Andy C. answered 05/05/18
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Per the normal table 83.4% = 0.8340 has a z-score of exactly z=0.97
0.97 = (X-85)/20
0.97*20 + 85 = X
X = 104.4 <--- cutoff passing grade
So Prob( X<=104.4) = Prob ( z < (104.4-85)/20) = Prob (z < 0.97) = 0.8340 = 83.4%