Questions about the solution to this multinomial distribuition problem

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In summary: Historically, 70% of the calls are answered in two rings or less, 25% are answered in three or four rings, and the remaining calls require five rings or more. Suppose you call this manufacturer 10 times and assume all calls are independent.a) What is the probability that eight calls are answered in two rings or less, one call is answered in three or four rings, and one call requires five rings or more?b) What is the conditional distribution of the number of calls requiring five rings or more given that eight calls are answered in two rings or less?c) Are the number of calls answered in two rings or less and the number of
  • #1
s3a
818
8

Homework Statement


The problem along with its solution are attached as TheProblemAndSolution.jpg.

Homework Equations


Multinomial distribution formula.

The Attempt at a Solution


I'm perfectly fine with part a.

As for part b, does the final answer need the ##f_{Z|8}(0) = 0.694##, ##f_{Z|8}(1) = 0.0277## and what I'm assuming should have been ##f_{Z|8}(2) = 0.027## parts of the solution for part b, or just the ##f_{Z|8}(z) = f_{Z|X = 8}(z) = 2! 0.25^{2-z} 0.05^z / {0.3^2 z! (2-z)!}## part of the solution for part b?

As for part c, the final answer should say "##f_{XZ}(x,z) ≠ f_X(x) f_Z(z)##, and therefore X and Z are dependent" instead of "##f_{XZ}(x,z) ≠ f_X(x) f_Y(y)##, and therefore X and Y are dependent", right?

Any confirmation or rejection as to whether my comments are true or not would be greatly appreciated!
 

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  • #2
s3a said:

Homework Statement


The problem along with its solution are attached as TheProblemAndSolution.jpg.

Homework Equations


Multinomial distribution formula.

The Attempt at a Solution


I'm perfectly fine with part a.

As for part b, does the final answer need the ##f_{Z|8}(0) = 0.694##, ##f_{Z|8}(1) = 0.0277## and what I'm assuming should have been ##f_{Z|8}(2) = 0.027## parts of the solution for part b, or just the ##f_{Z|8}(z) = f_{Z|X = 8}(z) = 2! 0.25^{2-z} 0.05^z / {0.3^2 z! (2-z)!}## part of the solution for part b?

As for part c, the final answer should say "##f_{XZ}(x,z) ≠ f_X(x) f_Z(z)##, and therefore X and Z are dependent" instead of "##f_{XZ}(x,z) ≠ f_X(x) f_Y(y)##, and therefore X and Y are dependent", right?

Any confirmation or rejection as to whether my comments are true or not would be greatly appreciated!

Your jpeg file is unreadable on my computer---it is fuzzy and almost microscopic but un-enlargable. If you would actually take the time to type out the problem I would be willing to look at it.
 
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  • #3
Alright, here is the typed version (with minor text-formatting differences). (Hopefully, I made no typos of my own. - I copied everything, including what I suspect is wrong, in the same way it is written in the solution.):

Problem statement:


P3:

To evaluate the technical support from a computer manufacturer, the number of rings before a call is answered by a service representative is tracked. Historically, 70% of the calls are answered in two rings or less, 25% are answered in three or four rings, and the remaining calls require five rings or more. Suppose you call this manufacturer 10 times and assume all calls are independent.

a) What is the probability that eight calls are answered in two rings or less, one call is answered in three or four rings, and one call requires five rings or more?

b) What is the conditional distribution of the number of calls requiring five rings or more given that eight calls are answered in two rings or less?

c) Are the number of calls answered in two rings or less and the number of calls requiring five rings or more independent random variables? Why?

Solution:
Part a:
Let X, Y, and Z denote the number of calls answered in two rings or less, three or four rings, and five rings or more, respectively. The joint mass function is extension of binomial with p=0.7, q=0.25, r=0.05, where n=10.

##f_{XYZ}(z,y,z) = n! / {x! y! z!} p^x q^y r^z##

P(X = 8, Y = 1, Z = 1) = ##f_{XYZ}(8,1,1) = 10! / (8! 1! 1!) 0.7^8 0.25^1 0.05^1## = 0.0649

Answer: P(X = 8, Y = 1, Z = 1) = 0.0649

Part b:
##f_{XZ}(x,z) = 10! / (x! z! (10 - x - z)!) 0.7^x 0.25^{10-x-z} 0.05^z## ⇒ ##f_{XZ}(8,0) = 0.1617##, ##f_{XZ}(8,1) = 0.0645##, ##f_{XZ}(8,0) = 0.0063##

##f_{X}(x) = 10! / ( x! (10-x)! ) 0.7^x 0.3^{10-x}## ⇒ ##f_{X}(8) = 0.233##

##f_{Z|8}(z) = f_{XZ}(8,z) / f_{X}(8) = [10! / ( 8! z! (10-8-z)! 0.7^8 0.25^(10-8-z) 0.05^z ) ] / [ 10! / ( 8! (10-8)! 0.7^8 0.3^{10-8} ) ] = [(10-8)! 0.25^{10-8-z} 0.05^z] / [(10-8-z)! 0.3^{10-8}] = [2! 0.25^{z-2} 0.05^z] / [(2-z)! 0.3^2]##

Answer: ##f_{Z|8} = [2! 0.25^{2-z} 0.05^z] / [0.3^2 z! (2 - z)!]## : ##f_{Z|8}(0) = 0.694##, ##f_{Z|8}(1) = 0.277##, ##f_{Z|8}(0) = 0.027##

Part c:
##f_{Z}(z) = 10! / [ (10-z)! 0.95^{10-z} 0.05^z]## and knowing ##f_{XZ}(x,z)## & ##f_X(x)## from part b we can easily show that

Answer: ##f_{XZ}(x,z) ≠ f_X(x) f_Y(y)## and therefore X and Y are dependent.
 
  • #4
For part (b), I agree that it should be fine to answer with just the generic fZ|X=8(z) expression. If they wanted numbers they should have said so.
I also agree with you on part c.
Btw, your pic was a bit small, but perfectly readable on my screen and I could enlarge it on my laptop+Windows with Ctrl-Shift-+. I guess it depends on the device/OS.
 
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  • #5
Thanks, haruspex!
 
  • #6
s3a said:
Alright, here is the typed version (with minor text-formatting differences). (Hopefully, I made no typos of my own. - I copied everything, including what I suspect is wrong, in the same way it is written in the solution.):

Problem statement:


P3:

To evaluate the technical support from a computer manufacturer, the number of rings before a call is answered by a service representative is tracked. Historically, 70% of the calls are answered in two rings or less, 25% are answered in three or four rings, and the remaining calls require five rings or more. Suppose you call this manufacturer 10 times and assume all calls are independent.

a) What is the probability that eight calls are answered in two rings or less, one call is answered in three or four rings, and one call requires five rings or more?

b) What is the conditional distribution of the number of calls requiring five rings or more given that eight calls are answered in two rings or less?

c) Are the number of calls answered in two rings or less and the number of calls requiring five rings or more independent random variables? Why?

Solution:
Part a:
Let X, Y, and Z denote the number of calls answered in two rings or less, three or four rings, and five rings or more, respectively. The joint mass function is extension of binomial with p=0.7, q=0.25, r=0.05, where n=10.

##f_{XYZ}(z,y,z) = n! / {x! y! z!} p^x q^y r^z##

P(X = 8, Y = 1, Z = 1) = ##f_{XYZ}(8,1,1) = 10! / (8! 1! 1!) 0.7^8 0.25^1 0.05^1## = 0.0649

Answer: P(X = 8, Y = 1, Z = 1) = 0.0649

Part b:
##f_{XZ}(x,z) = 10! / (x! z! (10 - x - z)!) 0.7^x 0.25^{10-x-z} 0.05^z## ⇒ ##f_{XZ}(8,0) = 0.1617##, ##f_{XZ}(8,1) = 0.0645##, ##f_{XZ}(8,0) = 0.0063##

##f_{X}(x) = 10! / ( x! (10-x)! ) 0.7^x 0.3^{10-x}## ⇒ ##f_{X}(8) = 0.233##

##f_{Z|8}(z) = f_{XZ}(8,z) / f_{X}(8) = [10! / ( 8! z! (10-8-z)! 0.7^8 0.25^(10-8-z) 0.05^z ) ] / [ 10! / ( 8! (10-8)! 0.7^8 0.3^{10-8} ) ] = [(10-8)! 0.25^{10-8-z} 0.05^z] / [(10-8-z)! 0.3^{10-8}] = [2! 0.25^{z-2} 0.05^z] / [(2-z)! 0.3^2]##

Answer: ##f_{Z|8} = [2! 0.25^{2-z} 0.05^z] / [0.3^2 z! (2 - z)!]## : ##f_{Z|8}(0) = 0.694##, ##f_{Z|8}(1) = 0.277##, ##f_{Z|8}(0) = 0.027##

Part c:
##f_{Z}(z) = 10! / [ (10-z)! 0.95^{10-z} 0.05^z]## and knowing ##f_{XZ}(x,z)## & ##f_X(x)## from part b we can easily show that

Answer: ##f_{XZ}(x,z) ≠ f_X(x) f_Y(y)## and therefore X and Y are dependent.

I agree with your answers, numbers and all.

I still cannot get your pic to enlarge on my HP notebook computer, but the machine is getting old now and its graphics are not great.

However, the fact is that posting images is discouraged in PF, unless (for example) you need to show images, etc.

Note added in edit:
If you examine your solution to (b) you can see that ##(Z|X=8) = \text{Binomial}(n=2,p=1/6)##. This is not an accident; it is an easy exercise to show
[tex] (X_1, X_2, \ldots, X_r) \sim \text{Multinom}(n; p_1, p_2, \ldots, p_r)\\
\text{implies}\\
(X_2, X_3, \ldots, X_r|X_1 = k_1) \sim \text{Multinom} \left( n-k_1;
\frac{p_2}{1-p_1}, \ldots, \frac{p_r}{1-p_1} \right) [/tex]
 
Last edited:

What is a multinomial distribution problem?

A multinomial distribution problem is a statistical problem in which the outcomes of a random experiment can be classified into multiple categories, and the probability of each category is known or can be calculated. This type of problem often involves multiple independent trials and is useful for analyzing data with more than two possible outcomes.

What are the key components of a multinomial distribution?

The key components of a multinomial distribution are the number of categories, the probabilities associated with each category, and the number of trials or experiments. These components help to determine the overall probability of each category occurring in a given experiment or trial.

How do you calculate the probability of a specific outcome in a multinomial distribution?

To calculate the probability of a specific outcome in a multinomial distribution, you can use the formula P(x1,x2,...,xn) = n!/(x1!x2!...xn!) * p1^x1 * p2^x2 * ... * pn^xn, where n is the number of categories, xi is the number of times category i occurs, and pi is the probability of category i occurring.

What is the difference between a multinomial distribution and a binomial distribution?

A multinomial distribution is used for experiments with more than two possible outcomes, while a binomial distribution is used for experiments with only two possible outcomes. Additionally, a multinomial distribution allows for multiple independent trials, while a binomial distribution only considers one trial.

How can a multinomial distribution be applied in real life?

A multinomial distribution can be applied in real life for analyzing data with multiple categories, such as survey responses or election results. It can also be used in genetics to analyze the distribution of genotypes in a population. In business, it can be used for market research to analyze consumer preferences among different products or brands.

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