Question 927344
mean score of 60 and a standard deviation of 12.  {{{z = blue(x - 60)/blue(12)}}}  
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P(x > 65) = P(z > 5/12) = normalcdf(.4167,100) = .3384  0r 33.84%
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P(x < 57) = P(z < -3/12)= normalcdf(-100, -.25) = .4013  0r 40.13%

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P(x < 55) = P(z < -5/12) = normalcdf(-100, -.4167)= .3384  
200(.384)= 67.68,  68 got a grade of less than 55
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12invNorm(.30) + 60 = X
X = 12(-.5244) + 60 = 53.7.  54 is the lowest passing score