Question
Plant Types: A,B,C,D,E
Number of Seedlings: 100,140,80,60,40
A farmer conducted an inventory on his two sqaure kilometers of rainforest; the results are in the table. He plans on randomly selecting a seedling to experiment on when he goes back to the forest. What is the probability that he will randomly pick a type B plant? Write your answer in the simplest form of a fraction.
2. Company X tested its lights to see how long they would last. On a production line, 2,500 bulbs were taken and tested. The department in charge reported that out of 2,500 bulbs, there were 2,200 bulbs that burnt out after more then 1,500 hours of use. Lucy purchased a light bulb from Compant X. What is the likelihood that it will burn for more than 1,500 hours? Write your answer in decimal form.
3. The letter tiles C,A,R,E,and S are placed in a box. Without looking, joelle picks a letter tile from the box and records the results before placing it back. He repeats the prodecure 100 times and observes the following results: outcomes: C,A,R,E,S frequency: 20,25,30,10,15
Based on the observed frequency, what is the probbaility that jorelle will pick letter R? The answer should be in percent form.
4. Outcomes: 2,4,6,8,10,12 Frequency: 10,9,6,15,13,8
Jede tossed a cube with faces numbered with 2,4,6,8,10, and 12. The results are recorded in the table. What outcome value is the largest discrepancy between expermiental and the expected probabiity of this experiment?
5. Outcomes: 2,4,6,8,10,12 Frequency: 10,9,6,15,13,8
Jede tossed a cube with faces numbered with 2,4,6,8,10, and 12. The results are recorded in the table. What outcome value is the largest discrepancy between expermiental and the expected probabiity of this experiment? Write your answer in 3 decimal places, rounded to the nearest thousandth.
Number of Seedlings: 100,140,80,60,40
A farmer conducted an inventory on his two sqaure kilometers of rainforest; the results are in the table. He plans on randomly selecting a seedling to experiment on when he goes back to the forest. What is the probability that he will randomly pick a type B plant? Write your answer in the simplest form of a fraction.
2. Company X tested its lights to see how long they would last. On a production line, 2,500 bulbs were taken and tested. The department in charge reported that out of 2,500 bulbs, there were 2,200 bulbs that burnt out after more then 1,500 hours of use. Lucy purchased a light bulb from Compant X. What is the likelihood that it will burn for more than 1,500 hours? Write your answer in decimal form.
3. The letter tiles C,A,R,E,and S are placed in a box. Without looking, joelle picks a letter tile from the box and records the results before placing it back. He repeats the prodecure 100 times and observes the following results: outcomes: C,A,R,E,S frequency: 20,25,30,10,15
Based on the observed frequency, what is the probbaility that jorelle will pick letter R? The answer should be in percent form.
4. Outcomes: 2,4,6,8,10,12 Frequency: 10,9,6,15,13,8
Jede tossed a cube with faces numbered with 2,4,6,8,10, and 12. The results are recorded in the table. What outcome value is the largest discrepancy between expermiental and the expected probabiity of this experiment?
5. Outcomes: 2,4,6,8,10,12 Frequency: 10,9,6,15,13,8
Jede tossed a cube with faces numbered with 2,4,6,8,10, and 12. The results are recorded in the table. What outcome value is the largest discrepancy between expermiental and the expected probabiity of this experiment? Write your answer in 3 decimal places, rounded to the nearest thousandth.
Answers
GPT-4o mini
Here are the answers to your questions:
1. **Probability of randomly picking a type B plant**:
- Total number of seedlings = 100 (A) + 140 (B) + 80 (C) + 60 (D) + 40 (E) = 420
- Number of type B plants = 140
- Probability of picking type B = Number of type B / Total number of seedlings = 140 / 420 = 1/3.
- **Answer**: \( \frac{1}{3} \)
2. **Likelihood that a purchased bulb will last more than 1,500 hours**:
- Total bulbs tested = 2,500
- Bulbs that lasted more than 1,500 hours = 2,200
- Probability = Number of bulbs lasting more than 1,500 hours / Total bulbs = 2,200 / 2,500 = 0.88.
- **Answer**: 0.88
3. **Probability that Joelle will pick letter R**:
- Total outcomes = 100 (sum of frequencies)
- Frequency of R = 30
- Probability of R = Frequency of R / Total outcomes = 30 / 100 = 0.30.
- To convert to percent = 0.30 × 100 = 30%.
- **Answer**: 30%
4. & 5. **Discrepancy between experimental and expected probabilities**:
- Expected probability of each outcome (assuming uniformly distributed outcomes) is \( \frac{1}{6} \) for each (since there are 6 outcomes).
- The frequencies are as follows:
- For 2: 10/100 = 0.10 (expected: 1/6 ≈ 0.1667)
- For 4: 9/100 = 0.09 (expected: 1/6 ≈ 0.1667)
- For 6: 6/100 = 0.06 (expected: 1/6 ≈ 0.1667)
- For 8: 15/100 = 0.15 (expected: 1/6 ≈ 0.1667)
- For 10: 13/100 = 0.13 (expected: 1/6 ≈ 0.1667)
- For 12: 8/100 = 0.08 (expected: 1/6 ≈ 0.1667)
- Discrepancies:
- For 2: |0.10 - 0.1667| = 0.0667
- For 4: |0.09 - 0.1667| = 0.0767
- For 6: |0.06 - 0.1667| = 0.1067
- For 8: |0.15 - 0.1667| = 0.0167
- For 10: |0.13 - 0.1667| = 0.0367
- For 12: |0.08 - 0.1667| = 0.0867
- The largest discrepancy is for outcome 6, which is 0.1067.
- **Answer**: 0.107 (rounded to three decimal places)
So, the answers are summarized as follows:
1. \( \frac{1}{3} \)
2. 0.88
3. 30%
4. & 5. 0.107
1. **Probability of randomly picking a type B plant**:
- Total number of seedlings = 100 (A) + 140 (B) + 80 (C) + 60 (D) + 40 (E) = 420
- Number of type B plants = 140
- Probability of picking type B = Number of type B / Total number of seedlings = 140 / 420 = 1/3.
- **Answer**: \( \frac{1}{3} \)
2. **Likelihood that a purchased bulb will last more than 1,500 hours**:
- Total bulbs tested = 2,500
- Bulbs that lasted more than 1,500 hours = 2,200
- Probability = Number of bulbs lasting more than 1,500 hours / Total bulbs = 2,200 / 2,500 = 0.88.
- **Answer**: 0.88
3. **Probability that Joelle will pick letter R**:
- Total outcomes = 100 (sum of frequencies)
- Frequency of R = 30
- Probability of R = Frequency of R / Total outcomes = 30 / 100 = 0.30.
- To convert to percent = 0.30 × 100 = 30%.
- **Answer**: 30%
4. & 5. **Discrepancy between experimental and expected probabilities**:
- Expected probability of each outcome (assuming uniformly distributed outcomes) is \( \frac{1}{6} \) for each (since there are 6 outcomes).
- The frequencies are as follows:
- For 2: 10/100 = 0.10 (expected: 1/6 ≈ 0.1667)
- For 4: 9/100 = 0.09 (expected: 1/6 ≈ 0.1667)
- For 6: 6/100 = 0.06 (expected: 1/6 ≈ 0.1667)
- For 8: 15/100 = 0.15 (expected: 1/6 ≈ 0.1667)
- For 10: 13/100 = 0.13 (expected: 1/6 ≈ 0.1667)
- For 12: 8/100 = 0.08 (expected: 1/6 ≈ 0.1667)
- Discrepancies:
- For 2: |0.10 - 0.1667| = 0.0667
- For 4: |0.09 - 0.1667| = 0.0767
- For 6: |0.06 - 0.1667| = 0.1067
- For 8: |0.15 - 0.1667| = 0.0167
- For 10: |0.13 - 0.1667| = 0.0367
- For 12: |0.08 - 0.1667| = 0.0867
- The largest discrepancy is for outcome 6, which is 0.1067.
- **Answer**: 0.107 (rounded to three decimal places)
So, the answers are summarized as follows:
1. \( \frac{1}{3} \)
2. 0.88
3. 30%
4. & 5. 0.107