When to use transformation for standard deviation of random variable Z = kX? I'm studying statistics using khan academy AP stats class.
One example says:
Each of a certain item at a factory gets inspected by 4 employees. The time it takes each employee to inspect the item has a mean of 30 seconds and a standard deviation of 6 seconds. Furthermore, the amount of time it takes a given employee to inspect an item is not impacted by how long it takes another employee to inspect that item.
Let T be the total amount of time it takes it takes 4 employees to inspect a randomly selected item.
What is the standard deviation of the total amount of time it takes 4 employees to inspect a randomly selected item?
Based on the formula σ_Z=kσ_X if σ is a constant and Z is defined as Z=kX, and the definition of standard deviation being the square root of the variance, I think I should be able to calculate the standard deviation using either of the following methods:
Method 1: 4*6 seconds = 24 seconds.
Method 2: σ_T^2 = 6^2 + 6^2 +6^2 +6^2 = 144 -->  σ = 12.
I have the answer, so I know method 1 doesn't work and method two does work. I understand method 2, but I don't understand why method 1 fails. When would you use method 1? Why isn't this the same as multiplying a constant by the standard deviation of the time it takes an employee to inspect an item?
THANK YOU!
 A: It's the nature of the variance. If you have n i.i.d. (independent and identical) random variables then
$$Var\left(\sum_{i=1}^n X_i\right)=n\cdot Var(X_i)$$.
Let's say you have two i.i.d. random variables and you calculate the variance of the sum of both. Then
$$Var\left( X_1+X_2\right)= Var(X_1)+2\cdot Cov(X_1,X_2)+Var(X_2)$$.
Since the random variables are independent the covariance is $0$. And the variances are identical.
$=2\cdot Var(X_1)$
To obtain the standard deviation we take the square root: $\sigma_{X_1+X_2}=\sqrt 2\cdot \sigma_{X_1}$
To summarize: If you have $n$ independent random variables you add the variances to obtain the variance of the sum of the random variables. To obtain the standard deviaton of the sum you take the square root.

... transforming a random variable by scaling it by a positive integer
is different than combining identical independent random variables

At this case you have n variables, which are mutually dependent. Consider the two variables case again.
$$Var\left( X_1+X_2\right)= Var(X_1)+2\cdot Cov(X_1,X_2)+Var(X_2)$$, where $X_1=X_2$
$Cov(X_1,X_1)$ is equal to $Var(X_1)$, since $X_1$ depends fully on $X_1$.
$$Var\left( X_1+X_1\right)= Var(X_1)+2\cdot Var(X_1)+Var(X_1)$$
$$Var\left( 2\cdot X_1\right)= 4\cdot Var(X_1)$$
This is how it works for the scaling factor $2$. The resulting factor is the square of 2. Then for the scaling factor n we have
$$Var\left( n\cdot X_1\right)= n^2\cdot Var(X_1)$$
