Manhattan Chapter 21 (Standard Deviation) Question 12 [Percentiles]

Jane scored in the 68th percentile and John scored in the 32nd percentile.

Quantity A: The proportion of the class that received a score less than John
Quantity B: the proportion of the class that scored equal to or greater than Jane’s score

Here the correct answer is C, they are the same. I wanted to clarify that when calculating percentiles the range is typically between 0-99. So to calculate quantity A it would be 32-0. Then for B to calculate B in the answer guide it explains "Jane scored in the 68th percentile so 68% of the class scored worse than she did. Since 100-68-32, 32% of the class scored equal or greater than Jane.

My guess is that the number 100 in their calculation comes into play from 99th percentile +1, since the maximum percentile is 99, but hypothetically the 100th percentile doesn’t exist.

I wanted to check to see that if they did not include the fact that for quantity B that it is greater than or equal the calculation would have been 99-68. Likewise if they had said for quantity A the proportion of scores less than or equal to John the calculation would have been 32-0+1 = 33%.

68th percentile → Jane scored strictly more than 68% of the students. Then it cannot be the case that 32% of students could score more than or equal to Jane’s score (note the emphasis).