# Probability and Statistics (Normal Distribution)

Having trouble with the last part of this question. Not sure how the man would divide his pile of vouchers? It seems that you could interpret this question in a lot of ways. Any tips would be appreciated. Thanks :)

Here are some hints to get you going...note that the last part of the question is forcing you to come up with some assumptions, so there isn't likely one answer:

The weight of the "reference pile" ($W_r$) used by the company will be distributed:

$$P_r\sim \mathcal{N}(50\times 36,50\times 2.5^2)=\mathcal{N}(1800,312.5)\implies \sigma_{W_r}\approx17.7$$

Therefore, their lower 5% cutoff is simply $1800-1.64\sigma\approx1,771$

Now, the man is making three piles which will contain$N_1,N_2,N_3$ cards such that $N_1+N_2+N_3=150$.

The weights of the piles will be $W_1,W_2,W_3$, where $W_i\sim \mathcal{N}(36N_i,2.5^2N_i)$

Thus, you need to come up with an assumption on the random process that generates the piles, then the probability of getting all three prizes is:

$$P(\cap_{i=1}^3W_i\geq1,771)=\sum P(N_1=n_1,N_2=n_2,N_3=n_3)P(\cap_{i=1}^3W_i\geq1,771|N_1=n_1,N_2=n_2,N_3=n_3)$$

The key to this problem is coming up with a model for $(N_1,N_2,N_3)$ that is both reasonable and simple....otherwise you'll have a real task ahead of you.

Welcome to the world of probabilistic modelling.