381 - Making the Grade
Solution Description : Ad hoc simulation problem Suppose Mr. Chips has 5 students and has given 3 tests. The following table shows the grades, number of bonuses and days absent, plus the computed average (with lowest test dropped), the adjusted average (with bonus), the unadjusted grade and the adjusted grade (with attendance).  1. mean = (100.0+80.0+65.0+40.0+60.0)/5 = 69.0 2. The standard deviation sd of a list of numbers  is: sd =sqrt ((100.0-69.0)^2 + (80.0-69.0)^2+(65.0-69.0)^2+( = 20.1 If the calculated standard deviation is less than 1 then Mr. Chips uses 1 in place of the standard deviation for grade calculation. (i.e. if sd < 1 then set sd = 1) 3. For every two bonus points accrued by a student Mr. Chips increases their computed average by 3 percentage points. Thus, if students have one bonus point, their averages are not bumped at all. If they have 4 or 5 bonus points, their averages are bumped by 6 percentage points, and so on. So, AdjAvg = avg + (Bns / 2)*3. 4. The cutoff points for grades are: an average  one sd above the mean is an A, an average the mean but < one sd above the mean is a B, an average one sd below the mean but < the mean is a C, and an average < one sd below the mean is a D. a. If AdjAvg >= (mean + sd) then, grade=A b. If (mean) > AdjAvg >= mean then, grade=B c. If mean > AdjAvg >= (mean - sd) then, grade=C d. If (mean-sd) > AdjAvg then, grade = D 5. After find grade, for every 4 absences, students lose one letter grade (from A to B, B to C, C to D, and D to F). For example, if they have 9 absences they will lose two letter grades. Students cannot get a grade lower than F. If students have perfect attendance(i.e. if absence =0 ), they gain one letter grade; although they cannot get a grade higher than an A. |
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300 - Maya Calendar