Normal distribution

For b)ii)a) why can I just do the probability of each one to the power of 12 if it’s normal distribution? And how would I do b)ii)b)?

biiA) You know P(X>450) = 0.95. You dont need to know the underlying distribution, just treat it as n=12 binomial distribution with a probability of p=0.95. There are only two outcomes for each sample, its underweight or not. Or more simply just note that each p=0.95 and each cheese pack is independent, so the joint is ...
biiB) You should know the distribution (mean and variance) of the mean weight of 12 samples? Then its just the usual cumulative probability.

Do you know what they are doing here and why?

Its as per the previous post and similar to the questions you were doing yesterday. If you draw 12 samples and use that to estimate the mean of a normal distribution, the mean is teh original mean and the standard deviation is sigma/sqrt(12)

I just don’t understand why they are finding the sd/variance if it was given as 10/10^2 ? And also why they are standardising with 475 as mu?

When you estimate the mean with 12 samples, that is N(475, 15.2/sqrt(12)). Do you understand that?
Then its just the usual cumulative for P(X>470).
Why not write down what you would do and upload that.

Oh so that’s the sd of the 12 together. Do I have to keep standardising things or can I just use those values?

Yes. The question simply wants you to apply the N(mu, sigma^2/n) formula. Its something like 3 marks, the mark scheme makes it seem harder than it actually is.