Answered

At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying tabulation.
y
p(x, y) 0 1 2
x 0 0.10 0.03 0.01
1 0.08 0.20 0.06
2 0.05 0.14 0.33
(a) Given that X = 1, determine the conditional pmf of Yi.e., pY|X(0|1), pY|X(1|1), pY|X(2|1). (Round your answers to four decimal places.)
y 0 1 2
pY|X(y|1)
(b) Given that two hoses are in use at the self-service island, what is the conditional pmf of the number of hoses in use on the full-service island? (Round your answers to four decimal places.)
y 0 1 2
pY|X(y|2)

Sagot :

MrRoyal

Answer:

(a)

[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|1)} &{0.2353}&{0.5882} & {0.1765}\ \end{array}[/tex]

(b)

[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|2)} &{0.0962}&{0.2692} & {0.6346}\ \end{array}[/tex]

Step-by-step explanation:

Given

[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {0} &{0.1}&{0.03} & {0.01} & {0.14} & {1} &{0.08}&{0.20} & {0.06} &{0.34}& {2} &{0.05}&{0.14} & {0.33} &{0.52 } \ \end{array}[/tex]

Solving (a): Find PMF of y if x = 1

This implies that, we consider the dataset of the row where x = 1 i.e.

[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {1} &{0.08}&{0.20} & {0.06} &{0.34} \ \end{array}[/tex]

The PMF of y given x is calculated using:

[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{P(x)}[/tex]

When x = 1

[tex]P(x) = 0.34[/tex] --- the sum of the rows

So, we have:

[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{0.34}[/tex]

For i = 0 to 2, we have:

[tex]i = 0[/tex]

[tex]P(y=0|x)=\frac{P(y=0\ \&\ x = 1)}{0.34}[/tex]

[tex]P(y=0|x)=\frac{0.08}{0.34}[/tex]

[tex]P(y=0|x)=0.2353[/tex]

[tex]i =1[/tex]

[tex]P(y=1|x)=\frac{P(y=1\ \&\ x = 1)}{0.34}[/tex]

[tex]P(y=1|x)=\frac{0.20}{0.34}[/tex]

[tex]P(y=1|x)=0.5882[/tex]

[tex]i = 2[/tex]

[tex]P(y=2|x)=\frac{P(y=2\ \&\ x = 1)}{0.34}[/tex]

[tex]P(y=2|x)=\frac{0.06}{0.34}[/tex]

[tex]P(y=2|x)=0.1765[/tex]

So, the PMF of y given that x = 1 is:

[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|1)} &{0.2353}&{0.5882} & {0.1765}\ \end{array}[/tex]

Solving (b): The interpretation of (b) is to find the PMF of y if x = 2

This implies that, we consider the dataset of the row where x = 2 i.e.

[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {2} &{0.05}&{0.14} & {0.33} &{0.52 }\ \end{array}[/tex]

When [tex]x = 2[/tex]

[tex]P(x) = 0.52[/tex]

So, we have:

[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{0.52}[/tex]

For i = 0 to 2, we have:

[tex]i = 0[/tex]

[tex]P(y=0|x)=\frac{P(y=0\ \&\ x = 2)}{0.52}[/tex]

[tex]P(y=0|x)=\frac{0.05}{0.52}[/tex]

[tex]P(y=0|x)=0.0962\\[/tex]

[tex]i =1[/tex]

[tex]P(y=1|x)=\frac{P(y=1\ \&\ x = 2)}{0.52}[/tex]

[tex]P(y=1|x)=\frac{0.14}{0.52}[/tex]

[tex]P(y=1|x)=0.2692[/tex]

[tex]i = 2[/tex]

[tex]P(y=2|x)=\frac{P(y=2\ \&\ x = 2)}{0.52}[/tex]

[tex]P(y=2|x)=\frac{0.33}{0.52}[/tex]

[tex]P(y=2|x)=0.6346[/tex]

[tex]P(y=0|x)=0.0962\\[/tex]

[tex]P(y=1|x)=0.2692[/tex]

[tex]P(y=2|x)=0.6346[/tex]

So, the PMF of y given that x = 1 is:

[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|2)} &{0.0962}&{0.2692} & {0.6346}\ \end{array}[/tex]

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.