Questions: At the Spring Bazaar Charity Fundraiser, you play a game in which you roll two dice (all proceeds go to the charity). The game costs 1 to play. If you roll a sum of 2, you win 3.75. If you roll a sum greater than 2 and at most 8, you win 1. If you roll a sum of 11, you win 0.5. If you roll a sum of 12, you must donate an additional 1.75 to the charity. If you roll any other sum, the game is over. Complete the table to find the expected value of this game from the point of view of the Fundraising committee. (Simplify all answers. Enter probabilities as reduced fractions, and enter all dollar amounts rounded to two decimal places. Winning values should be positive and losing values should be negative.) Event Value () Probability Value × Probability () Sum of 2 1/36 2 < sum ≤ 8 Sum of 11 + 0.50 1/18 0.03 Sum of 12 1/36 All others Expected value of the game from the point of view of the fundraising committee: (answer if a loss is expected.)

Solution Steps

To find the expected value of the game from the perspective of the fundraising committee, we consider the net gains for each possible outcome and their associated probabilities. The expected value \( E \) is calculated as follows:

\[ E = (-2.75) \times \frac{1}{36} + 0 \times \frac{25}{36} + 0.5 \times \frac{1}{18} + 2.75 \times \frac{1}{36} + 1 \times \frac{7}{36} \]

Calculating each term:

• For the sum of 2: \[ -2.75 \times \frac{1}{36} = -0.0763888888888889 \]
• For sums greater than 2 and at most 8: \[ 0 \times \frac{25}{36} = 0 \]
• For the sum of 11: \[ 0.5 \times \frac{1}{18} = 0.027777777777777776 \]
• For the sum of 12: \[ 2.75 \times \frac{1}{36} = 0.0763888888888889 \]
• For all other sums: \[ 1 \times \frac{7}{36} = 0.19444444444444445 \]

Adding these values together gives: \[ E = -0.0763888888888889 + 0 + 0.027777777777777776 + 0.0763888888888889 + 0.19444444444444445 = 0.22 \]

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \sum (x_i - E)^2 \cdot P(x_i) \]

Where \( x_i \) are the values and \( P(x_i) \) are the probabilities. Thus, we compute:

\[ \sigma^2 = (-2.75 - 0.22)^2 \times \frac{1}{36} + (0 - 0.22)^2 \times \frac{25}{36} + (0.5 - 0.22)^2 \times \frac{1}{18} + (2.75 - 0.22)^2 \times \frac{1}{36} + (1 - 0.22)^2 \times \frac{7}{36} \]

Calculating each term:

• For the sum of 2: \[ (-2.75 - 0.22)^2 \times \frac{1}{36} = (-2.97)^2 \times \frac{1}{36} = 8.8209 \times \frac{1}{36} = 0.2445 \]
• For sums greater than 2 and at most 8: \[ (0 - 0.22)^2 \times \frac{25}{36} = (0.22)^2 \times \frac{25}{36} = 0.0484 \times \frac{25}{36} = 0.0335 \]
• For the sum of 11: \[ (0.5 - 0.22)^2 \times \frac{1}{18} = (0.28)^2 \times \frac{1}{18} = 0.0784 \times \frac{1}{18} = 0.0044 \]
• For the sum of 12: \[ (2.75 - 0.22)^2 \times \frac{1}{36} = (2.53)^2 \times \frac{1}{36} = 6.4009 \times \frac{1}{36} = 0.1778 \]
• For all other sums: \[ (1 - 0.22)^2 \times \frac{7}{36} = (0.78)^2 \times \frac{7}{36} = 0.6084 \times \frac{7}{36} = 0.1180 \]

Adding these values together gives: \[ \sigma^2 = 0.2445 + 0.0335 + 0.0044 + 0.1778 + 0.1180 = 0.5782 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{0.5782} \approx 0.76 \]

Final Answer