Question

In a group of 10 ​kittens, 5 are female. Two kittens are chosen at random. ​a) Create a probability model for the number of male kittens chosen. ​b) What is the expected number of male kittens​ chosen? ​c) What is the standard​ deviation?

Answer 1

Answer:

(b) Expected number of male kitten E(x) = 1

(c) Standard deviation is ±$\frac{2}{3}$.

Step-by-step explanation:

Given that,

Number of kittens in a group are 10. Out of these 5 are female.

To find:- (a) create a probability model of male kitten chosen.

    So,  total number of kitten = 10

           number of female kitten = 5    

then, Number of male kitten = $10-5= 5$

Now total number of ways to choosing two kittens from group of 10 = $^{10} C_{2}$

                                                                                                                 = $\frac{10!}{2!\times8!}$

                                                                                                                 = $45$

for choosing (i) No male P(x=0) = P(Two female) =  $\frac{^{5} C_{2}}{45}$

                                                                           = $\frac{2}{9}$

                     (ii) 1 male P(x=1) = P(1 male and 1 female) = $\frac{^{5} C_{1}\times^{5} C_{1}}{45}$ =  $\frac{25}{45}$ =$\frac{5}{9}$

                     (iii) 2 male P(x=2)= P(2 male) =$\frac{^{5} C_{2}}{45}$ =  $\frac{2}{9}$

      $x_{i}$                             0                     1                      2

P(X=$x_{i}$)                           $\frac{2}{9}$                      $\frac{5}{9}$                      $\frac{2}{9}$

(b) what is expected number of male kittens chosen ?

    Expected number of male kitten E(x) =  $o\times \frac{2}{9} + 1\times \frac{5}{9} + 2\times\frac{2}{9}$

                                                                  = $1$

(c) what is standard deviation ?

                              $\sigma(x) = \sqrt{ (E(x^{2} )-(Ex)^{2})$

No,                          

                               $Ex^{2} =o\times \frac{2}{9} + 1\times \frac{5}{9} + 4\times\frac{2}{9}$

                                       = $\frac{13}{9}$

                               $\sigma(x)=\sqrt{\frac{13}{9} - 1 }$$=\sqrt{\frac{4}{9} }$  

                                       = ± $\frac{2}{3}$