To find the mean and standard deviation of the number of games won by the basketball team, let's define the parameters for our problem:
- The probability of winning any given game, [tex]\( p \)[/tex], is 0.58.
- The number of games played, [tex]\( n \)[/tex], is 6.
- The number of games won, [tex]\( X \)[/tex], follows a binomial distribution with parameters [tex]\( n \)[/tex] and [tex]\( p \)[/tex].
The mean [tex]\((\mu_x)\)[/tex] of a binomial distribution can be calculated using the formula:
[tex]\[ \mu_x = n \cdot p \][/tex]
Substituting the given values:
[tex]\[ \mu_x = 6 \cdot 0.58 \][/tex]
[tex]\[ \mu_x = 3.48 \][/tex]
The standard deviation [tex]\((\sigma_x)\)[/tex] of a binomial distribution can be calculated using the formula:
[tex]\[ \sigma_x = \sqrt{n \cdot p \cdot (1 - p)} \][/tex]
Again, substituting the given values:
[tex]\[ \sigma_x = \sqrt{6 \cdot 0.58 \cdot (1 - 0.58)} \][/tex]
[tex]\[ \sigma_x \approx 1.21 \][/tex]
So, we find that the mean number of games won, [tex]\(\mu_x\)[/tex], is approximately 3.48, and the standard deviation, [tex]\(\sigma_x\)[/tex], is approximately 1.21.
Therefore, the correct answer is:
[tex]\(\mu_x = 3.48, \sigma_x = 1.21\)[/tex]
