A student gets 96 points on a test. The class average is 84, with a standard deviation of 4 points. What percentage of the class scored lower than this student? I understand the 96-84/4 part to get a z-score of 3. What I don’t understand is that if I don’t have access to an area z-score table, how can I get the area and then get the student-specific domain under that student’s score? There is no easy way. The actual mathematical formula is simple, but difficult to do by hand. I suggest you just have to know the values of 1,2,3 std from the mean. Just as it is better to know the trig values for common angles. Without more information, I would use a normal curve distribution and notice that the resulting score is 3 standard deviations from the mean and the 96% earning student is in the 99.9 percentile.
From the average on a normal curve, the 1st standard deviation is 34.1, then 13.6, then 2.1 percent with leaves only 0.1% beyond the 3rd standard deviation.
This is my answer and I stand by it😆 The answer is 99.85%. There is 0.15% on either side of the normal curve beyond three standard deviations.