Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
### Step-by-Step Solution:
1. Tabulating the Data:
- The prizes and their respective counts are given as follows:
- \[tex]$1 drink: 44 prizes - \$[/tex]5 meal: 25 prizes
- \[tex]$10 gift card: 15 prizes - \$[/tex]20 gift card: 10 prizes
- \[tex]$100 gift card: 5 prizes 2. Total Number of Prizes: - Total number of prizes: \( 44 + 25 + 15 + 10 + 5 = 100 \) 3. Computing the Mean Value of Prizes: - The mean value of prizes is calculated by dividing the total value of prizes by the total number of prizes. - Total value of the prizes: \( 44 \times 1 + 25 \times 5 + 15 \times 10 + 10 \times 20 + 5 \times 100 \) - Total value: \( 44 + 125 + 150 + 200 + 500 = 1019 \) - Mean value: \( \frac{1019}{100} \approx 10.29 \) 4. Computing the Median Value of Prizes: - To find the median, we need to list all the prizes and find the middle value. - Sorted prizes: - \$[/tex]1 (44 times), \[tex]$5 (25 times), \$[/tex]10 (15 times), \[tex]$20 (10 times), \$[/tex]100 (5 times)
- Since the total number of prizes is 100 (an even number), the median is the average of the 50th and 51st values.
- Both the 50th and 51st values are \[tex]$5 (since the first 44 are \$[/tex]1 prizes and the next 25 are \[tex]$5 prizes). - Median value: \( \$[/tex]5 \)
5. Recomputing Without the \[tex]$100 Gift Card: - If the \$[/tex]100 gift card is removed, the counts change as follows:
- \[tex]$1 drink: 44 prizes - \$[/tex]5 meal: 25 prizes
- \[tex]$10 gift card: 15 prizes - \$[/tex]20 gift card: 10 prizes
- \[tex]$100 gift card: 0 prizes - New total number of prizes: \( 44 + 25 + 15 + 10 + 0 = 94 \) - Total value of the prizes without \$[/tex]100 gift card: [tex]\( 44 \times 1 + 25 \times 5 + 15 \times 10 + 10 \times 20 + 0 \times 100 \)[/tex]
- New total value: [tex]\( 44 + 125 + 150 + 200 + 0 = 519 \)[/tex]
- New mean value: [tex]\( \frac{519}{94} \approx 5.52 \)[/tex]
6. Recomputing Median Without the \[tex]$100 Gift Card: - Sorted prizes without \$[/tex]100:
- \[tex]$1 (44 times), \$[/tex]5 (25 times), \[tex]$10 (15 times), \$[/tex]20 (10 times)
- Since the total number of prizes is 94 (an even number), the median is the average of the 47th and 48th values.
- Both the 47th and 48th values are \[tex]$5 (since the first 44 are \$[/tex]1 prizes and the next 25 are \[tex]$5 prizes). - Median value remains: \( \$[/tex]5 \)
7. Effect of \[tex]$100 Gift Card: - Mean Value Comparison: - Mean with \$[/tex]100 gift card: [tex]\( 10.29 \)[/tex]
- Mean without \[tex]$100 gift card: \( 5.52 \) - Difference in mean: \( 10.29 - 5.52 = 4.77 \) - So, the \$[/tex]100 gift card increases the mean value of the prizes.
- Median Value Comparison:
- Median with \[tex]$100 gift card: \( 5 \) - Median without \$[/tex]100 gift card: [tex]\( 5 \)[/tex]
- Difference in median: [tex]\( 5 - 5 = 0 \)[/tex]
- So, the \[tex]$100 gift card does not change the median value of the prizes. ### Conclusion: - The \$[/tex]100 gift card increases the mean value of the prizes.
- The \[tex]$100 gift card does not affect the median value of the prizes. Hence, the correct interpretation of the \$[/tex]100 gift card's effect on the measure of center of the data is:
- It increases the mean value of the prizes.
1. Tabulating the Data:
- The prizes and their respective counts are given as follows:
- \[tex]$1 drink: 44 prizes - \$[/tex]5 meal: 25 prizes
- \[tex]$10 gift card: 15 prizes - \$[/tex]20 gift card: 10 prizes
- \[tex]$100 gift card: 5 prizes 2. Total Number of Prizes: - Total number of prizes: \( 44 + 25 + 15 + 10 + 5 = 100 \) 3. Computing the Mean Value of Prizes: - The mean value of prizes is calculated by dividing the total value of prizes by the total number of prizes. - Total value of the prizes: \( 44 \times 1 + 25 \times 5 + 15 \times 10 + 10 \times 20 + 5 \times 100 \) - Total value: \( 44 + 125 + 150 + 200 + 500 = 1019 \) - Mean value: \( \frac{1019}{100} \approx 10.29 \) 4. Computing the Median Value of Prizes: - To find the median, we need to list all the prizes and find the middle value. - Sorted prizes: - \$[/tex]1 (44 times), \[tex]$5 (25 times), \$[/tex]10 (15 times), \[tex]$20 (10 times), \$[/tex]100 (5 times)
- Since the total number of prizes is 100 (an even number), the median is the average of the 50th and 51st values.
- Both the 50th and 51st values are \[tex]$5 (since the first 44 are \$[/tex]1 prizes and the next 25 are \[tex]$5 prizes). - Median value: \( \$[/tex]5 \)
5. Recomputing Without the \[tex]$100 Gift Card: - If the \$[/tex]100 gift card is removed, the counts change as follows:
- \[tex]$1 drink: 44 prizes - \$[/tex]5 meal: 25 prizes
- \[tex]$10 gift card: 15 prizes - \$[/tex]20 gift card: 10 prizes
- \[tex]$100 gift card: 0 prizes - New total number of prizes: \( 44 + 25 + 15 + 10 + 0 = 94 \) - Total value of the prizes without \$[/tex]100 gift card: [tex]\( 44 \times 1 + 25 \times 5 + 15 \times 10 + 10 \times 20 + 0 \times 100 \)[/tex]
- New total value: [tex]\( 44 + 125 + 150 + 200 + 0 = 519 \)[/tex]
- New mean value: [tex]\( \frac{519}{94} \approx 5.52 \)[/tex]
6. Recomputing Median Without the \[tex]$100 Gift Card: - Sorted prizes without \$[/tex]100:
- \[tex]$1 (44 times), \$[/tex]5 (25 times), \[tex]$10 (15 times), \$[/tex]20 (10 times)
- Since the total number of prizes is 94 (an even number), the median is the average of the 47th and 48th values.
- Both the 47th and 48th values are \[tex]$5 (since the first 44 are \$[/tex]1 prizes and the next 25 are \[tex]$5 prizes). - Median value remains: \( \$[/tex]5 \)
7. Effect of \[tex]$100 Gift Card: - Mean Value Comparison: - Mean with \$[/tex]100 gift card: [tex]\( 10.29 \)[/tex]
- Mean without \[tex]$100 gift card: \( 5.52 \) - Difference in mean: \( 10.29 - 5.52 = 4.77 \) - So, the \$[/tex]100 gift card increases the mean value of the prizes.
- Median Value Comparison:
- Median with \[tex]$100 gift card: \( 5 \) - Median without \$[/tex]100 gift card: [tex]\( 5 \)[/tex]
- Difference in median: [tex]\( 5 - 5 = 0 \)[/tex]
- So, the \[tex]$100 gift card does not change the median value of the prizes. ### Conclusion: - The \$[/tex]100 gift card increases the mean value of the prizes.
- The \[tex]$100 gift card does not affect the median value of the prizes. Hence, the correct interpretation of the \$[/tex]100 gift card's effect on the measure of center of the data is:
- It increases the mean value of the prizes.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.