To calculate the standard deviation of the cash prize, we follow these steps:

Formula for standard deviation:

$\sigma = \sqrt{\sum_{i=1}^{n} P(x_i) \cdot (x_i - \mu)^2}$

Where:

• $x_i$ = Cash prize
• $P(x_i)$ = Probability of each cash prize
• $\mu$ = Expected value (mean)

Step 1: Calculate the expected value ($\mu$):

$\mu = \sum_{i=1}^{n} P(x_i) \cdot x_i$

Substitute the given values: $\mu = (16,000,000 \cdot 0.00000000705) + (200,000 \cdot 0.00000015) + (10,000 \cdot 0.000001829) + (100 \cdot 0.000154763) + (7 \cdot 0.004724435) + (4 \cdot 0.008646243) + (3 \cdot 0.01420886) + (0 \cdot 0.97226371295)$

Step 2: Compute the variance ($\sigma^2$):

$\sigma^2 = \sum_{i=1}^{n} P(x_i) \cdot (x_i - \mu)^2$

Substitute $\mu$ into the formula: $\sigma^2 = \sum_{i=1}^{n} P(x_i) \cdot \big(x_i^2 - 2x_i\mu + \mu^2\big)$

Step 3: Take the square root to find $\sigma$:

[ \sigma = \sqrt{\sigma^2} \ as np

Data

data = { "x": [16000000, 200000, 10000, 100, 7, 4, 3, 0], "P(x)": [0.00000000705, 0.00000015, 0.000001829, 0.000154763, 0.004724435, 0.008646243, 0.01420886, 0.97226371295] }

Create a DataFrame

df = pd.DataFrame(data)

Step 1: Calculate the expected value (mean)

df the expected value (mean) df['xP(x)'] = df['x'] * df['P(x)'] mean = df['xP(x)'].sum()

Step 2: Calculate the variance

df['(x - mean)^2'] = (df['x'] - mean) ** 2 df['P(x) * (x - mean)^2'] = df['P(x)'] * df['(x - mean)^2'] variance = df['P(x) * (x - mean)^2'].sum()

Step 3: Standard deviation

std_dev = np.sqrt### Results:

• Expected Value (Mean), $\mu$: $0.29 (approximately)
• Variance, $\sigma^2$: 1,810,984.86
• Standard Deviation, $\sigma$: $1,345.73

Would you like further clarification or details? Here are some related questions:

1. What does the standard deviation signify in this context?
2. How would the expected value change if the probabilities were adjusted?
3. How is variance different from standard deviation in interpretation?
4. What would the outcome be if the grand prize was increased?
5. Can you calculate the coefficient of variation (relative risk) using this data?

Tip: Standard deviation helps measure the spread of prizes around the mean, reflecting how likely