What are the cutoffs for B?

Question

Suppose exam scores form an approximately normal distribution that has μ = 500 points and σ = 100 points. Letter grades on the exam were distributed as follows: Ds made up 16% of the exam, Cs 30%, Bs 17.5%, and As 2.5%.

 

(a) If 730 students scored 730 points or more, how many students took the exam?

 

(b) What are the cutoffs for a B?

Answer 1

Taylor,

 

The solution involves transforming an exam s-score to its corresponding z-score so you can look up values in the standardized Normal probability function.  Therefore,

 

Let            S =  the actual number of points actually scored in the exam (a random variable).

                Z =  the corresponding z-value of S given a mean and standard deviation (hence, also a random variable).

N(μ, σ)        = the general Normal probability distribution corresponding to mean μ and standard deviation σ.

N(500, 100) = the actual specific Normal distribution of the s-score random variable. 

N(0, 1)        = the standardized Normal Distribution given by a table.

               M = the number of students who took the exam.

 

Recall that  Z = (S - μ) / σ  = (S - 500)/100  is the relationship between S and Z.

 

Part (a):

S = 730 , so Z = (730 -500)/100 = 230/100 = 2.3     So 730 is actually 2.3 σ's to the right of the mean.

 

The area under the curve N(0, 1) between [2.3, + ∞)  is  0.0107 in its right-hand tail.

So only  1.07% of the students scored 730 or higher on this exam. Round it to 1 % .

Then M= 1.0/0.01 = 100.  So 100 students took the exam in a class.

 

Notice that if you don't round it, then you get a ridiculous M (93.46) since M must be an integer (a count of students). Or some very large number that clears the fraction and might represent a state-wide or national test.

 

Part (b):

The percentage of students who got any letter grade is given and represents the area under the N(0, 1) curve between the lower and upper boundary point corresponding to the letter.  The boundary points are the z-scores.  Remember that the table only gives the right half of the N(0, 1) because it is symmetrical about zero.  Then you convert them to the corresponding actual s-scores. 

 

Thus, if 2.5% got A's, then the area between [z*, the max possible] = 0.025 and z* = 1.96

So 1.96 = (S-500)/100 implies that S = 696 points.  The teacher may round it to 700 or above to get an A grade.

 

Thus if 17.5% got B's, then 20.0% (17.5% B's and 2.5% A's) got B or better.  Then, 20% of the area to the right of C begins at z* = 0.84.  So. 0.84 = (S - 500)/100 implies that S = 584 for the start of the B's.

Then B's correspond to the S-score range [584, 695], with A's above that.

 

z* for C + B + A is 0.00 because these grades correspond to 50% of the letter grades issued. So C = [500, 583].

 

Now the D's represent 16% of the area and it is in the left part of the curve, which is the negative of the right side. So 16% on the right side is z = 0.41, but because we're below the mean, flip the sign, so z* = - 0.41.

Then -0.41 = (S - 500)/100 implies that S = 459.  Hence a D = [459, 499].

 

Then every score below 459 is an "F".