A
group of statistics students took a midterm exam. The class mean was 72
and the standard deviation was 3.2. The instructor decided to "curve"
the grades based on their normal distribution. A grade of C was given if
students were within one standard deviation of the mean. B’s and D’s
were between one and two standard deviations, and A’s and F’s were
outside of two standard deviations.
Part a)
Using the rules of rounding and
the normal curve, establish a grade scale for this exam. All grades
should be rounded to a whole number.
A grade of A is between 72 + 2(3.2) = 72 + 6.4 = 78.4 ≈ 78 and
100. [grade of A is above 2 standard deviations of the mean].
Thus a grade of A is between 78 and 100.
A grade of B is between 72 + 3.2 = 75.2 ≈ 75 and 72 + 2(3.2) = 72 + 6.4 = 78.4 ≈ 78. [grade of B is between 1 standard deviations and 2 standard deviations of the mean].
Thus, a grade of B is between 75 and 78.
A grade of C is
between 72 - 3.2 = 68.8 ≈ 69 and 72 + 3.2 = 75.2 ≈ 75. [grade of C is within 1 standard deviations of the mean].
Thus, a grade of C is between 69 and 75.
A grade of D is between 72 - 3.2 = 68.8 ≈ 69 and 72 - 2(3.2) = 72 - 6.4 = 65.6 ≈ 66. [grade of D is between 1 standard deviations and 2 standard deviations of the mean]
Thus, a grade of D is between 66 and 69
A
grade of F is between 0 and 72 - 6.4 = 65.6 ≈ 66. [grade of F is below 2 standard deviations of the mean].
Thus, a grade of D is between 0 and 66.
Part b)
What letter grade would a
score of 66 receive?
Let the number of points of 66 from the mean be a, then
72 - a(3.2) = 66
3.2a = 72 - 66 = 6
a = 6 / 3.2 = 1.875
Thus, 66 is between 1 standard deviation and 2 standard deviation of the mean.
Therefore, a score of 66 would receive a letter grade of D.
Part c)
What letter grade would a score of 77 receive?
Let the number of points of 77 from the mean be b, then
72 + b(3.2) = 77
3.2b = 77 - 72 = 5
b = 5 / 3.2 = 1.5625
Thus, 77 is between 1 standard deviation and 2 standard deviation of the mean.
Therefore, a score of 77 would receive a letter grade of B.
Part d)
If there are 50 students in the class, how many scored a C?
By the empirical rule, 68% of a dataset is within 1 standard deviation.
Given that a grade of C was given if students were within 1 standard deviation of the mean, then the number of students that scored C is given by 68% of 50 = 0.68(50) = 34.
Therefore, 34 students scored a C.
Part e)
If
there are 40 students in the class, how many scored an A?
By the empirical rule, 95% of a dataset is within 2 standard deviation.
Given that A’s and F’s were outside of two standard deviations, then the number of students that scored A's and F's is given by 40 - 95% of 40 = 40 - 0.95(40) = 40 - 38 = 2.
Therefore, 2 / 2 = 1 student scored an A.
