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

Answer:

$(a)\ g(x) = \frac{2}{3}(x+1)$

$(b)\ h(y) = \frac{1}{3}[1 + 4y]$

$(c)$ $P(x>0.5) =\frac{5}{12}$

Step-by-step explanation:

Given

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

Solving (a): The marginal density of X

This is calculated as:

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

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

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

Integrate

$g(x) = \frac{2}{3}(xy+y^2)|\limits^{1}_{0}$

Substitute 1 and 0 for y

$g(x) = \frac{2}{3}[(x*1+1^2) - (x*0 + 0^2)}$

$g(x) = \frac{2}{3}[(x+1)}$

Solving (b): The marginal density of Y

This is calculated as:

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

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

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

Integrate

$h(y) = \frac{2}{3}(\frac{x^2}{2} + 2xy)|\limits^{1}_{0}$

Substitute 1 and 0 for x

$h(y) = \frac{2}{3}[(\frac{1^2}{2} + 2y*1) - (\frac{0^2}{2} + 2y*0) ]$

$h(y) = \frac{2}{3}[(\frac{1}{2} + 2y)]$

$h(y) = \frac{1}{3}[1 + 4y]$

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

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

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

Integrate w.r.t. x

$P(x>0.5) =\frac{2}{3}\int\limits^1_0 (\frac{x^2}{2} + 2xy) |^{0.5}_0\, dy$

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

$P(x>0.5) =\frac{2}{3}\int\limits^1_0 (0.125 + y), dy$

$P(x>0.5) =\frac{2}{3}(0.125y + \frac{y^2}{2})|^{1}_{0}$

$P(x>0.5) =\frac{2}{3}[(0.125*1 + \frac{1^2}{2}) - (0.125*0 + \frac{0^2}{2})]$

$P(x>0.5) =\frac{2}{3}[(0.125 + \frac{1}{2})]$

$P(x>0.5) =\frac{2}{3}[(0.125 + 0.5]$

$P(x>0.5) =\frac{2}{3} * 0.625$

$P(x>0.5) =\frac{2 * 0.625}{3}$

$P(x>0.5) =\frac{1.25}{3}$

Express as a fraction, properly

$P(x>0.5) =\frac{1.25*4}{3*4}$

$P(x>0.5) =\frac{5}{12}$