X is a normally distributed random variable with mean 85 and standard deviation 17. What is the probability that X is less than 55?
You're given that X~N(85, 17), and need to find P(X<55).
If you're doing this by a TI-83 or TI-84, you do normalcdf(-1E99, 55, 85, 17) to arrive at an answer of 0.0388. (Remember that -1E99 is a simulation of negative infinity used for these "less than" problems.)
If you do this by hand, however, you'll need a z-table which operates on a standard normal distribution, N(0,1). However, your X is N(85,17), so you need to standardize it with a z score!
z = (x - mu) / sd
z = (55 - 85) / 17
z = -1.76
I rounded to two decimal places since most z-tables accommodate z-scores rounded to two-decimal places. Since this score is negative, this is a score below average, and your answer is below 0.5. If your z-table is in two pages, look at the first page with the negative z-scores. Look in the -1.7 row and the 0.06 column. This row and this column intersect at 0.0392, you answer.
Of course, the calculator is more accurate than a z-table because the z-table can only hold so much printed information. Manually, your answer is 0.0392, and by calculator, your answer is 0.0388. It seems that your answer choices were found by table, because the only option that makes sense of those four is 0.0388, or B.