Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
To determine which statement is true, let's carefully examine both theoretical and experimental probabilities of students participating in swimming.
### Theorical Probability
1. Theoretical Probability of Swimming:
- Each student flips a coin. Since flipping a fair coin results in heads (swimming) or tails (hiking) with an equal chance of 1/2 each, the theoretical probability for swimming is indeed:
[tex]\[ P(\text{swimming}) = \frac{1}{2} = 0.5 \][/tex]
### Experimental Probability
2. Experimental Probability of Swimming:
- We are given that 23 students chose swimming out of 50 students.
- The experimental probability is calculated as:
[tex]\[ P(\text{swimming})_{exp} = \frac{23}{50} = 0.46 \][/tex]
### Statement Comparison
Now, let's verify the statements:
1. Statement 1:
- "The theoretical probability of swimming, [tex]\( P(\text{swimming}) \)[/tex], is [tex]\( \frac{1}{2} \)[/tex], but the experimental probability is [tex]\( \frac{23}{27} \)[/tex]."
- This statement is not correct as the experimental probability is [tex]\( \frac{23}{50} \)[/tex], not [tex]\( \frac{23}{27} \)[/tex].
2. Statement 2:
- "The theoretical probability of swimming, [tex]\( P(\text{swimming}) \)[/tex], is [tex]\( \frac{23}{27} \)[/tex], but the experimental probability is [tex]\( \frac{1}{2} \)[/tex]."
- This statement is also incorrect because the theoretical probability should be [tex]\( \frac{1}{2} \)[/tex]. Additionally, the experimental probability is [tex]\( \frac{23}{50} \)[/tex], not [tex]\( \frac{1}{2} \)[/tex].
3. Statement 3:
- "The theoretical probability of swimming, [tex]\( P(\text{swimming}) \)[/tex], is [tex]\( \frac{23}{50} \)[/tex], but the experimental probability is [tex]\( \frac{1}{2} \)[/tex]."
- This statement contains an accurate theoretical probability of [tex]\( \frac{1}{2} \)[/tex]. However, it incorrectly states the experimental probability. Actually, the theoretical probability is [tex]\( \frac{1}{2} \)[/tex] and the experimental probability is [tex]\( \frac{23}{50} \)[/tex] or 0.46.
Upon reviewing all options, none of the statements seem entirely accurate. Thus, it would be better to conclude which hypothesis matches the original given data and calculations.
#### Correct Review (From Given Data and Calculations):
The theoretical probability of swimming is [tex]\( \frac{1}{2} \)[/tex], and the experimental probability is [tex]\( \frac{23}{50} \)[/tex], which means that none of the original statements hold true. However, the closest accurate relation that matches given theoretical and experimental data is:
Theoretical probability of swimming, [tex]\( P(\text{swimming}) \)[/tex], is [tex]\( \frac{1}{2} \)[/tex], but the experimental probability is [tex]\( \frac{23}{50} \)[/tex].
Thus, the appropriate interpretation of both probabilities while based on given selected options confirms:
Theoretical probability [tex]\( P(\text{swimming}) = 0.5 \)[/tex], and Experimental probability [tex]\( P(\text{swimming})_{exp} = 0.46 \)[/tex].
### Theorical Probability
1. Theoretical Probability of Swimming:
- Each student flips a coin. Since flipping a fair coin results in heads (swimming) or tails (hiking) with an equal chance of 1/2 each, the theoretical probability for swimming is indeed:
[tex]\[ P(\text{swimming}) = \frac{1}{2} = 0.5 \][/tex]
### Experimental Probability
2. Experimental Probability of Swimming:
- We are given that 23 students chose swimming out of 50 students.
- The experimental probability is calculated as:
[tex]\[ P(\text{swimming})_{exp} = \frac{23}{50} = 0.46 \][/tex]
### Statement Comparison
Now, let's verify the statements:
1. Statement 1:
- "The theoretical probability of swimming, [tex]\( P(\text{swimming}) \)[/tex], is [tex]\( \frac{1}{2} \)[/tex], but the experimental probability is [tex]\( \frac{23}{27} \)[/tex]."
- This statement is not correct as the experimental probability is [tex]\( \frac{23}{50} \)[/tex], not [tex]\( \frac{23}{27} \)[/tex].
2. Statement 2:
- "The theoretical probability of swimming, [tex]\( P(\text{swimming}) \)[/tex], is [tex]\( \frac{23}{27} \)[/tex], but the experimental probability is [tex]\( \frac{1}{2} \)[/tex]."
- This statement is also incorrect because the theoretical probability should be [tex]\( \frac{1}{2} \)[/tex]. Additionally, the experimental probability is [tex]\( \frac{23}{50} \)[/tex], not [tex]\( \frac{1}{2} \)[/tex].
3. Statement 3:
- "The theoretical probability of swimming, [tex]\( P(\text{swimming}) \)[/tex], is [tex]\( \frac{23}{50} \)[/tex], but the experimental probability is [tex]\( \frac{1}{2} \)[/tex]."
- This statement contains an accurate theoretical probability of [tex]\( \frac{1}{2} \)[/tex]. However, it incorrectly states the experimental probability. Actually, the theoretical probability is [tex]\( \frac{1}{2} \)[/tex] and the experimental probability is [tex]\( \frac{23}{50} \)[/tex] or 0.46.
Upon reviewing all options, none of the statements seem entirely accurate. Thus, it would be better to conclude which hypothesis matches the original given data and calculations.
#### Correct Review (From Given Data and Calculations):
The theoretical probability of swimming is [tex]\( \frac{1}{2} \)[/tex], and the experimental probability is [tex]\( \frac{23}{50} \)[/tex], which means that none of the original statements hold true. However, the closest accurate relation that matches given theoretical and experimental data is:
Theoretical probability of swimming, [tex]\( P(\text{swimming}) \)[/tex], is [tex]\( \frac{1}{2} \)[/tex], but the experimental probability is [tex]\( \frac{23}{50} \)[/tex].
Thus, the appropriate interpretation of both probabilities while based on given selected options confirms:
Theoretical probability [tex]\( P(\text{swimming}) = 0.5 \)[/tex], and Experimental probability [tex]\( P(\text{swimming})_{exp} = 0.46 \)[/tex].
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
