In the figure attached standard normal distribution can be seen. The area under the curve from = 0 to the left is 0.5. You have to find the area between = 0 and Z, where Z is computed as Z = (x - mean)/standard deviation and x is the point of your interest, in this case x = 67. This procedure is made in this way because the table show the case when mean = 0 and standard deviation = 1. So, Z = (67 - 59)/8 = 1 and area = 0.3413. In consequence, the probability of a student to get less than 67 is 0.3413 + 0.5 = 0.8413 (the area under the curve from the left to Z = 1). In terms of students, 0.8413*5000 = 4207
Divide 6 1/3 by 3/4 first, convert 6 1/3 to an improper fraction 6 1/3 --> 19/3 make the denominators match 19/3 --> 76/12 3/4 --> 9/12 then multiply 76/12 by the reciprocal of 9/12 (which is 12/9) cross reduce, you get 76/9 convert back to a mixed number 76/9 --> 8 4/9 it would take 9 loads
