Solution

 

b. $$P(\textrm{Both orchestra representatives are women})=p(x=2)= \dfrac{1}{15}

c.

$$\begin{align*} \textrm{Mean, }\mu&= \sum_x p(x) x\\ &= \dfrac{7}{15} (0) + \dfrac{8}{15}(1) + \dfrac{1}{15}(2)\\ &=\dfrac{2}{3}\\ \textrm{Variance, } \sigma^2&= \sum_x p(x) x^2 - \mu^2\\ &=\dfrac{7}{15} (0) ^2+ \dfrac{8}{15}(1)^2 + \dfrac{1}{15}(2)^2 - \left(\dfrac{2}{3}\right)^2\\ &= \dfrac{16}{45} \end{align*}$$

Answer

b. $\dfrac{1}{15}$ ~~~~~c. $\mu= \dfrac{2}{3}, ~ \sigma^2= \dfrac{16}{45}$

Visit J-Loaded.com

For more.....