Respuesta :
Answer:
(b) Expected number of male kitten E(x) = 1
(c) Standard deviation is ±[tex]\frac{2}{3}[/tex].
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 = [tex]10-5= 5[/tex]
Now total number of ways to choosing two kittens from group of 10 = [tex]^{10} C_{2}[/tex]
= [tex]\frac{10!}{2!\times8!}[/tex]
= [tex]45[/tex]
for choosing (i) No male P(x=0) = P(Two female) = [tex]\frac{^{5} C_{2}}{45}[/tex]
= [tex]\frac{2}{9}[/tex]
(ii) 1 male P(x=1) = P(1 male and 1 female) = [tex]\frac{^{5} C_{1}\times^{5} C_{1}}{45}[/tex] = [tex]\frac{25}{45}[/tex] =[tex]\frac{5}{9}[/tex]
(iii) 2 male P(x=2)= P(2 male) =[tex]\frac{^{5} C_{2}}{45}[/tex] = [tex]\frac{2}{9}[/tex]
[tex]x_{i}[/tex] 0 1 2
P(X=[tex]x_{i}[/tex]) [tex]\frac{2}{9}[/tex] [tex]\frac{5}{9}[/tex] [tex]\frac{2}{9}[/tex]
(b) what is expected number of male kittens chosen ?
Expected number of male kitten E(x) = [tex]o\times \frac{2}{9} + 1\times \frac{5}{9} + 2\times\frac{2}{9}[/tex]
= [tex]1[/tex]
(c) what is standard deviation ?
[tex]\sigma(x) = \sqrt{ (E(x^{2} )-(Ex)^{2})[/tex]
No,
[tex]Ex^{2} =o\times \frac{2}{9} + 1\times \frac{5}{9} + 4\times\frac{2}{9}[/tex]
= [tex]\frac{13}{9}[/tex]
[tex]\sigma(x)=\sqrt{\frac{13}{9} - 1 }[/tex][tex]=\sqrt{\frac{4}{9} }[/tex]
= ± [tex]\frac{2}{3}[/tex]
