What is the largest number of families that can have at least 4 members according to Markov’s Inequality?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












I was doing a question on Markov's Inequality where I have to give the answer and the question is stated as follows :



A town of 30 families, the average household size is 2.5. What is the largest number of families that can have at least 4 members according to Markov’s Inequality?



According to this formula P(X >= a) = µ/a



here µ=2.5 and and a=4 thus P(X >= a) = 0.625



that is 30*0.625=18.75 and since 18.75 is not a integer as number of family can't be in fraction therefore answer should be 18 but the question is showing incorrect answer.



I don't understand what am I doing wrong.



Thanks in advance.




probability inequality




share|cite|improve this question





















  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    yesterday










  • Here's a MathJax tutorial :)
    – Shaun
    yesterday






  • 1




    @Shaun Thank you I will use this next time.
    – Udolf
    yesterday














up vote
0
down vote

favorite












I was doing a question on Markov's Inequality where I have to give the answer and the question is stated as follows :



A town of 30 families, the average household size is 2.5. What is the largest number of families that can have at least 4 members according to Markov’s Inequality?



According to this formula P(X >= a) = µ/a



here µ=2.5 and and a=4 thus P(X >= a) = 0.625



that is 30*0.625=18.75 and since 18.75 is not a integer as number of family can't be in fraction therefore answer should be 18 but the question is showing incorrect answer.



I don't understand what am I doing wrong.



Thanks in advance.




probability inequality




share|cite|improve this question





















  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    yesterday










  • Here's a MathJax tutorial :)
    – Shaun
    yesterday






  • 1




    @Shaun Thank you I will use this next time.
    – Udolf
    yesterday












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I was doing a question on Markov's Inequality where I have to give the answer and the question is stated as follows :



A town of 30 families, the average household size is 2.5. What is the largest number of families that can have at least 4 members according to Markov’s Inequality?



According to this formula P(X >= a) = µ/a



here µ=2.5 and and a=4 thus P(X >= a) = 0.625



that is 30*0.625=18.75 and since 18.75 is not a integer as number of family can't be in fraction therefore answer should be 18 but the question is showing incorrect answer.



I don't understand what am I doing wrong.



Thanks in advance.




probability inequality




share|cite|improve this question













I was doing a question on Markov's Inequality where I have to give the answer and the question is stated as follows :



A town of 30 families, the average household size is 2.5. What is the largest number of families that can have at least 4 members according to Markov’s Inequality?



According to this formula P(X >= a) = µ/a



here µ=2.5 and and a=4 thus P(X >= a) = 0.625



that is 30*0.625=18.75 and since 18.75 is not a integer as number of family can't be in fraction therefore answer should be 18 but the question is showing incorrect answer.



I don't understand what am I doing wrong.



Thanks in advance.





probability inequality





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday









Shaun

7,31592870




7,31592870









asked yesterday









Udolf

355




355











  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    yesterday










  • Here's a MathJax tutorial :)
    – Shaun
    yesterday






  • 1




    @Shaun Thank you I will use this next time.
    – Udolf
    yesterday
















  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    yesterday










  • Here's a MathJax tutorial :)
    – Shaun
    yesterday






  • 1




    @Shaun Thank you I will use this next time.
    – Udolf
    yesterday















Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
yesterday




Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
yesterday












Here's a MathJax tutorial :)
– Shaun
yesterday




Here's a MathJax tutorial :)
– Shaun
yesterday




1




1




@Shaun Thank you I will use this next time.
– Udolf
yesterday




@Shaun Thank you I will use this next time.
– Udolf
yesterday










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










You don't need Markov for this. Since the average is $2.5$ the total number of people must be $30times 2.5 = 75$. If you assume that "family" must mean at least $1$ person then, if there are $N$ members with at least $4$ then, even assuming all the rest have only $1$ we compare $4N+(30-N)$ to $75$ to see that $N≤15$. After all, if $N=16$ we'd have at least $4times 16+1times (30-16)>75$ people.



Is $N=15$ possible? Sure. If all the other families have exactly $1$ member and each of those $15$ have exactly $4$ then we'd have $4times 15+1times (30-15)=75$ people as desired.



Of course, if you define a family to have at least $2$ people then the answer drops considerably. In the same spirit, your answer of $18$ would be possible if families were allowed to have $0$ members.



Note: if you insist on using Markov, note that you can subtract $1$ from each family in order to accommodate the "at least one person per family" rule. Then we get $mu=1.5, a=3$ and we conclude that $P(X≥a)≤ frac 1.53=.5$ (Where, now, $X$ denotes the number of people in a family, less one). That leads us to consider $30times .5=15$ as before.






share|cite|improve this answer























  • Thanks for such a descriptive answer.
    – Udolf
    yesterday










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2872390%2fwhat-is-the-largest-number-of-families-that-can-have-at-least-4-members-accordin%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










You don't need Markov for this. Since the average is $2.5$ the total number of people must be $30times 2.5 = 75$. If you assume that "family" must mean at least $1$ person then, if there are $N$ members with at least $4$ then, even assuming all the rest have only $1$ we compare $4N+(30-N)$ to $75$ to see that $N≤15$. After all, if $N=16$ we'd have at least $4times 16+1times (30-16)>75$ people.



Is $N=15$ possible? Sure. If all the other families have exactly $1$ member and each of those $15$ have exactly $4$ then we'd have $4times 15+1times (30-15)=75$ people as desired.



Of course, if you define a family to have at least $2$ people then the answer drops considerably. In the same spirit, your answer of $18$ would be possible if families were allowed to have $0$ members.



Note: if you insist on using Markov, note that you can subtract $1$ from each family in order to accommodate the "at least one person per family" rule. Then we get $mu=1.5, a=3$ and we conclude that $P(X≥a)≤ frac 1.53=.5$ (Where, now, $X$ denotes the number of people in a family, less one). That leads us to consider $30times .5=15$ as before.






share|cite|improve this answer























  • Thanks for such a descriptive answer.
    – Udolf
    yesterday














up vote
2
down vote



accepted










You don't need Markov for this. Since the average is $2.5$ the total number of people must be $30times 2.5 = 75$. If you assume that "family" must mean at least $1$ person then, if there are $N$ members with at least $4$ then, even assuming all the rest have only $1$ we compare $4N+(30-N)$ to $75$ to see that $N≤15$. After all, if $N=16$ we'd have at least $4times 16+1times (30-16)>75$ people.



Is $N=15$ possible? Sure. If all the other families have exactly $1$ member and each of those $15$ have exactly $4$ then we'd have $4times 15+1times (30-15)=75$ people as desired.



Of course, if you define a family to have at least $2$ people then the answer drops considerably. In the same spirit, your answer of $18$ would be possible if families were allowed to have $0$ members.



Note: if you insist on using Markov, note that you can subtract $1$ from each family in order to accommodate the "at least one person per family" rule. Then we get $mu=1.5, a=3$ and we conclude that $P(X≥a)≤ frac 1.53=.5$ (Where, now, $X$ denotes the number of people in a family, less one). That leads us to consider $30times .5=15$ as before.






share|cite|improve this answer























  • Thanks for such a descriptive answer.
    – Udolf
    yesterday












up vote
2
down vote



accepted







up vote
2
down vote



accepted






You don't need Markov for this. Since the average is $2.5$ the total number of people must be $30times 2.5 = 75$. If you assume that "family" must mean at least $1$ person then, if there are $N$ members with at least $4$ then, even assuming all the rest have only $1$ we compare $4N+(30-N)$ to $75$ to see that $N≤15$. After all, if $N=16$ we'd have at least $4times 16+1times (30-16)>75$ people.



Is $N=15$ possible? Sure. If all the other families have exactly $1$ member and each of those $15$ have exactly $4$ then we'd have $4times 15+1times (30-15)=75$ people as desired.



Of course, if you define a family to have at least $2$ people then the answer drops considerably. In the same spirit, your answer of $18$ would be possible if families were allowed to have $0$ members.



Note: if you insist on using Markov, note that you can subtract $1$ from each family in order to accommodate the "at least one person per family" rule. Then we get $mu=1.5, a=3$ and we conclude that $P(X≥a)≤ frac 1.53=.5$ (Where, now, $X$ denotes the number of people in a family, less one). That leads us to consider $30times .5=15$ as before.






share|cite|improve this answer















You don't need Markov for this. Since the average is $2.5$ the total number of people must be $30times 2.5 = 75$. If you assume that "family" must mean at least $1$ person then, if there are $N$ members with at least $4$ then, even assuming all the rest have only $1$ we compare $4N+(30-N)$ to $75$ to see that $N≤15$. After all, if $N=16$ we'd have at least $4times 16+1times (30-16)>75$ people.



Is $N=15$ possible? Sure. If all the other families have exactly $1$ member and each of those $15$ have exactly $4$ then we'd have $4times 15+1times (30-15)=75$ people as desired.



Of course, if you define a family to have at least $2$ people then the answer drops considerably. In the same spirit, your answer of $18$ would be possible if families were allowed to have $0$ members.



Note: if you insist on using Markov, note that you can subtract $1$ from each family in order to accommodate the "at least one person per family" rule. Then we get $mu=1.5, a=3$ and we conclude that $P(X≥a)≤ frac 1.53=.5$ (Where, now, $X$ denotes the number of people in a family, less one). That leads us to consider $30times .5=15$ as before.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited yesterday


























answered yesterday









lulu

34.7k13971




34.7k13971











  • Thanks for such a descriptive answer.
    – Udolf
    yesterday
















  • Thanks for such a descriptive answer.
    – Udolf
    yesterday















Thanks for such a descriptive answer.
– Udolf
yesterday




Thanks for such a descriptive answer.
– Udolf
yesterday












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2872390%2fwhat-is-the-largest-number-of-families-that-can-have-at-least-4-members-accordin%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more