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| Preparatory Class | No Preparatory Class | Totals | |
|---|---|---|---|
| Passed Exams | 14 | 11 | 25 |
| Did Not Pass Exams | 3 | 6 | 9 |
| Totals | 17 | 17 | 34 |
m= 14, n= 17
Use a calculator
Round to 2 decimal place(s)
| Preparatory Class | No Preparatory Class | Totals | |
|---|---|---|---|
| Passed Exams | 14 | 11 | 25 |
| Did Not Pass Exams | 3 | 6 | 9 |
| Totals | 17 | 17 | 34 |
To find the probability, first let n denote the number of all students that did not take the preparatory class. We will also let m be the number of those students who neither took the class nor passed the exams. P(not passed | no class) = m/n Note that the second column of the contingency table shows the total number of students who did not take the preparatory classes, 17. The second row of this column represents the number of students from this group who did not pass the exams. P(not passed | no class) = 6/17 ≈ 0.35
| Preparatory Class | No Preparatory Class | Totals | |
|---|---|---|---|
| Passed Exams | 14 | 11 | 25 |
| Did not Pass Exams | 3 | 6 | 9 |
| Totals | 17 | 17 | 34 |
Similar to Parts A and B, let's use the given table to calculate the probability. Note that there are 11 students that passed the exams after not attending the class. Additionally, there are 17 students in total who did not take the class. Let's calculate the probability! P(passed | no class) = 11/17 ≈ 0.647 Now we can see that about 6 out of 10 students who do not take the class pass the exam. From Part A, we can conclude that about 8 out of 10 students pass the exams after taking the class. Therefore, the class appears to be beneficial and the student makes a good decision.