33 students in a college Physics 101 course recently took a mid-term exam. 8 stu

Question

33 students in a college Physics 101 course recently took a mid-term exam. 8 students earned an A, 8 students earned a B, 6 students earned a C, 5 students earned a D and 6 students earned an F on the exam. The students were queried regarding the number of hours they had devoted to studying for the exam. 7 of the students who earned an A, 6 of the students who earned a B, 4 of the students who earned a C, 1 of the students who earned a D, and 1 of the students who earned an F reported that they had devoted more than 8 hours to studying for the exam. The remaining students reported that they had devoted no more than 8 hours to studying for the exam.

A.

0.05

B.

0.18

C.

0.36

D.

0.82

A.

0.42

B.

0.50

C.

0.58

D.

0.71

0.36

B.

0.05

C.

0.18

D.

0.64

4.      What is the probability of a randomly selected student having earned an A or a B on the exam given they devoted more than 8 hours to studying for the exam?

A.

0.21

B.

0.48

C.

0.68

D.

0.32

A.

0.05

B.

0.18

C.

0.36

D.

0.82

Explanation / Answer

1. Students(F) = 6; Students(Total) = 33. Students(F)/Students(Total) = 6/33 = .182 = (B) 2. Students(>8hrs) = 7 (>8hrs and A) + 6 (>8hrs and B)+ 4 (>8hrs and C) + 1 (>8hrs and D) + 1 (>8hrs and F) = 7+6+4+1+1 = 19; Students(Total) = 33. Students(>8hrs)/Students(Total) = 19/33 = .576 = .58 = (C) 3. Students(F and 6 Students(>8hrs) = see part 1 -> 19. (Students(A and >8hrs) + Students(B and >8hrs))/Students(>8hrs) = (7+6)/19 = 13/19 = .684 = .68 = (C)

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