To find the mean and standard deviation for the number of heads when a biased coin is tossed 15 times, we need to apply the properties of a binomial distribution.
1. Mean (μ) of X:
The mean of a binomial distribution is given by:
[tex]\[
\mu_x = n \cdot p
\][/tex]
where [tex]\( n \)[/tex] is the number of trials (tosses), and [tex]\( p \)[/tex] is the probability of success (getting heads in this case).
For this problem:
- [tex]\( n = 15 \)[/tex] (number of tosses)
- [tex]\( p = \frac{2}{3} \)[/tex] (probability of getting heads)
Substituting these values into the formula, we get:
[tex]\[
\mu_x = 15 \cdot \frac{2}{3}
\][/tex]
Simplifying this expression:
[tex]\[
\mu_x = 15 \cdot 0.6667 \approx 10
\][/tex]
2. Standard Deviation (σ) of X:
The standard deviation of a binomial distribution is given by:
[tex]\[
\sigma_x = \sqrt{n \cdot p \cdot (1 - p)}
\][/tex]
Again, using [tex]\( n = 15 \)[/tex] and [tex]\( p = \frac{2}{3} \)[/tex]:
Substituting these values into the formula, we get:
[tex]\[
\sigma_x = \sqrt{15 \cdot \frac{2}{3} \cdot \left(1 - \frac{2}{3}\right)}
\][/tex]
Simplifying inside the square root:
[tex]\[
\sigma_x = \sqrt{15 \cdot \frac{2}{3} \cdot \frac{1}{3}}
\][/tex]
[tex]\[
\sigma_x = \sqrt{15 \cdot \frac{2}{9}}
\][/tex]
[tex]\[
\sigma_x = \sqrt{\frac{30}{9}}
\][/tex]
[tex]\[
\sigma_x = \sqrt{3.\overline{3}}
\approx 1.826
\][/tex]
Based on the calculations, the mean and standard deviation for the number of heads when the coin is tossed 15 times are:
[tex]\[
\mu_x = 10, \quad \sigma_x = 1.826
\][/tex]
Therefore, the correct option is:
[tex]\[
\boxed{\mu_x = 10, \sigma_x = 1.8}
\][/tex]
