To solve this problem, we need to calculate the mean ($\mu_X$) and standard deviation ($\sigma_X$) for a binomial distribution.

Binomial Distribution Parameters

1. Number of trials ($n$): 6 (6 games in the tournament)
2. Probability of success ($p$): 0.58 (58% chance of winning each game)

The formulas for the mean and standard deviation of a binomial distribution are: $\mu_X = n \cdot p$ $\sigma_X = \sqrt{n \cdot p \cdot (1-p)}$

Step 1: Calculate the Mean ($\mu_X$)

$\mu_X = n \cdot p = 6 \cdot 0.58 = 3.48$

Step 2: Calculate the Standard Deviation ($\sigma_X$)

$\sigma_X = \sqrt{n \cdot p \cdot (1-p)}$ $\sigma_X = \sqrt{6 \cdot 0.58 \cdot (1 - 0.58)}$ $\sigma_X = \sqrt{6 \cdot 0.58 \cdot 0.42} = \sqrt{1.4604} \approx 1.21$

Final Answer

$\mu_X = 3.48, \, \sigma_X = 1.21$

The correct option is: Mu Subscript x Baseline = 3.48, Sigma Subscript x Baseline = 1.21

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Would you like further clarification or details? Here are 5 related questions to explore:

1. How does the independence of games affect the calculation of probability?
2. What happens to the mean and standard deviation if the number of games increases to 10?
3. Can we calculate the probability of winning at least 4 games in this tournament?
4. How do we interpret the mean and standard deviation in practical terms for this problem?
5. What is the difference between binomial and normal approximations in this context?

Tip: For larger $n$, you can approximate the binomial distribution with a normal distribution using $\mu$ and $\sigma$.