Answer :
Certainly! Let's dive into the detailed calculations for each part of the problem using the provided probability density function for \(Z\).
### Problem Analysis
The random variable \(Z\) is uniformly distributed on the interval \([0, 5]\). This gives us a constant density function \(f_Z(z)\):
[tex]\[ f_Z(z) = \begin{cases} \frac{1}{5}, & \text{if } 0 \leq z \leq 5 \\ 0, & \text{otherwise} \end{cases} \][/tex]
Let's consider \(k = 2.5\) as a middle point for simplicity.
### Part (a): Expected Profit of the Game
To calculate the expected profit, we need to consider the probability of losing or winning based on the value of \(Z\):
1. Probability of Losing (\(Z \leq k\)):
[tex]\[ P(Z \leq k) = \frac{k}{5} = \frac{2.5}{5} = 0.5 \][/tex]
2. Probability of Winning (\(Z > k\)):
[tex]\[ P(Z > k) = 1 - P(Z \leq k) = 1 - 0.5 = 0.5 \][/tex]
The expected profit, therefore, can be computed as:
[tex]\[ E[\text{Profit}] = 10 \cdot P(Z > k) - 10 \cdot P(Z \leq k) = 10 \cdot 0.5 - 10 \cdot 0.5 = 5 - 5 = 0 \][/tex]
Hence, the expected profit of the game is \(\boxed{0}\).
### Part (b): Variance of the Profit
The variance of the profit involves the variances of indicator functions for winning and losing:
1. Variance of Indicator Function (\(Z > k\)):
[tex]\[ \text{Var}[I(Z > k)] = P(Z > k) \cdot (1 - P(Z > k)) = 0.5 \cdot 0.5 = 0.25 \][/tex]
2. Variance of Indicator Function (\(Z \leq k\)):
[tex]\[ \text{Var}[I(Z \leq k)] = P(Z \leq k) \cdot (1 - P(Z \leq k)) = 0.5 \cdot 0.5 = 0.25 \][/tex]
The total variance of the profit is:
[tex]\[ \text{Var}[\text{Profit}] = 10^2 \cdot \text{Var}[I(Z > k)] + 10^2 \cdot \text{Var}[I(Z \leq k)] = 100 \cdot 0.25 + 100 \cdot 0.25 = 25 + 25 = 50 \][/tex]
Hence, the variance of the profit is \(\boxed{50}\).
### Part (c): Probability of Gaining 20 Dollars After 10 Games
To calculate the probability of gaining 20 dollars after 10 games, we can model the number of wins \(X\) in 10 games as a binomial random variable with parameters \(n = 10\) and \(p = 0.5\):
1. To gain 20 dollars, one must win 6 games:
[tex]\[ X \sim \text{Binomial}(n=10, p=0.5) \][/tex]
2. The probability of winning exactly 6 games out of 10 can be computed using the binomial probability mass function:
[tex]\[ P(X = 6) = \binom{10}{6} p^6 (1-p)^{10-6} = \binom{10}{6} (0.5)^6 (0.5)^4 = \binom{10}{6} (0.5)^{10} \][/tex]
Upon calculating, the result is:
[tex]\[ P(X = 6) \approx 0.205078125 \][/tex]
Hence, the probability of gaining 20 dollars after playing the game 10 times is approximately [tex]\(\boxed{0.205}\)[/tex].
### Problem Analysis
The random variable \(Z\) is uniformly distributed on the interval \([0, 5]\). This gives us a constant density function \(f_Z(z)\):
[tex]\[ f_Z(z) = \begin{cases} \frac{1}{5}, & \text{if } 0 \leq z \leq 5 \\ 0, & \text{otherwise} \end{cases} \][/tex]
Let's consider \(k = 2.5\) as a middle point for simplicity.
### Part (a): Expected Profit of the Game
To calculate the expected profit, we need to consider the probability of losing or winning based on the value of \(Z\):
1. Probability of Losing (\(Z \leq k\)):
[tex]\[ P(Z \leq k) = \frac{k}{5} = \frac{2.5}{5} = 0.5 \][/tex]
2. Probability of Winning (\(Z > k\)):
[tex]\[ P(Z > k) = 1 - P(Z \leq k) = 1 - 0.5 = 0.5 \][/tex]
The expected profit, therefore, can be computed as:
[tex]\[ E[\text{Profit}] = 10 \cdot P(Z > k) - 10 \cdot P(Z \leq k) = 10 \cdot 0.5 - 10 \cdot 0.5 = 5 - 5 = 0 \][/tex]
Hence, the expected profit of the game is \(\boxed{0}\).
### Part (b): Variance of the Profit
The variance of the profit involves the variances of indicator functions for winning and losing:
1. Variance of Indicator Function (\(Z > k\)):
[tex]\[ \text{Var}[I(Z > k)] = P(Z > k) \cdot (1 - P(Z > k)) = 0.5 \cdot 0.5 = 0.25 \][/tex]
2. Variance of Indicator Function (\(Z \leq k\)):
[tex]\[ \text{Var}[I(Z \leq k)] = P(Z \leq k) \cdot (1 - P(Z \leq k)) = 0.5 \cdot 0.5 = 0.25 \][/tex]
The total variance of the profit is:
[tex]\[ \text{Var}[\text{Profit}] = 10^2 \cdot \text{Var}[I(Z > k)] + 10^2 \cdot \text{Var}[I(Z \leq k)] = 100 \cdot 0.25 + 100 \cdot 0.25 = 25 + 25 = 50 \][/tex]
Hence, the variance of the profit is \(\boxed{50}\).
### Part (c): Probability of Gaining 20 Dollars After 10 Games
To calculate the probability of gaining 20 dollars after 10 games, we can model the number of wins \(X\) in 10 games as a binomial random variable with parameters \(n = 10\) and \(p = 0.5\):
1. To gain 20 dollars, one must win 6 games:
[tex]\[ X \sim \text{Binomial}(n=10, p=0.5) \][/tex]
2. The probability of winning exactly 6 games out of 10 can be computed using the binomial probability mass function:
[tex]\[ P(X = 6) = \binom{10}{6} p^6 (1-p)^{10-6} = \binom{10}{6} (0.5)^6 (0.5)^4 = \binom{10}{6} (0.5)^{10} \][/tex]
Upon calculating, the result is:
[tex]\[ P(X = 6) \approx 0.205078125 \][/tex]
Hence, the probability of gaining 20 dollars after playing the game 10 times is approximately [tex]\(\boxed{0.205}\)[/tex].
