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

Contest Prizes

[tex]\[
\begin{tabular}{|c|c|}
\hline
\textbf{Prize} & \textbf{Number of Prizes} \\
\hline
\$1 drink & 44 \\
\hline
\$5 meal & 25 \\
\hline
\$5 gift card & 15 \\
\hline
\$10 gift card & 10 \\
\hline
\$20 gift card & 5 \\
\hline
\$100 gift card & 1 \\
\hline
\end{tabular}
\][/tex]

How does the \$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.

Sagot :

Let's analyze the given data to understand how the \[tex]$100 gift card affects the measures of the center, specifically the mean and the median. Data Provided: - \$[/tex]1 drink: 44 prizes
- \[tex]$5 meal: 25 prizes - \$[/tex]5 gift card: 15 prizes
- \[tex]$10 gift card: 10 prizes - \$[/tex]20 gift card: 5 prizes
- \[tex]$100 gift card: 1 prize Step-by-step Calculation: a) Calculating the Mean: The mean value of the prizes is the weighted average of all the prizes. We calculate it by summing up the product of each prize amount and its frequency, then dividing by the total number of prizes. \[ \text{Mean} = \frac{\sum (\text{Prize Amount} \times \text{Frequency})}{\sum \text{Frequency}} \] Plugging in the values: \[ \text{Sum of Products} = (1 \times 44) + (5 \times 25) + (5 \times 15) + (10 \times 10) + (20 \times 5) + (100 \times 1) \] \[ \text{Sum of Frequencies} = 44 + 25 + 15 + 10 + 5 + 1 = 100 \] \[ \text{Mean} = \frac{(1 \cdot 44) + (5 \cdot 25) + (5 \cdot 15) + (10 \cdot 10) + (20 \cdot 5) + (100 \cdot 1)}{100} \] \[ \text{Mean} = \frac{44 + 125 + 75 + 100 + 100 + 100}{100} = \frac{544}{100} = 5.44 \] Therefore, the mean value of the prizes is 5.44. The \$[/tex]100 gift card contributes significantly to increasing the mean value due to its large amount relative to other prizes.

b) Calculating the Median:

To determine the median, we need to find the middle value(s) of the ordered list of prizes.

An ordered list (sorted non-decreasingly) of all the prizes with their respective frequencies would be:

[tex]\[ \text{(44 \$1 prizes)}, \text{(40 \$5 prizes)}, \text{(10 \$10 prizes)}, \text{(5 \$20 prizes)}, \text{(1 \$100 prize)} \][/tex]

The total number of prizes is 100, which is even. Therefore, the median is the average of the 50th and 51st values in this ordered list.

- The first 44 values are all \[tex]$1. - The next 25 values (from 45th to 69th) are all \$[/tex]5.

Since the 50th and 51st values lie in the middle of the 25 \[tex]$5 values, both 50th and 51st values are \$[/tex]5.

[tex]\[ \text{Median} = \frac{\text{50th value} + \text{51st value}}{2} = \frac{5 + 5}{2} = 5 \][/tex]

Therefore, the median value of the prizes is 5. Since the \[tex]$100 prize does not change which values are in the middle, it does not affect the median. Conclusion: - The \$[/tex]100 gift card increases the mean value of the prizes to 5.44.
- The \[tex]$100 gift card does not affect the median value, which remains 5. Answer: The \$[/tex]100 gift card increases the mean value of the prizes.