The question of this problem is asking to find the expected value for the following:
Roll a four-sided die, double your result, then add $7$.
(We're using a four-sided die, not actually rolling one).
This is the explanation shown in class, which had me confused (it's discrete uniform distribution);
$E(X)=a+b/2$, where $a=1$ and $b=4$
and then,
$y=2x+7$ because I have to double and then add seven
But from here, it goes like this;
$E(Y) = E(2x+7)$
which is then equal to;
= $2E(x)+7$
Can someone explain this part to me? Does this part even work?
And then from my understanding, since I want to find the expected value of a four-sided die, would it make sense to do this;
$(1*(1/4)) + (2*(1/4)) + (3*(1/4)) + (4*(1/4)) = 2.5$ since each of the numbers are all equally likely to occur.
If I am to use the $2E(x)+7$ from above, is this the correct equation;
$2(2.5)+7$ ?
$(1*(1/4)) + (2*(1/4)) + (3*(1/4)) + (4*(1/4)) = 2.5$ is completely correct.
As $E[X] = \sum xP(x)$ and is since this is the uniform distribution.
$P(x) = \frac 14 $if $x\in\{1,2,3,4\}$ and $0$ otherwise.
$E[X] = \frac 14 \sum_\limits{x=1}^4 x$
With uniform distribution, we can take the a short cut of adding the highest and the lowest and dividing by 2.
As for the next part $E[2X+7] = 2E[X] + 7$ you can do this because of what is called "linearity of expectation."
which means that $E[X+Y] = E[X]+E[Y]$ and $E[aX] = aE[X]$
It is good to ask if this is operation is acceptable.
and we can can go back to the definition:
$E[2X+7] = \frac 14 \sum_\limits{x=1}^4 (2x + 7) = \frac 14 (2\sum_\limits{x=1}^4 x) + 7$
and see that this property indeed holds.