[See] The problem above allows you to come up with an estimate of the proportion p. Assuming the real proportion p was really equal to your sample proportion
Question:
The problem above allows you to come up with an estimate of the proportion p. Assuming the real proportion p was really equal to your sample proportion p-hat (in other words, using the sample proportion as the value for p), use the normal approximation to binomial to calculate the probability that out of a sample of 31 students, at least 5 spent more than 60 minutes per week doing homework.
| Minuets on | |
| homework per week | Test Average |
| 120 | 91 |
| 60 | 70 |
| 30 | 65 |
| 60 | 100 |
| 75 | 86 |
| 35 | 64 |
| 50 | 90 |
| 25 | 72 |
| 45 | 85 |
| 15 | 45 |
| 160 | 96 |
| 60 | 80 |
| 75 | 83 |
| 30 | 78 |
| 180 | 97 |
| 35 | 76 |
| 25 | 85 |
| 45 | 93 |
| 75 | 70 |
| 50 | 84 |
| 40 | 70 |
| 30 | 64 |
| 300 | 100 |
| 60 | 80 |
| 0 | 64 |
| 50 | 80 |
| 120 | 89 |
| 15 | 70 |
| 0 | 72 |
| 150 | 90 |
| 30 | 72 |
Deliverable: Word Document 