The expected probability of each outcome when rolling the cube is 1/6 or approximately 16.67%.
The largest discrepancy between the experimental and expected probability is for the outcome of 8, which has a frequency of 15. The experimental probability for 8 is 15/61 or approximately 24.59%.
Therefore, the largest discrepancy is 24.59% - 16.67% = 7.92%, rounded to the nearest whole number is approximately 8%.
Use the table to answer the question.
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 is the largest discrepancy between the experimental and the expected probability of this experiment? The answer needs to be in percent form to the nearest whole number.
11 answers
Use the table to answer the question.
Outcomes
Frequency
2
10
4
6
8
10
15
6
9
13
8
12
Jede tossed a cube with faces numbered with 2, 4, 6, 8, 10, and 12. The results are recorded in the table. What is the smallest discrepancy between the experimental and the expected probability of this experiment? Write your answer in 3 decimal places, rounded to the nearest thousandth.
Outcomes
Frequency
2
10
4
6
8
10
15
6
9
13
8
12
Jede tossed a cube with faces numbered with 2, 4, 6, 8, 10, and 12. The results are recorded in the table. What is the smallest discrepancy between the experimental and the expected probability of this experiment? Write your answer in 3 decimal places, rounded to the nearest thousandth.
To find the smallest discrepancy between the experimental and expected probability, we first need to calculate the expected probability for each outcome. Since there are 6 sides on a cube, each outcome has an expected probability of 1/6 or approximately 0.167.
Now we can calculate the experimental probability for each outcome based on the frequencies given in the table.
For outcome 2: Experimental probability = 10/61 ≈ 0.164.
For outcome 4: Experimental probability = 6/61 ≈ 0.098.
For outcome 6: Experimental probability = 9/61 ≈ 0.148.
For outcome 8: Experimental probability = 15/61 ≈ 0.246.
For outcome 10: Experimental probability = 13/61 ≈ 0.213.
For outcome 12: Experimental probability = 8/61 ≈ 0.131.
To find the smallest discrepancy, we need to compare the experimental and expected probabilities for each outcome:
For outcome 2: Discrepancy = |0.164 - 0.167| ≈ 0.003.
For outcome 4: Discrepancy = |0.098 - 0.167| ≈ 0.069.
For outcome 6: Discrepancy = |0.148 - 0.167| ≈ 0.019.
For outcome 8: Discrepancy = |0.246 - 0.167| ≈ 0.079.
For outcome 10: Discrepancy = |0.213 - 0.167| ≈ 0.046.
For outcome 12: Discrepancy = |0.131 - 0.167| ≈ 0.036.
The smallest discrepancy is 0.003, which occurs for outcome 2.
Now we can calculate the experimental probability for each outcome based on the frequencies given in the table.
For outcome 2: Experimental probability = 10/61 ≈ 0.164.
For outcome 4: Experimental probability = 6/61 ≈ 0.098.
For outcome 6: Experimental probability = 9/61 ≈ 0.148.
For outcome 8: Experimental probability = 15/61 ≈ 0.246.
For outcome 10: Experimental probability = 13/61 ≈ 0.213.
For outcome 12: Experimental probability = 8/61 ≈ 0.131.
To find the smallest discrepancy, we need to compare the experimental and expected probabilities for each outcome:
For outcome 2: Discrepancy = |0.164 - 0.167| ≈ 0.003.
For outcome 4: Discrepancy = |0.098 - 0.167| ≈ 0.069.
For outcome 6: Discrepancy = |0.148 - 0.167| ≈ 0.019.
For outcome 8: Discrepancy = |0.246 - 0.167| ≈ 0.079.
For outcome 10: Discrepancy = |0.213 - 0.167| ≈ 0.046.
For outcome 12: Discrepancy = |0.131 - 0.167| ≈ 0.036.
The smallest discrepancy is 0.003, which occurs for outcome 2.
Use the table to answer the question.
Outcomes
2
4
6
8
10
12
Frequency
10
6
15
13
8
9
Jede tossed a cube with faces numbered with 2, 4, 6, 8, 10, and 12. The results are recorded in the table. What is the smallest discrepancy between the experimental and the expected probability of this experiment? Write your answer in 3 decimal places, rounded to the nearest thousandth.
Outcomes
2
4
6
8
10
12
Frequency
10
6
15
13
8
9
Jede tossed a cube with faces numbered with 2, 4, 6, 8, 10, and 12. The results are recorded in the table. What is the smallest discrepancy between the experimental and the expected probability of this experiment? Write your answer in 3 decimal places, rounded to the nearest thousandth.
To find the smallest discrepancy between the experimental and expected probability in this case:
First, we need to calculate the total number of trials, which is the sum of all frequencies given in the table:
Total number of trials = 10 + 6 + 15 + 13 + 8 + 9 = 61
Next, we calculate the expected probability for each outcome:
Since there are 6 faces on the cube, each outcome has an expected probability of 1/6 ≈ 0.167.
Now, we calculate the experimental probability for each outcome based on the frequencies given in the table:
Experimental probability for 2: 10/61 ≈ 0.164
Experimental probability for 4: 6/61 ≈ 0.098
Experimental probability for 6: 15/61 ≈ 0.246
Experimental probability for 8: 13/61 ≈ 0.213
Experimental probability for 10: 8/61 ≈ 0.131
Experimental probability for 12: 9/61 ≈ 0.148
To find the smallest discrepancy, we compare the experimental and expected probabilities for each outcome:
For outcome 2: |0.164 - 0.167| = 0.003
For outcome 4: |0.098 - 0.167| = 0.069
For outcome 6: |0.246 - 0.167| = 0.079
For outcome 8: |0.213 - 0.167| = 0.046
For outcome 10: |0.131 - 0.167| = 0.036
For outcome 12: |0.148 - 0.167| = 0.019
The smallest discrepancy is 0.003, which occurs for outcome 2.
First, we need to calculate the total number of trials, which is the sum of all frequencies given in the table:
Total number of trials = 10 + 6 + 15 + 13 + 8 + 9 = 61
Next, we calculate the expected probability for each outcome:
Since there are 6 faces on the cube, each outcome has an expected probability of 1/6 ≈ 0.167.
Now, we calculate the experimental probability for each outcome based on the frequencies given in the table:
Experimental probability for 2: 10/61 ≈ 0.164
Experimental probability for 4: 6/61 ≈ 0.098
Experimental probability for 6: 15/61 ≈ 0.246
Experimental probability for 8: 13/61 ≈ 0.213
Experimental probability for 10: 8/61 ≈ 0.131
Experimental probability for 12: 9/61 ≈ 0.148
To find the smallest discrepancy, we compare the experimental and expected probabilities for each outcome:
For outcome 2: |0.164 - 0.167| = 0.003
For outcome 4: |0.098 - 0.167| = 0.069
For outcome 6: |0.246 - 0.167| = 0.079
For outcome 8: |0.213 - 0.167| = 0.046
For outcome 10: |0.131 - 0.167| = 0.036
For outcome 12: |0.148 - 0.167| = 0.019
The smallest discrepancy is 0.003, which occurs for outcome 2.
The sample space of an experiment is S = \{1, 2, 3, 4, 5\} If each outcome is equally likely to occur, which graph models the expected probability of each outcome? (1 point)
Since each outcome is equally likely to occur in the sample space S = {1, 2, 3, 4, 5}, the expected probability for each outcome is 1/5 or 0.2.
The graph that models the expected probability of each outcome when each outcome is equally likely to occur is a uniform distribution graph, where each outcome has the same probability represented by a constant level of 0.2 across the outcomes 1, 2, 3, 4, and 5.
The graph that models the expected probability of each outcome when each outcome is equally likely to occur is a uniform distribution graph, where each outcome has the same probability represented by a constant level of 0.2 across the outcomes 1, 2, 3, 4, and 5.
Use the tables to answer the question.
Simon's Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A
5
15
Bowl B
16
4
Clark's Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A
45
55
Bowl B
60
40
Bowls A and B contain a number of white and red balls. Simon and Clark repeatedly selected a ball from both bowls and recorded the results in a table. Whose results will give you a better indication about the proportion of white and red balls in each bowl? Explain your answer.
Simon's Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A
5
15
Bowl B
16
4
Clark's Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A
45
55
Bowl B
60
40
Bowls A and B contain a number of white and red balls. Simon and Clark repeatedly selected a ball from both bowls and recorded the results in a table. Whose results will give you a better indication about the proportion of white and red balls in each bowl? Explain your answer.
To determine whose results will give a better indication of the proportion of white and red balls in each bowl, we need to analyze the data provided by Simon and Clark.
Simon's results:
- From Bowl A, Simon selected 5 white balls and 15 red balls.
- From Bowl B, Simon selected 16 white balls and 4 red balls.
Clark's results:
- From Bowl A, Clark selected 45 white balls and 55 red balls.
- From Bowl B, Clark selected 60 white balls and 40 red balls.
Comparing the results:
- Simon's results from Bowl A and Bowl B show a fairly large difference in the number of white and red balls selected, indicating that the proportions are not balanced.
- Clark's results from Bowl A and Bowl B show a more balanced distribution of white and red balls selected, with a more significant number of white balls selected in Bowl B and a more significant number of red balls selected in Bowl A.
Therefore, Clark's results will give a better indication of the proportion of white and red balls in each bowl because they show a more balanced distribution and provide more reliable information about the contents of the bowls. Simon's results have larger discrepancies, and the proportions are less clear from his data.
Simon's results:
- From Bowl A, Simon selected 5 white balls and 15 red balls.
- From Bowl B, Simon selected 16 white balls and 4 red balls.
Clark's results:
- From Bowl A, Clark selected 45 white balls and 55 red balls.
- From Bowl B, Clark selected 60 white balls and 40 red balls.
Comparing the results:
- Simon's results from Bowl A and Bowl B show a fairly large difference in the number of white and red balls selected, indicating that the proportions are not balanced.
- Clark's results from Bowl A and Bowl B show a more balanced distribution of white and red balls selected, with a more significant number of white balls selected in Bowl B and a more significant number of red balls selected in Bowl A.
Therefore, Clark's results will give a better indication of the proportion of white and red balls in each bowl because they show a more balanced distribution and provide more reliable information about the contents of the bowls. Simon's results have larger discrepancies, and the proportions are less clear from his data.
Use the tables to answer the question.
Simon's Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A
5
15
Bowl B
16
4
Clark's Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A
45
55
Bowl B
60
40
Bowls A and B contain a number of white and red balls. Simon and Clark repeatedly selected a ball from both bowls and recorded the results in a table. Whose results will give you a better indication about the proportion of white and red balls in each bowl? Explain your answer.
Neither; they conducted the same number of trials.
Clark's experiment. He conducted a greater number of trials than Simon.
Simon's experiment. He conducted a greater number of trials than Clark.
Both results can be used, as they performed the experiment at random.
Simon's Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A
5
15
Bowl B
16
4
Clark's Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A
45
55
Bowl B
60
40
Bowls A and B contain a number of white and red balls. Simon and Clark repeatedly selected a ball from both bowls and recorded the results in a table. Whose results will give you a better indication about the proportion of white and red balls in each bowl? Explain your answer.
Neither; they conducted the same number of trials.
Clark's experiment. He conducted a greater number of trials than Simon.
Simon's experiment. He conducted a greater number of trials than Clark.
Both results can be used, as they performed the experiment at random.
To determine whose results will give a better indication of the proportion of white and red balls in each bowl, we need to consider the reliability of the data provided by Simon and Clark.
Simon's Results:
- Simon selected a total of 20 balls from each bowl, with 5 white and 15 red balls from Bowl A, and 16 white and 4 red balls from Bowl B.
- Simon's results show a larger difference in the number of white and red balls selected from each bowl, indicating imbalanced proportions.
Clark's Results:
- Clark selected a total of 100 balls from each bowl, with 45 white and 55 red balls from Bowl A, and 60 white and 40 red balls from Bowl B.
- Clark's results show a more balanced distribution of white and red balls selected from each bowl, with a clearer representation of the proportions.
Explanation:
- Clark's results will give a better indication of the proportion of white and red balls in each bowl because he conducted a larger number of trials (100 per bowl) compared to Simon's 20 trials per bowl. More trials reduce the impact of randomness and provide a more accurate representation of the true proportions.
Therefore, the correct response is: Clark's experiment. He conducted a greater number of trials than Simon.
Simon's Results:
- Simon selected a total of 20 balls from each bowl, with 5 white and 15 red balls from Bowl A, and 16 white and 4 red balls from Bowl B.
- Simon's results show a larger difference in the number of white and red balls selected from each bowl, indicating imbalanced proportions.
Clark's Results:
- Clark selected a total of 100 balls from each bowl, with 45 white and 55 red balls from Bowl A, and 60 white and 40 red balls from Bowl B.
- Clark's results show a more balanced distribution of white and red balls selected from each bowl, with a clearer representation of the proportions.
Explanation:
- Clark's results will give a better indication of the proportion of white and red balls in each bowl because he conducted a larger number of trials (100 per bowl) compared to Simon's 20 trials per bowl. More trials reduce the impact of randomness and provide a more accurate representation of the true proportions.
Therefore, the correct response is: Clark's experiment. He conducted a greater number of trials than Simon.