Question 1180818
a. z(0.05)=-1.645, so temp is -1.645 C.
b. 90th percentile is +1.282 C.
z(0.04)=-1.75 C. and z(0.97)=+1.88 C.
Those are the lower and upper limits respectively. 
use 2nd VARS3invnorm(0.05,0,1)ENTER, where 0.05 would be the 5th percentile, the 0, 1 are mean and sd of the standard normal distribution.