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When exam scores are low, students often ask the teacher whether he or she is going to "curve" the grades. The hope is that by curving a low score on the exam, the students will wind up getting a higher letter grade than might otherwise be expected. The term curving grades, or grading on a curve, comes from the bell curve of the normal distribution. If we assume that scores for a large number of students are distributed normally (as with SAT scores) and we also assume that the class average should be a "C," then a teacher might award grades as listed in the table below.

 

(A)  1.5 standard deviations above the mean or higher

(B)   0.5 to 1.5 standard deviations above the mean

(C)   within 0.5 standard deviation of the mean

(D)   0.5 to 1.5 standard deviations below the mean

(F)   1.5 standard deviations below the mean or lower

 

Suppose a teacher curved grades using the bell curve as in the table above and the grades were indeed normally distributed. What percent of students would get a grade of "A"? Round your answer as a percentage to one decimal place.
%

What percent of students would get a grade of "B"? Round your answer as a percentage to one decimal place. Suggestion: To find the percentage of students getting a grade of "B," subtract the percentage of students 0.5 standard deviation or less above the mean from the percentage of students 1.5 standard deviations or less above the mean.
%

Guest Mar 1, 2018
 #1
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0

coolHI how are you doing.coolcoolcoolcoolcoolcoolcoolcoolcoolcoolcool

Guest Mar 1, 2018
 #2
avatar+12565 
+2

Here is a bell curve with % and standard deviations shown

 

It does not have 1/2 standard deviations listed, but the graph is near linear in this region so we will interprolate the values for 1/2 SD

For an 'A' , 1 1/2 SD would encompass

  .15  + 2.35 + 13.5/2 % of the students = 9 %  (rounded to 1 decimal point)

 

B would encompass  the first 1/2 SD  34/2  plus 1/2 of the next SD   13.5/2  = 24% (rounded)

ElectricPavlov  Mar 1, 2018
 #3
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+1

Thanks for your help and explanation! I tried 9 percent for students getting an A, my homework said wrong answer. Also for grade B 24 percent was wrong too. I'm I missing information for this problem?

 

Correct answers ended up being 6.7% for an A.       100- 93.32= 6.68 round one place 6.7 %

For B it's    93.32 - 69.1463 = 24.17  round one place so   24.2%

Guest Mar 1, 2018
edited by Guest  Mar 1, 2018
edited by Guest  Mar 1, 2018
 #4
avatar+1101 
+3

To convert the standard deviations to percentage of AUC, use a Z-chart or calculator.

 

Using a Z-chart, find the 1.5 and 0.5 Sdvs area values, then subtract to find the area range.

The area from 0.5 to 1.5 Sdv is 0.9332 of the area (above 1.5) minus 0.6915 of the area (below 0.5) giving 24.2% of the total area.

 

Here are the percentages for the other deviations.

 

(A)  1.5 standard deviations above the mean or higher

       (1-0.9332) = 0.0668                          6.7%

 

(B)   0.5 to 1.5 standard deviations above the mean 

        (0.9332-0.6915) = 0.2417              24.2%

 

(C)   within 0.5 standard deviation of the mean

       (0.6925 - 0.3085) = 0.384              38.4%

 

(D)   0.5 to 1.5 standard deviations below the mean

       (0.3085 - 0.0668) = 0.2417           24.2%

 

(F)   1.5 standard deviations below the mean or lower

        (1-0.9332) = 0.0668                      6.7%

 

Note: when the standard deviations are below the mean the values are negative, use the “negative” or “left of mean” chart (or adjust by subtracting from 1).  

 

 

GA

GingerAle  Mar 2, 2018
edited by GingerAle  Mar 2, 2018
 #5
avatar+12565 
+1

Thanx for the correction GA !     ~EP

ElectricPavlov  Mar 2, 2018

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