Distributions of sampling statistics problem (not a hw problem) for the following question:

the lifetime of a particular brand of cutting blade has a mean of 40 hours and sd of 8 hrs. when one blade fails, it is immediately replaced by a new identical blade. suppose there are 36 blades on hand. use CLT to find approximate probabilities:
  a. what is the probability that sample average of lifetimes of the 36 blades exceeds 60 hrs?
  b. compute the probability that all these 36 blades will be used up in less than 1250 hrs.

let n=36 and sd = standard deviation.
can someone please explain why

a follows N(u, sd^2/n) and b follows N(n*u, n*sd^2)

 A: Let random variables $X_1, X_2, \dots, X_n$ be the lifetime of the first blade, the second,and so on. Here $n=36$.
The $X_i$ are independent, and therefore the variance of their sum is the sum of their variances. Thus the sum $X_1+\cdots+X_n$ of the lifetimes has variance $8^2+8^2+\cdots +8^2$ ($n$ terms). This sum is $(n)(8^2)$, and hence the variance of the sum is $(n)(8^2)$. The standard deviation of the sum is therefore$(\sqrt{n})(8)$.  
We also use an informal version of the Central Limit Theorem. The sum of $n$ independent identically distributed "nice" random variables is "nearly normal" if $n$ is large enough.   
The numerical questions: (a) The sample average exceeds $60$ precisely if the sample sum exceeds $(36)(60)$. So in a sense problems (a) and (b) are very much alike. 
But there is a slightly different approach to (a) which is probably a bit better. Let $S$ be the sample sum discussed above. Then the sample average $Y$ is equal to $\dfrac{S}{n}$. The random variable $Y$ is also nearly normal. It has mean $40$. The variance of $\dfrac{S}{n}$ is $\dfrac{1}{n^2}$ times the variance of $S$ calculated above. It follows that $Y$ has variance $\dfrac{(n)(8^2)}{n^2}$, that is, $\dfrac{8^2}{n}$. Since $n=36$, $Y$ has standard deviation $\dfrac{8}{6}$.
So we want the probability that a normal random variable with mean $40$ and standard deviation $\dfrac{8}{6}$ exceeds $60$. Since $60$ is a ridiculous number of standard deviation units away from $40$, the probability is nearly $0$.
(b) The mean of the sum is $1440$. We want the probability that a nearly normal random variable with standard deviation $48$ is less than $1250$. This is a standard calculation. Pretty unlikely!
Remark: You may be asking why is the variance of an independent sum equal to the sum of the variances. Here is a proof for a sum of two random variables $X$ and $Y$. The proof readily extends to longer sums. 
Let $X$ and $Y$ be independent, with variances $\sigma^2_X$ and $\sigma^2_Y$. Recall that $\text{Var}(W)=E(W^2)-(E(W))^2$. Apply this with $W=X+Y$. We have
$$E((X+Y)^2)=E(X^2+2XY++Y^2)=E(X^2)+2E(XY)+E(Y^2)=E(X^2)+2E(X)E(Y)+E(Y^2).\tag{$!$}$$
(For the fact that $E(XY)=E(X)E(Y)$ we used independence.)  Also.
$$(E(X+Y))^2=(E(X)+E(Y))^2=(E(X))^2+2E(X)E(Y)+(E(Y))^2.\tag{$2$}$$
Using $(1)$ and $(2)$ and rearranging a bit, we find that
$$\text{Var}(X+Y)=E(X^2)-(E(X))^2 +E(Y^2)-(E(Y))^2.$$
But the above is just $\sigma^2_X+\sigma^2_Y$.
