# 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?

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1. 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.

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2. 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😆

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3. The answe is 99.85%. There is .15% on each side of the normal curve past three standard deviations.

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