To determine the mean and the standard deviation of the number of games the basketball team might win in a six-game tournament, we must understand that [tex]\(X\)[/tex] follows a binomial distribution. This is because each game has two possible outcomes (win or lose), and the games are played independently.
The parameters for a binomial distribution [tex]\(X\)[/tex] are:
- [tex]\(n\)[/tex]: the number of trials (games), which is 6.
- [tex]\(p\)[/tex]: the probability of success on each trial (probability of winning a game), which is 0.58.
Step 1: Calculate the Mean ([tex]\(\mu_X\)[/tex])
The mean of a binomial distribution [tex]\(X \sim \text{Binomial}(n, p)\)[/tex] is given by:
[tex]\[
\mu_X = n \cdot p
\][/tex]
Substitute [tex]\(n = 6\)[/tex] and [tex]\(p = 0.58\)[/tex]:
[tex]\[
\mu_X = 6 \cdot 0.58 = 3.48
\][/tex]
Step 2: Calculate the Standard Deviation ([tex]\(\sigma_X\)[/tex])
The standard deviation of a binomial distribution [tex]\(X \sim \text{Binomial}(n, p)\)[/tex] is given by:
[tex]\[
\sigma_X = \sqrt{n \cdot p \cdot (1-p)}
\][/tex]
Substitute [tex]\(n = 6\)[/tex], [tex]\(p = 0.58\)[/tex], and [tex]\(1-p = 1-0.58 = 0.42\)[/tex]:
[tex]\[
\sigma_X = \sqrt{6 \cdot 0.58 \cdot 0.42}
\][/tex]
After calculating the above expression:
[tex]\[
\sigma_X \approx 1.21
\][/tex]
So, the mean and standard deviation of [tex]\(X\)[/tex] are:
[tex]\[
\mu_X = 3.48 \quad \text{and} \quad \sigma_X \approx 1.21
\][/tex]
Thus, the correct choice from the given options is:
[tex]\[
\mu_x=3.48, \sigma_x=1.21
\][/tex]
