# Central Limit Theorem, Samples

I've part of a solution I need help grasping. First, here's the question:

"Let the average lifetime of certain electronic components be 6 months, and the standard deviation of the lifetime equal to 2 months. Suppose now that the components are independent, but that the distribution of their lifetime is unknown, and we only known the mean and standard deviation of the lifetime. We consider a machine made up of n components of this type placed in parallel. Furthermore, only one component is active at a time (standby redundancy). Use the central limit theorem to find the smallest value of n for which the probability that the machine functions during at least 15 years is greater than 90%."

Now the solution we were given is:

"Let S(n)= T1+T2+...+Tn, where Ti denotes a lifetime. If n is large enough, by the central limit theorem, we can write that S(n) ~~ N(n*6,n*4). We seek n(min) such that P[S(n)>= 180]>0.90..." (By the way ~~ "means approximately equal to")

So, my question

Why are both our mean and standard deviation multiplied by n in N(6*n,4*n)? I know, of course, why 6 and 4 are there (as they represent mu and sigma squared respectively). Is it something to do with the fact that we take n samples?

-
Hint: the mean of a sum is the sum of the means. The variance of a sum of independent random variables os the sum of the variances. The standard deviation is NOT being multiplied by $n$ as you think. –  Dilip Sarwate Apr 23 '12 at 20:43
What Dilip is saying is that n variables with mean 6 will have mean 6n for their sum (adding 6 n times. For the variance you use the same argument invoking independence (otherwise covariance terms would enter in). Since 2 is the standard deviation for any of the Tis they each have variance 4. –  Michael Chernick May 21 '12 at 18:00