I am trying to figure out which approach is appropriate for the next problem. I throw the ball into the bin with 30% probability of succes. If successful, I get 5 points and 0 if I fail. The task is to find the variance of the points earned after 10 throws.
My first approach is to calculate variance for number of successes and then multiply it by points: $Var[X] = np(1 - p) = 10*0.3*0.7 = 2.1*5 = 10.5$. Another approach is to calculate variance for single throw and then, assuming that throws are independent:
$Var[Xi] = E[X^2] - (E[X])^2 = 7.5 - 2.25 = 5.25$.
$Var[X] = 10*Var[Xi] = 52.5$.
Since there is a difference in results, I would appreciate any clarification or advice on this!
You might have gotten the answer but I still want to clarify something. $Var[Throws] = np(1−p) = 10 * 0.3 * 0.7 = 2.1$ and not $Var[X] = 2.1 * 5 = 10.5, (X = earnings)$. Instead the variance of the earnings is different as it is transformed by a linear factor. The variance represents an area of where most of the values lie, it is not a figure in the event (like $\mu$, which you can multiply by $5$ to get the total earnings.)
Lets look over the scenario again. The $\mu$ throws is $np = 10 * 0.3 = 3$. This means on average you would expect to land $3$ in $10$ throws. Multiply this by 5 and you get your total $\mu$ earnings $= 5 * 3 = 15$. $E[a * X] = a * E[X]$. Again, this because $\mu$ can be considered as physical value in your binomial set, so any transformations can be applied directly.
However, since $Var[X]$ is an area the transformations applied to it differ. $$Var[a * X + b] = a^2 * Var[X]$$
This proof comes from the $\mu$ as it directly related to the mean.
$Var(aX+b)=E[(aX+b)^2]−[(aX+b)]^2=(a^2X^2+2abX+b^2)−[aE(X)+E(b)]^2=^2(^2)+2()+^2−[^2()^2+2()+^2]=^2(^2)−^2[()]^2=^2[(^2)−()^2]=^2Var().$
Where $a$, again, is your winning
$$. So, $$a^2 * Var[X] = 5^2 * np(1 - p) = 25 * 2.1 = 52.50$$ Hence, you theoretically have an area of winning a total $52.50.