S3 - Combinations of random variables question.

Sorry for the double posting, I put this in the exams section by accident.

Hi, I'm stuck on part c) for this question:

In the vegetable section of a local supermarket, leeks are on sale either loose (and unprepared) or prepared in packs of 6. The weights of unprepared leeks are modelled by the random variable X which has the Normal distribution with mean 260 grams and standard deviation 23 grams. The prepared leeks have had 40% of their weight removed, so that their weights, Y, are modelled by Y=0.6X.
(a) Find the probability that a randomly chosen unprepared leek weighs less than 290 grams.
(b) Find the probability that a randomly chosen prepared leek weighs more than 155 grams.
(c) Find the probability that the total weight of 4 randomly chosen prepared leeks in a pack is less than 913 grams.

For part b) I found the distribution for Y to be Y ~ N (56, 190.44) and so I thought for part C, I would multiply the mean and variance by 4 to give me the new distribution.

But instead, the answer requires me to multiply it by 6 - the number of leeks in a pack and then simply works out P(X < 913). Why do they use 6 instead of 4? And where do they take into account the fact that we're working out the probability of the total weight of 4 prepared leeks?

Thank you for you kind help,

Maulik.