Question

a professor records the following final grades in one course. construct a frequency table for the grades.
a a a b b b b b b b b c
c c c c c c d d d d d f

complete the table.
(type an integer or decimal rounded to the nearest tenth as needed.)
grade frequency relative frequency cumulative frequency
a 3 12.5 % 3
b □ □% □

Explanation:

Step1: Count Frequency of B

Count the number of 'B's in the grade list. The grades are: A, A, A, B, B, B, B, B, B, B, B, C, C, C, C, C, C, D, D, D, D, D, F. So B appears 8 times.

Step2: Calculate Total Number of Grades

Sum all frequencies: 3 (A) + 8 (B) + 6 (C) + 5 (D) + 1 (F) = 23? Wait, wait, let's recount. Wait the original list: First row: A, A, A, B, B, B, B, B, B, B, B, C (that's 12 elements). Second row: C, C, C, C, C, C, D, D, D, D, D, F (another 12? Wait no, maybe I miscounted. Wait the first line: A A A B B B B B B B B C (12 elements: 3 A, 8 B, 1 C? Wait no, 3 A, then B: let's count Bs: from index 3 to 10 (assuming 0-based), that's 8 Bs (3,4,5,6,7,8,9,10: 8 elements). Then C: index 11. Then second line: C, C, C, C, C, C (indices 12-17: 6 Cs), D, D, D, D, D (indices 18-22: 5 Ds), F (index 23). Wait total elements: 3 + 8 + 1 + 6 + 5 + 1? No, wait first row: A(3), B(8), C(1) → 12. Second row: C(5), D(5), F(1)? Wait no, the second row is: C, C, C, C, C, C, D, D, D, D, D, F. So C: 6, D:5, F:1. So total: 3 (A) + 8 (B) + 6 (C) + 5 (D) + 1 (F) = 23? Wait but the relative frequency of A is 3 / total = 12.5%, so 3 / total = 0.125 → total = 3 / 0.125 = 24. Ah! So total number of grades is 24. Because 3 / 24 = 0.125 = 12.5%. So that's the key. So total n = 24. So let's recalculate:

Total grades: n = 24 (since 3/24 = 0.125).

Now, frequency of B: count the Bs. Let's list all grades:

First row: A, A, A, B, B, B, B, B, B, B, B, C → B: 8 (positions 3-10, 8 elements)

Second row: C, C, C, C, C, C, D, D, D, D, D, F → Wait no, maybe the first row is 12 elements: A(3), B(8), C(1) → 12. Second row: C(5), D(5), F(1)? No, that can't be. Wait the problem's grade list:

"A A A B B B B B B B B C

C C C C C C D D D D D F"

Wait let's count each grade:

A: 3

B: let's count the Bs: in the first line, after 3 A's, how many B's? Let's see: "B B B B B B B B" → 8 B's (since 3 A's, then 8 B's: 3+8=11, then 1 C: 12th element. Then second line: "C C C C C C" → 6 C's (12-17: 6), then "D D D D D" → 5 D's (18-22: 5), then "F" → 1 (23). Wait 3+8+1+6+5+1=24. Yes! So total 24. So:

A: 3, B: 8, C: 1+6=7? Wait no, first line: C is 1, second line: C is 6 → total C: 7? Wait no, first line: A(3), B(8), C(1) → 12. Second line: C(5), D(5), F(1)? No, the second line is "C C C C C C D D D D D F" → C:6, D:5, F:1. So first line: A(3), B(8), C(1) → 12. Second line: C(6), D(5), F(1) → 12. So total C: 1+6=7, D:5, F:1. So total: 3+8+7+5+1=24. Correct.

So frequency of B: count the Bs. In the first line: after 3 A's, the B's are "B B B B B B B B" → 8 B's. So frequency of B is 8.

Relative frequency of B: (frequency of B) / total = 8 / 24 = 1/3 ≈ 0.333... = 33.3% (rounded to nearest tenth).

Cumulative frequency for B: cumulative frequency is the sum of frequencies up to and including B. So frequency of A (3) + frequency of B (8) = 11.

Step3: Verify

Total n=24.

Frequency of B: 8.

Relative frequency: 8/24 = 0.333... ≈ 33.3%.

Cumulative frequency: 3 (A) + 8 (B) = 11.

Answer:

For grade B:

Frequency: 8

Relative frequency: 33.3%

Cumulative frequency: 11

So the table cells for B are:

Frequency: 8

Relative frequency: 33.3%

Cumulative frequency: 11