Mathematics, 31.03.2020 04:41 adrianvigil6400

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?

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Answer from: browndalton55

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

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 = $^{10} C_{2}$

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

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) = $o\times \frac{2}{9} + 1\times \frac{5}{9} + 2\times\frac{2}{9}$

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

No,

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

= $\frac{13}{9}$

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

= ± $\frac{2}{3}$

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In a group of 10 kittens, 5 are female. Two kittens are chosen at random. a) Create a probability mo...