Question

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, and X and Y denote the number of pages with moderate and high graphics output in the sample. Determine: a. fxy(x, y) b. fx(x) c. E(X) d. fyß(y) e. E(Y | X = 3) g. Are X and Y independent?

Answer 1

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)  $f_{XY}(x, y)$  from the given data in the question;

However; the probability mass function can be expressed via the relation:

$f_{XY}(x,y) = \dfrac{(^{30} _x ) ( ^{10} _y ) (^{60} _ {4-x-y} ) }{ ( ^{100}_4)}$

We can now have a table shown as :

$X|Y$                0           1               2              3              4          Total    $f_X(x)$

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 $F_Y(y)$   0.6516   0.2996   0.0460      0.0028  0.0001    1

b) $f_X(x)$

The marginal distribution definition of $f_X(x)$$= P(X=x)$

$f_X(x)$ $= \sum P(X=x, Y=y)$

From the table above ; the corresponding values of $f_X(x)$  are :

X           0           1         2          3           4

$f_X(x)$    0.234   0.419  0.268   0.073    0.007

( since $f_X(x)$ represent the vertical column)

c) E(X)

By using the expression $E(x) = \sum ^4 _{x= 0} x f_X(x)$

we have:

E(X) = $0*0.234+1*0.419+ 2*0.268+3*0.073+4*0.007$

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 :

$P(A|B) = \dfrac{ P(A,B) }{ P(B) }$

The conditional probability for the mass function is then:

$f_{Y|X=3}(y) = \dfrac{f_{XY}(3,y)}{f_{X}(x)}$

where;

$f_X(3) = 0.0725$

values of $f_{XY} (3,y)$ for every y ∈ (0,1,2,3,4)

Therefore; the mass function is:

$Y|{_X_3}:\left[\begin{array}{ccccc}0&1&2&3&4\\0.857&0.143&0&0&0\\ \end{array}\right]$

e) E(Y | X = 3)

By using the expression $E(Y|X=3) = \sum ^4 _{y= 0} y f_{y \beta} \ (y|x)$

we have:

⇒ $0 * 0.857 + 1*0.143 +0 +0+0$

= 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 $f_{XY}(x,y) = f_X(x)f_{Y}(y)$ ; then we can say that X and Y are independent.

From the above previous table :

$f_{(XY)} (0.4) = 0.0001$

$f_X (0)$ = 0.1244 + 0.087268+0.02031+ 0.001836 + 0.0001

$f_X (0)$  = 0.234

$f_X (4)=0.0001 +0+0 \\ \\ = 0.001$

$f_{X}(0) f_Y(4) = 0.234*0.0001$

$f_{X}(0) f_Y(4) = 0.00002$

We conclude that $f_{(XY)} (0.4) \neq f_X(0) f_Y(y)$; As such X and Y are said to be  non - independent.