A student scores 96 on a test. The class average is 84, with a standard deviation of 4 points. What percentage of the class scored below this student?
I understand the 96-84/4 part to obtain a z-score of 3. What I do not understand is, if I do not have access to a z-score to area table, how can I obtain the area and then obtain the area specific to students under this student's score?
3 answers
There is no easy way. The actual math formula is simple, but hard to do by hand. I suggest that you just have to know the values for 1,2,3 std from the mean. Just as it is best if you just know the trig values for common angles.
Without more information I'd use a normal curve distribution and notice the earned score is 3 standard deviations from the mean and the student earning a 96% is in the 99.9 percentile.
From the mean on a normal curve the 1st standard deviation is 34.1, then 13.6, then 2.1 percentages with leaves only .1% beyond the 3rd standard deviation.
That's my answer and I'm sticking to it😆
From the mean on a normal curve the 1st standard deviation is 34.1, then 13.6, then 2.1 percentages with leaves only .1% beyond the 3rd standard deviation.
That's my answer and I'm sticking to it😆
The answe is 99.85%. There is .15% on each side of the normal curve past three standard deviations.
2 answers