The problem I have with this is calculating the variance of the weight of the 15 books.
To me, X is the RV for the weight of 1 book, where the mean = 12, and variance = 15 (root 15 squared).
Let Y be the weight of 15 books...
$Y = X + X + X + ... + X$
$Y = 15X$
$4E(Y) = E(X + X + X + ... + X)$
$E(Y) = E(15X) $
$E(Y) = 15 E(X)$ {this makes perfect sense}
Then I used the property $Var(aX) = a^2 Var(X)$...
Hence,
$Y = 15X$
$Var(Y) = Var(15X)$
$Var(Y) = 15^2 * Var(X)$
However, I also know that with random variables, variances add linearly (if RVs are independent)
Hence,
$Var(X + X + X + X .... + X) = Var(X) + Var(X) + .. + Var(X) =$ 15 * Var(X)
Obviously, both answers are starkly different(answer key says 2nd answer). Both make sense to me. My big question is what is the difference between the two?
$15^2 \times Var(X) != 15 \times Var(X)$.
I feel like the answer has something to do with independence? I really must get to the bottom of the difference.
The weight of fifteen books may not be fifteen times the weight of one book, because the books do not have the same weight. So the variance we seek is that of the sum of fifteen independent and identically distributed random variables.
Identical distribution means identical variances, and the covariance of independent book weights is zero. So the Bilinearity of Covariance gives us:
$$\begin{align}\mathsf {Var}(\sum_{i=1}^{15} X_i)&=\mathsf{Cov}(\sum_{i=1}^{15}X_i, \sum_{j=1}^{15} X_j)\\&=\sum_{k=1}^{15}\mathsf{Var}(X_k)+2\sum_{i=1}^{14}\sum_{j=i+1}^{15}\mathsf{Cov}(X_i,X_j)\\[1ex]&=15\,\mathsf {Var}(X_1)+0\end{align}$$
Where as the variance of fifteen times the weight of one book will be much larger.
$$\begin{align}\mathsf{Var}(15X_1)&=\mathsf{Cov}(15X_1,15X_1)\\[1ex]&=225\,\mathsf{Var}(X_1)\end{align}$$