Answer:
Step-by-step explanation:
Given that:
A small-business Web site contains 100 pages and 60%, 30%, and 10% of the pages contain low, moderate, and high graphic content, respectively.
. A sample of four pages is selected without replacement,
Let X and Y denote the number of pages with moderate and high graphics output in the sample
We are meant to determine
a) [tex]f_{XY}(x, y)[/tex] from the given data in the question;
However; the probability mass function can be expressed via the relation:
[tex]f_{XY}(x,y) = \dfrac{(^{30} _x ) ( ^{10} _y ) (^{60} _ {4-x-y} ) }{ ( ^{100}_4)}[/tex]
We can now have a table shown as :
[tex]X|Y[/tex] 0 1 2 3 4 Total [tex]f_X(x)[/tex]
0 0.1244 0.0873 0.02031 0.0018 0.0001 0.234
1 0.2618 0.13542 0.02066 0.00092 0 0.419
2 0.1964 0.0666 0.00499 0 0 0.268
3 0.0621 0.01035 0 0 0 0.073
4 0.0069 0 0 0 0 0.007
Total [tex]F_Y(y)[/tex] 0.6516 0.2996 0.0460 0.0028 0.0001 1
b) [tex]f_X(x)[/tex]
The marginal distribution definition of [tex]f_X(x)[/tex][tex]= P(X=x)[/tex]
[tex]f_X(x)[/tex] [tex]= \sum P(X=x, Y=y)[/tex]
From the table above ; the corresponding values of [tex]f_X(x)[/tex] are :
X 0 1 2 3 4
[tex]f_X(x)[/tex] 0.234 0.419 0.268 0.073 0.007
( since [tex]f_X(x)[/tex] represent the vertical column)
c) E(X)
By using the expression [tex]E(x) = \sum ^4 _{x= 0} x f_X(x)[/tex]
we have:
E(X) = [tex]0*0.234+1*0.419+ 2*0.268+3*0.073+4*0.007[/tex]
E(X) = 0 + 0.419 + 0.536 + 0.218 + 0.028
E(X) = 1.202
d) fyß(y)
Using the thesis of conditional Probability; we have :
[tex]P(A|B) = \dfrac{ P(A,B) }{ P(B) }[/tex]
The conditional probability for the mass function is then:
[tex]f_{Y|X=3}(y) = \dfrac{f_{XY}(3,y)}{f_{X}(x)}[/tex]
where;
[tex]f_X(3) = 0.0725[/tex]
values of [tex]f_{XY} (3,y)[/tex] for every y ∈ (0,1,2,3,4)
Therefore; the mass function is:
[tex]Y|{_X_3}:\left[\begin{array}{ccccc}0&1&2&3&4\\0.857&0.143&0&0&0\\ \end{array}\right][/tex]
e) E(Y | X = 3)
By using the expression [tex]E(Y|X=3) = \sum ^4 _{y= 0} y f_{y \beta} \ (y|x)[/tex]
we have:
⇒ [tex]0 * 0.857 + 1*0.143 +0 +0+0[/tex]
= 0.143
The value of E(Y | X = 3) = 0.143
g) Are X and Y independent?
To Check if X and Y independent; Let assume if [tex]f_{XY}(x,y) = f_X(x)f_{Y}(y)[/tex] ; then we can say that X and Y are independent.
From the above previous table :
[tex]f_{(XY)} (0.4) = 0.0001[/tex]
[tex]f_X (0)[/tex] = 0.1244 + 0.087268+0.02031+ 0.001836 + 0.0001
[tex]f_X (0)[/tex] = 0.234
[tex]f_X (4)=0.0001 +0+0 \\ \\ = 0.001[/tex]
[tex]f_{X}(0) f_Y(4) = 0.234*0.0001[/tex]
[tex]f_{X}(0) f_Y(4) = 0.00002[/tex]
We conclude that [tex]f_{(XY)} (0.4) \neq f_X(0) f_Y(y)[/tex]; As such X and Y are said to be non - independent.
