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The expected value, often denoted as \( E(X) \), represents the average outcome you can expect from an experiment if it were repeated many times. In this exercise, the expected value is calculated based on the different monetary wins from rolling a die. Each possible monetary outcome (0, 1, or 10 dollars) is multiplied by its probability, and these products are then summed to find the expected average winnings per roll.
For our die example, the calculation is as follows:

• The probability of winning nothing (\(0\)) is \( \frac{1}{2} \),
• The probability of winning \(\\(1\) is \( \frac{1}{3} \),
• The probability of winning \(\\)10\) is \( \frac{1}{6} \).

Therefore, the expected value becomes:
\[ E(X) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{3} + 10 \cdot \frac{1}{6} = 2 \text{ dollars} \].
This tells us that, on average, you can expect to win \(\$2\) per roll.

Variance provides a measure of how much the outcomes of a probability distribution spread out from the expected value. A higher variance means a wider spread of possible outcomes, showing more unpredictability.
The variance for this dice-rolling game (\( Var(X) \)), is computed using:
\[ Var(X) = E(X^2) - [E(X)]^2 \].
You start by determining \( E(X^2) \) which involves squaring each outcome, multiplying by the probability, and summing:

• 0 squared times \( \frac{1}{2} \),
• 1 squared times \( \frac{1}{3} \),
• 10 squared times \( \frac{1}{6} \).

This results in \( E(X^2) = \frac{1}{3} + \frac{50}{3} = 17 \).
The variance is then computed as \( 17 - (2)^2 = 13 \).
A variance of \(13\) means there's a moderate spread in your winnings, indicating variability from rolling the dice.

The standard deviation is a way to quantify the amount of variation or dispersion in a set of values. It is simply the square root of the variance. For this exercise, since the variance \( Var(X) \) is 13, the standard deviation \( \sigma \) is given by:
\[ \sigma = \sqrt{13} \approx 3.6056 \].
This number shows how much your winnings are likely to fluctuate around the expected value (\(\$2\) per roll).
When you play multiple times, like twice or 40 times, the standard deviation helps understand how total winnings vary. By multiplying the variance by the number of plays and taking the square root, we see how the uncertainty grows with more rolls.

The normal approximation is a method used to approximate the probability of an outcome when dealing with large sample sizes. In this case, we're interested in the probability of winning at least \(\\(100\) in 40 rolls. Because we have 40 trials, the distribution of total winnings can be approximated using a normal distribution.
The total winnings have a mean \( \mu = 80 \) (calculated as \(40 \times 2\)) and a standard deviation \( \sigma \approx 22.8 \).
To find the probability of winning at least \(\\)100\), calculate the standardized \( Z \) score using:
\[ Z = \frac{X - \mu}{\sigma} \]
Where \(X = 100\), \(\mu = 80\), and \(\sigma = 22.8\), giving us:
\[ Z = \frac{100 - 80}{22.8} \approx 0.877 \].
Using the standard normal distribution table, this corresponds to a probability of about 0.19, meaning there is a 19% chance of winning at least \(\$100\) when playing 40 times.