Mathematics, 10.06.2020 17:57, cschellfamily

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?

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answered: 375025

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.

answered: Guest

the answer is b.

step-by-step explanation:

by the trigonometric ratio tangent, tangent equals opposite divided by adjacent.

therefore, tan(50) equals ac/3

multiply each side by 3 to get: ac=3tan(50)

the answer is b.

answered: Guest

step-by-step explanation:

consider an orange peel: if you want to try and lay it flat, you have to stretch it, we know exactly how things are being stretched or squashed at any given point. we have many different map projections because each has different patterns of some projections can even preserve certain features of the earth without