Homework Answers
Here as the distribution is normal
Mean score = 65
standard deviation = 4
I scored = 72
so here
standarized z score = (72 - 65)/4 = 1.75
so here as we look the percentile score related to z score = NORMSINV(1.75) = 0.96
so here there are 96% of students had a z score less than mine.
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