Question

A fast-food restaurant operates both a drive through facility and a walk-in facility. On a randomly selected day, let X and Y, respectively, be the proportions of the time that the drive-through and walk-in facilities are in use, and suppose that the joint density function of these random variables is,

f (x, y) ={2/3(x+2y) 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1
(a) Find the marginal density of X.
(b) Find the marginal density of Y .
(c) Find the probability that the drive-through facility is busy less than one-half of the time.

Answer 1

`(a)\ g(x) = (2)/(3)(x+1)`

`(b)\ h(y) = (1)/(3)[1 + 4y]`

`(c)`
`P(x>0.5) =(5)/(12)`

Explanation:

Given

`f(x,y) = \left \{ {{(2)/(3)(x+2y)\ \ 0\le x \le 1,\ 0\le y\le 1} \right.`

Solving (a): The marginal density of X

This is calculated as:

`g(x) = \int\limits^(\infty)_(-\infty) {f(x,y)} \, dy`

`g(x) = \int\limits^(1)_(0) {(2)/(3)(x + 2y)} \, dy`

`g(x) = (2)/(3)\int\limits^(1)_(0) {(x + 2y)} \, dy`

Integrate

`g(x) = (2)/(3)(xy+y^2)|\limits^(1)_(0)`

Substitute 1 and 0 for y

`g(x) = (2)/(3)[(x*1+1^2) - (x*0 + 0^2)}`

`g(x) = (2)/(3)[(x+1)}`

Solving (b): The marginal density of Y

This is calculated as:

`h(y) = \int\limits^(\infty)_(-\infty) {f(x,y)} \, dx`

`h(y) = \int\limits^(1)_(0) {(2)/(3)(x + 2y)} \, dx`

`h(y) = (2)/(3)\int\limits^(1)_(0) {(x + 2y)} \, dx`

Integrate

`h(y) = (2)/(3)((x^2)/(2) + 2xy)|\limits^(1)_(0)`

Substitute 1 and 0 for x

`h(y) = (2)/(3)[((1^2)/(2) + 2y*1) - ((0^2)/(2) + 2y*0) ]`

`h(y) = (2)/(3)[((1)/(2) + 2y)]`

`h(y) = (1)/(3)[1 + 4y]`

Solving (c): The probability that the drive-through facility is busy less than one-half of the time.

This is represented as:

`P(x>0.5)`

The solution is as follows:

`P(x>0.5) = P(0\le x\le 0.5,0\le y\le 1)`

Represent as an integral

`P(x>0.5) =\int\limits^1_0 \int\limits^(0.5)_0 {(2)/(3)(x + 2y)} \, dx dy`

`P(x>0.5) =(2)/(3)\int\limits^1_0 \int\limits^(0.5)_0 {(x + 2y)} \, dx dy`

Integrate w.r.t. x

`P(x>0.5) =(2)/(3)\int\limits^1_0 ((x^2)/(2) + 2xy) |^(0.5)_0\, dy`

`P(x>0.5) =(2)/(3)\int\limits^1_0 [((0.5^2)/(2) + 2*0.5y) -((0^2)/(2) + 2*0y)], dy`

`P(x>0.5) =(2)/(3)\int\limits^1_0 (0.125 + y), dy`

`P(x>0.5) =(2)/(3)(0.125y + (y^2)/(2))|^(1)_(0)`

`P(x>0.5) =(2)/(3)[(0.125*1 + (1^2)/(2)) - (0.125*0 + (0^2)/(2))]`

`P(x>0.5) =(2)/(3)[(0.125 + (1)/(2))]`

`P(x>0.5) =(2)/(3)[(0.125 + 0.5]`

`P(x>0.5) =(2)/(3) * 0.625`

`P(x>0.5) =(2 * 0.625)/(3)`

`P(x>0.5) =(1.25)/(3)`

Express as a fraction, properly

`P(x>0.5) =(1.25*4)/(3*4)`

`P(x>0.5) =(5)/(12)`