The table shows the probabilities of certain prizes in a restaurant's contest where the first 100 customers are winners.

| Prize | Number of Prizes |
|------------------|------------------|
| [tex]$1 drink | 44 |
| $[/tex]5 meal | 25 |
| [tex]$10 gift card | 15 |
| $[/tex]20 gift card | 10 |
| [tex]$100 gift card | 5 |

How does the $[/tex]100 gift card affect the measure of center of the data?

A. It increases the mean value of the prizes.
B. It decreases the mean value of the prizes.
C. It increases the median value of the prizes.
D. It decreases the median value of the prizes.



Answer :

### 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.