Question 1110607
The width of the interval is z(0.90)*sigma/sqrt(n)
this is 1.28*0.36/sqrt(15)=0.1190 or 0.12
The interval is (3.03, 3.27) units gms.

I need sigma is known and normal distribution.  They go together, and when I have them, n can be anything, so it is not required to be large.

The second choice that there is an 80% chance the interval is one of those... I like to say that if I construct 100 similar intervals for this sample size, 80 of them will contain the parameter.  I don't know which 80, so the probability can be used.  The important concept is not that there is an 80% chance the mean is in this interval.  It either is or it isn't, which is why we use confidence and not probability.

The last part requires 1.28*0.36/sqrt(n) to be < 0.1
0.2123<.01 n, squaring everything
21.23 is the sample size required, or n=22