so I am trying to solve this problem:
Let X and Y equal the respective numbers of hours a randomly selected child watches movies or cartoons on TV during a certain month. From experience, it is know that E[X] = 30 E[Y ] = 50, V ar(X) = 52, V ar(Y ) = 64; and Cov(X; Y ) = 14: Twenty-five children are selected at random. Let Z equal the total number of hours these 25 children watch TV movies or cartoons in the next month. Approximate P[1970 < Z < 2090]:
I believe that I am supposed to add X and Y and then solve for 1970< Z <2090.
My question is how do I add X and Y. Are the E|X| and E|Y| just added or...?
Thanks so much
Hint: Denote with $T$ the total hours that a child watches movies or cartoons. That is $$T=X+Y$$ Then you can use the following rule (linearity of expectation) $$E[T]=E[X+Y]=E[X]+[Y] \tag1$$ to find the expected value of each $T$ and the rule $$Var(T)=Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y) \tag2$$ to find the variance of $T$.
Finally you are supposed to use the fact that $Z$ (which is the sum of the $25$ $T$'s) follows approximately the normal distribution (that is however a supposition), i.e. that $$Z \sim N(μ=25\cdot Ε[T], σ^2=25 \cdot Var(T))$$ approximately.
Partial solution: You should find that $Z \sim N(μ=2000, σ^2=3600)$. Now you want to find the probability $$P(1970<Z<2090)=P(\frac{1970-2000}{60}<\frac{Z-μ}{σ}<\frac{2090-2000}{60})=Φ(1,5)-Φ(-0,5)$$