Hi Katie, this problem is easiest solved using software of some sort. Without software, this question can be solved using the Z table in the back of your textbook.
To solve with a ti-84: 2nd -> vars -> normalcdf(0.284,2.051, 0, 1) = 0.3681
To solve in with a Z-table, we will find our two values using the first column and row, before moving to the associated probability. So, we will first find the probability associated with temperatures below 0.284 degrees celsius. I start by going to 0.2 in the first column, then over to 0.08 in the first row. I meet these two values for a probability = 0.6118. Next we find the probability associated to the left of 2.051 by finding 2.05 on the z table and its probability = 0.9799.
0.9799 - 0.6118 = 0.368
The process behind the Z values is a bit complicated to understand in forms of words. I'm happy to meet with you to draw out and explain if need be!
