$A$ is a random variable with a mean of 14 and standard deviation of 1.2. $B=4A$ and $C=A_1+A_2+A_3+A_4+A_5$ where each $A_i$ is independent observation of $A$.
I am needing to find which, between $B$ and $C$, has the bigger mean and standard deviation.
We have been given direction to use a standardized version which may be a hint, but one that I do not understand: $$C=\frac{\hat A-\overline{\hat A}}{\sigma(\hat A)}$$ How may I apply this standardized version to solve? Any guidance would be appreciated!
Because expectation is linear, $E(B)=4\cdot E(A)$ and $E(C)=5\cdot E(A)$, so $C$ has a higher mean than $B$.
As to the standard deviation, if one random variable has a higher standard deviation than another, that variable also has a higher variance (squared standard deviation) than the other. Since $B$ is a constant multiple of $A$, its variance is 16 ($4^2$; not 4) times that of $A$, so its standard deviation is 4 times that of $A$. $C$ is a sum of iid $A_i$, so its variance is only 5 times that of $A$ and its standard deviation is $\sqrt5$ times that of $A$. Since $\sqrt5<4$, we conclude that $B$ has a higher standard deviation than $C$.