Normal distribution: Weight of a package of cookies Suppose the mass of a cookie is a normal random variable X. Let's say a cookie weighs $20g$ on
average with a standard deviation of $2g$. A packet contains exactly $25$ cookies, with the weight
of the packaging also being a normal random variable Y with mean $100g$ and standard
deviation $6g$. Assume all random variables are independent.
(i)  Calculate the variance of the weight of a packet of cookies.
(ii) If you buy $3$ packets, what is the probability that the total weight of
     of this purchase is less than $1.82$ kg.
My attempt:
(i) $6^{2}$ $=$ $36$
(ii) X ~ N($20,4$) and Y ~ N($100,36$). Let W be the total mass 
     of the package. So, $W=X+3Y~N(20+3(100),4+3(36))=N(320,112)$. We are 
     required to find $P_r[W<1.82]$ but the z score I am getting does not 
     make sense at all. I got a z score of 141.7366. I converted everyone to grams. Please help.

$(i) \ \ X \ $~$ \  N(20,4), \ \ Y \ $~$ \  N(100,36)$. A packet contains $25$ cookies, so the weight of the packet $W \ $ ~ $\ N(100+25\times20,36+25\times 4)=N(600,136).$
So the variance of the packet is $136$.     
$(ii)$ If you buy $3$ packets, their combined weight follows $K \ $~ $\ N(1800,408)$ and operating with this you get $\mathbb{P}(K<1820) =51.95\%$.
