I think the median is 8 or 11
Answer:
(a)![E[X+Y]=E[X]+E[Y]](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3DE%5BX%5D%2BE%5BY%5D)
(b)
Step-by-step explanation:
Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.
(a)We want to show that E[X + Y ] = E[X] + E[Y ].
When we have two random variables instead of one, we consider their joint distribution function.
For a function f(X,Y) of discrete variables X and Y, we can define
![E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).](https://tex.z-dn.net/?f=E%5Bf%28X%2CY%29%5D%3D%5Csum_%7Bx%2Cy%7Df%28x%2Cy%29%5Ccdot%20P%28X%3Dx%2C%20Y%3Dy%29.)
Since f(X,Y)=X+Y
![E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3D%5Csum_%7Bx%2Cy%7D%28x%2By%29P%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7Bx%2Cy%7DxP%28X%3Dx%2CY%3Dy%29%2B%5Csum_%7Bx%2Cy%7DyP%28X%3Dx%2CY%3Dy%29.)
Let us look at the first of these sums.
![\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].](https://tex.z-dn.net/?f=%5Csum_%7Bx%2Cy%7DxP%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7Bx%7Dx%5Csum_%7By%7DP%28X%3Dx%2CY%3Dy%29%5C%5C%5Ctext%7BTaking%20Marginal%20distribution%20of%20x%7D%5C%5C%3D%5Csum_%7Bx%7DxP%28X%3Dx%29%3DE%5BX%5D.)
Similarly,
![\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].](https://tex.z-dn.net/?f=%5Csum_%7Bx%2Cy%7DyP%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7By%7Dy%5Csum_%7Bx%7DP%28X%3Dx%2CY%3Dy%29%5C%5C%5Ctext%7BTaking%20Marginal%20distribution%20of%20y%7D%5C%5C%3D%5Csum_%7By%7DyP%28Y%3Dy%29%3DE%5BY%5D.)
Combining these two gives the formula:
Therefore:
![E[X+Y]=E[X]+E[Y] \text{ as required.}](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3DE%5BX%5D%2BE%5BY%5D%20%5Ctext%7B%20%20as%20required.%7D)
(b)We want to show that if X and Y are independent random variables, then:
By definition of Variance, we have that:
![Var(X+Y)=E(X+Y-E[X+Y]^2)](https://tex.z-dn.net/?f=Var%28X%2BY%29%3DE%28X%2BY-E%5BX%2BY%5D%5E2%29)
![=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)](https://tex.z-dn.net/?f=%3DE%5B%28X-%5Cmu_X%20%20%2BY-%20%5Cmu_Y%29%5E2%5D%5C%5C%3DE%5B%28X-%5Cmu_X%29%5E2%20%20%2B%28Y-%20%5Cmu_Y%29%5E2%2B2%28X-%5Cmu_X%29%28Y-%20%5Cmu_Y%29%5D%5C%5C%24Since%20we%20have%20shown%20that%20expectation%20is%20linear%24%5C%5C%3DE%28X-%5Cmu_X%29%5E2%20%20%2BE%28Y-%20%5Cmu_Y%29%5E2%2B2E%28X-%5Cmu_X%29%28Y-%20%5Cmu_Y%29%5D%5C%5C%3DE%5B%28X-E%28X%29%5D%5E2%20%20%2BE%5BY-%20E%28Y%29%5D%5E2%2B2Cov%20%28X%2CY%29)
Since X and Y are independent, Cov(X,Y)=0
Therefore as required:
3 yrs ago:
Alice - x
Jane - 3x
Fast forward 2 years... 1 year ago:
Sum of their ages = 62
Equation to solve: x + 3x = 62
4x = 62
----- -----
4 4
x = 15.5
From this, we can deterime that Alice is 15 1/2 years old.
Then, plug in 15.5 for x in 3x to find Jane's age.
15.5 × 3 = 46.5
From this, we can determine that Jane is 46 1/2 years old.
I hope this helps you to understand this problem better.
<span>Toni can carry five books in her backpack. You must start by subtracting the weight of the other required items. Toni's lunch which is one pound and gym clothes which are two pounds. This can be calculated by 18-1-2=15. Toni has 15 pounds left to carry books. She should divide this by the weight of each book, which is three pounds. 15/3=5. Therefore, Toni can carry five books.</span>
By cross-multiplying, we get
15x + 30 = 24
15x = -6
x = -3 / 5
