Hey alb41192.

The sum of independent normal distributions are normally distributed with the means added up and the variances added up.

So for three days you have a mean of 3*mu and a variance of 3*sigma^2.

This means you are looking at Z ~ (X - 3*mu)/SQRT(3*sigma^2).

Now this is for X1 = 3Y where this is for three days. If you want to do this for two days then you need to use X2 = 2Y.

The probabilities are P(X1 > 2000 OR X3 > 2000) and P(X_bar > 1800).

Recall that X_bar is equal to (X1 + X2 + ... + XN)/n which has mean mu and variance sigma^2/n.