Questions: For a multistate lottery, the following probability distribution represents the cash prizes of the lottery with their corresponding probabilities. Complete parts (a) through (d) below. x (cash prize, ) P(x) Grand prize 0.00000000588 200,000 0.00000027 10,000 0.000001592 100 0.000163221 7 0.004732029 4 0.007546146 3 0.01133647 0 0.97622026612 The expected profit from one 1 ticket is -0.71. (Round to the nearest cent as needed.) (b) If the grand prize is 18,000,000, what is the standard deviation of the cash prize? σx= 1384 (Round to the nearest dollar as needed.) What does this value suggest? A. This suggests the expected value is only accurate for a small number of people. B. This suggests there is a wide range of payouts. C. This suggests many people buy tickets. D. This suggests there is almost no variability.

Solution Steps

The expected value (mean) of the cash prizes is calculated using the formula:

\[ \text{Mean} = \sum (x \cdot P(x)) = 18000000 \times 5.88 \times 10^{-9} + 200000 \times 2.7 \times 10^{-7} + 10000 \times 1.592 \times 10^{-6} + 100 \times 0.000163221 + 7 \times 0.004732029 + 4 \times 0.007546146 + 3 \times 0.01133647 + 0 \times 0.97622026612 = 0.0 \]

Thus, the mean cash prize is \(0.0\).

The variance is calculated using the formula:

\[ \sigma^2 = \sum ((x - \text{Mean})^2 \cdot P(x)) = (18000000 - 0.0)^2 \times 5.88 \times 10^{-9} + (200000 - 0.0)^2 \times 2.7 \times 10^{-7} + (10000 - 0.0)^2 \times 1.592 \times 10^{-6} + (100 - 0.0)^2 \times 0.000163221 + (7 - 0.0)^2 \times 0.004732029 + (4 - 0.0)^2 \times 0.007546146 + (3 - 0.0)^2 \times 0.01133647 + (0 - 0.0)^2 \times 0.97622026612 = 1916081.0 \]

Thus, the variance is \(1916081.0\).

The standard deviation is the square root of the variance:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{1916081.0} = 1384.0 \]

Thus, the standard deviation of the cash prize is \(1384.0\).

The calculated standard deviation suggests that there is a wide range of payouts in the lottery, indicating variability in the cash prizes.

Final Answer