation) as well as the variance

It's mu* not mew.

Two ways of looking at $\mathbb{E}[X_i-\mu] = 0$. First, expectation is a linear operator therefore

$\mathbb{E}[X_i-\mu] = \underbrace{\mathbb{E}[X_i]}_{=\mu} - \underbrace{\mathbb{E}[\mu]}_{=\mu} = 0$

On the other hand, a more classical viewpoint is that if you consider the distribution $X_i$, it has the mean of $\mu$. Which means that subtracting $\mu$ from the distribution shifts all of the points in the distribution by $\mu$. Precisely, the mean is shifted by this too hence it ends up on 0.


For the other question, $\mathbb{E} \left[ \dfrac{X_i-\mu}{\sigma} \right]=\dfrac{1}{\sigma}\mathbb{E}[X_i-\mu]$. As for the variance, well the standard deviation of $X_i-\mu$ gets divided by $\sigma$ when you divide the distribution by it, so the variance is.....???


No need to panic about this stuff lol, it's just stats it's not gonna give you a heart attack!

\frac{1}{\sigma}\mathbb{E}[X_i-\mu] =0 for the variance tho idk can you do Var(Xi-mu)??? whats Var(mu)? and obv Var(Xi)= standard deviation^2 so dividing that by standard deviation would give just
\sigma what happens with the Var(mu)

Er... you're just confusing yourself. Slow down.

What is the distribution of $X_i - \mu$ ?

P.S. if you're going to use $\LaTeX$ then you need the 'tex' tags between your code.

$X_i - \mu$~N(0, this is the part i dont understand) how do i find Var(Xi-mu)

Well go back to the basic principles first to get it without doing any work. What does the variance (or even the standard deviation) tell you?

Well variance tells you how spread out the values of the independent variables would be

Close, it tells you how spread out the data is.

So then we know how spread out $X_i$'s data is. Now if you are going to shift all the data to the left or the right, so that every single piece of data is shifted by the same amount, is there going to be any change to how spread out it is?

Well if you shift all of them by the same amount there shouldn't be any change

Precisely. When you go from $X_i$ to $X_i-\mu$ all you do is shift the data by $\mu$. So the variance is...?

the same..... the variance stays the same, So Var(Xi-mu)=Var (Xi)= Variance

Exactly.

So then when it comes to the variance of $\dfrac{X_i-\mu}{\sigma}$ you simply take the standard deviation of $X_i-\mu$ and divide it by $\sigma$. So the new standard deviation is....? Hence the variance of $\dfrac{X_i-\mu}{\sigma}$ is.....??

variance would just be variance/standard deviation which is just standard deviation??? but the answer says its 1