For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

# Solved Examples on Apportionment

Pre-requisites:
(1.) Numbers and Notations
(2.) Fractions, Decimals and Percents

NOTE: Students may use the calculators to check their solutions.

Solve all questions.
Show all work.

(1.) To encourage her students to study the Holy Bible, Anna intends to distribute $10$ identical pieces of candy among her $4$ students based on the the number of pages they read last month.
The table below lists the total number of pages read by each student.
Student Pages
Peter $455$
Paul $755$
James $405$
John $630$

Determine the:
(a.) Standard Divisor
(b.) Standard Quota
(c.) Lower Quota for each student
(d.) Normal Quota for each student
(e.) Upper Quota for each student.

$(a.) \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 455 + 755 + 405 + 630 = 2245 \\[3ex] NIA = 10 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{2245}{10} \\[5ex] = 224.5 \\[3ex] (b.) \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Peter \\[3ex] = \dfrac{455}{224.5} \\[5ex] = 2.02672606 \\[3ex] SQ\:\:for\:\:Paul \\[3ex] = \dfrac{755}{224.5} \\[5ex] = 3.36302895 \\[3ex] SQ\:\:for\:\:James \\[3ex] = \dfrac{405}{224.5} \\[5ex] = 1.80400891 \\[3ex] SQ\:\:for\:\:John \\[3ex] = \dfrac{630}{224.5} \\[5ex] = 2.80623608 \\[3ex] (c.),\:\:(d.),\:\:(e.) \\[3ex]$
Student Standard Quota Lower Quota Normal Quota Upper Quota
Peter $2.02672606$ $2$ $2$ $3$
Paul $3.36302895$ $3$ $3$ $4$
James $1.80400891$ $1$ $2$ $2$
John $2.80623608$ $2$ $3$ $3$
$\Sigma LQ = 8$ $\Sigma NQ = 10$ $\Sigma UQ = 12$
(2.) To encourage her students to study the Holy Bible, Anna intends to distribute $30$ identical pieces of candy among her $4$ students based on the the number of pages they read last month.
The table below lists the total number of pages read by each student.
Student Pages
Joshua $455$
Judges $230$
Ruth $430$
Samuel $180$

Apportion the candy using Hamilton's Method

$\underline{First\:\:Step} \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 455 + 230 + 430 + 180 = 1295 \\[3ex] NIA = 30 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{1295}{30} \\[5ex] = 43.1666667 \\[3ex] \underline{Second\:\:Step} \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Joshua \\[3ex] = \dfrac{455}{43.1666667} \\[5ex] = 10.5405405 \\[3ex] SQ\:\:for\:\:Judges \\[3ex] = \dfrac{230}{43.1666667} \\[5ex] = 5.32818533 \\[3ex] SQ\:\:for\:\:Ruth \\[3ex] = \dfrac{430}{43.1666667} \\[5ex] = 9.96138996 \\[3ex] SQ\:\:for\:\:Samuel \\[3ex] = \dfrac{180}{43.1666667} \\[5ex] = 4.16988417 \\[3ex]$
Hamilton's Method - Part $1$
Student Standard Quota Lower Quota
Joshua $10.5405405$ $10$
Judges $5.32818533$ $5$
Ruth $9.96138996$ $9$
Samuel $4.16988417$ $4$
$\Sigma LQ = 28$
$28 \ne 30$
$30 - 28 = 2$
Balance = $2$

Hamilton's Method - Part $2$
Student Standard Quota Decimal Part of Standard Quota Assign
Joshua $10.5405405$ $0.5405405$ $1$
Judges $5.32818533$ $0.32818533$
Ruth $9.96138996$ $0.96138996$ $1$
Samuel $4.16988417$ $0.16988417$

Hamilton's Method of Apportionment
Student Lower Quota Extra Item Apportion
Joshua $10$ $1$ $11$
Judges $5$ $5$
Ruth $9$ $1$ $10$
Samuel $4$ $4$
$\Sigma Apportion = 30$
$NIA = 30$
$\Sigma Apportion = NIA$

As you can see:
Joshua was apportioned $11$ pieces of candy...the upper quota
Judges was apportioned $5$ pieces of candy...the lower quota
Ruth was apportioned $10$ pieces of candy...the upper quota
Samuel was apportioned $4$ pieces of candy...the lower quota
Each student was apportioned either the lower quota or the upper quota.
Therefore, Quota Rule is met.
Hamilton's method always satisfies the Quota Rule.
(3.) To encourage her students to study the Holy Bible, Anna intends to distribute $14$ identical pieces of candy among her $4$ students based on the the number of pages they read last month.
The table below lists the total number of pages read by each student.
Student Pages
Genesis $555$
Exodus $255$
Leviticus $355$
Numbers $455$

Apportion the candy using Jefferson's Method

$\underline{First\:\:Step} \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 555 + 255 + 355 + 455 = 1620 \\[3ex] NIA = 14 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{1620}{14} \\[5ex] = 115.714286 \\[3ex] \underline{Second\:\:Step} \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Genesis \\[3ex] = \dfrac{555}{115.714286} \\[5ex] = 4.79629628 \\[3ex] SQ\:\:for\:\:Exodus \\[3ex] = \dfrac{255}{115.714286} \\[5ex] = 2.2037037 \\[3ex] SQ\:\:for\:\:Leviticus \\[3ex] = \dfrac{355}{115.714286} \\[5ex] = 3.06790123 \\[3ex] SQ\:\:for\:\:Numbers \\[3ex] = \dfrac{455}{115.714286} \\[5ex] = 3.93209876 \\[3ex]$
Jefferson's Method - Part $1$
Student Standard Quota Lower Quota
Genesis $4.79629628$ $4$
Exodus $2.2037037$ $2$
Leviticus $3.06790123$ $3$
Numbers $3.93209876$ $3$
$\Sigma LQ = 12$
$12 \ne 14$
$14 - 12 = 2$
Balance = $2$

The balance is positive.
Try a smaller standard divisor
Let the modified divisor = $114$

$MD = 114 \\[3ex] \underline{Third\:\:Step - Part\:\:1} \\[3ex] Modified\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Modified\:\:Divisor} \\[7ex] MQ\:\:for\:\:Genesis \\[3ex] = \dfrac{555}{114} \\[5ex] = 4.86842105 \\[3ex] MQ\:\:for\:\:Exodus \\[3ex] = \dfrac{255}{114} \\[5ex] = 2.23684211 \\[3ex] MQ\:\:for\:\:Leviticus \\[3ex] = \dfrac{355}{114} \\[5ex] = 3.11403509 \\[3ex] MQ\:\:for\:\:Numbers \\[3ex] = \dfrac{455}{114} \\[5ex] = 3.99122807 \\[3ex]$
Jefferson's Method - First Try
Student Modified Quota Modified Lower Quota
Genesis $4.86842105$ $4$
Exodus $2.23684211$ $2$
Leviticus $3.11403509$ $3$
Numbers $3.99122807$ $3$
$\Sigma LQ = 12$
$12 \ne 14$
$14 - 12 = 2$
Balance = $2$

The balance is still positive.
Try another smaller standard divisor
Let the modified divisor = $113$

$MD = 113 \\[3ex] \underline{Third\:\:Step - Part\:\:2} \\[3ex] MQ\:\:for\:\:Genesis \\[3ex] = \dfrac{555}{113} \\[5ex] = 4.91150442 \\[3ex] MQ\:\:for\:\:Exodus \\[3ex] = \dfrac{255}{113} \\[5ex] = 2.25663717 \\[3ex] MQ\:\:for\:\:Leviticus \\[3ex] = \dfrac{355}{113} \\[5ex] = 3.14159292 \\[3ex] MQ\:\:for\:\:Numbers \\[3ex] = \dfrac{455}{113} \\[5ex] = 4.02654867 \\[3ex]$
Jefferson's Method - Second Try
Student Modified Quota Modified Lower Quota
Genesis $4.91150442$ $4$
Exodus $2.25663717$ $2$
Leviticus $3.14159292$ $3$
Numbers $4.02654867$ $4$
$\Sigma LQ = 13$
$13 \ne 14$
$14 - 13 = 1$
Balance = $1$

The balance is still positive.
Try another smaller standard divisor
Let the modified divisor = $112$

$MD = 112 \\[3ex] \underline{Third\:\:Step - Part\:\:3} \\[3ex] MQ\:\:for\:\:Genesis \\[3ex] = \dfrac{555}{112} \\[5ex] = 4.95535714 \\[3ex] MQ\:\:for\:\:Exodus \\[3ex] = \dfrac{255}{112} \\[5ex] = 2.27678571 \\[3ex] MQ\:\:for\:\:Leviticus \\[3ex] = \dfrac{355}{112} \\[5ex] = 3.16964286 \\[3ex] MQ\:\:for\:\:Numbers \\[3ex] = \dfrac{455}{112} \\[5ex] = 4.0625 \\[3ex]$
Jefferson's Method - Third Try
Student Modified Quota Modified Lower Quota
Genesis $4.95535714$ $4$
Exodus $2.27678571$ $2$
Leviticus $3.16964286$ $3$
Numbers $4.0625$ $4$
$\Sigma LQ = 13$
$13 \ne 14$
$14 - 13 = 1$
Balance = $1$

The balance is still positive.
Try another smaller standard divisor
Let the modified divisor = $112$

$MD = 111 \\[3ex] \underline{Third\:\:Step - Part\:\:4} \\[3ex] MQ\:\:for\:\:Genesis \\[3ex] = \dfrac{555}{111} \\[5ex] = 5 \\[3ex] MQ\:\:for\:\:Exodus \\[3ex] = \dfrac{255}{111} \\[5ex] = 2.2972973 \\[3ex] MQ\:\:for\:\:Leviticus \\[3ex] = \dfrac{355}{111} \\[5ex] = 3.1981982 \\[3ex] MQ\:\:for\:\:Numbers \\[3ex] = \dfrac{455}{111} \\[5ex] = 4.0990991 \\[3ex]$
Jefferson's Method - Fourth Try
Student Modified Quota Modified Lower Quota
Genesis $5$ $5$
Exodus $2.2972973$ $2$
Leviticus $3.1981982$ $3$
Numbers $4.0990991$ $4$
$\Sigma LQ = 14$
$14 = 14$
This works!

Jefferson's Method of Apportionment
Student Modified Lower Quota Apportion
Genesis $5$ $5$
Exodus $2$ $2$
Leviticus $3$ $3$
Numbers $4$ $4$
$\Sigma Apportion = 14$
$NIA = 14$
$\Sigma Apportion = NIA$

As you can see:
Looking at the standard quota (not the modified quota):
Genesis was apportioned $5$ pieces of candy...the upper quota
Exodus was apportioned $2$ pieces of candy...the lower quota
Leviticus was apportioned $3$ pieces of candy...the lower quota
Numbers was apportioned $4$ pieces of candy...the upper quota
Each student was apportioned either the lower quota or the upper quota.
Therefore, Quota Rule is met in this case.
However, Jefferson's method does not always satisfy the Quota Rule as seen in Example $1$ of Jefferson's Method of Apportionment

Student: Sir, do we need to draw all these tables?...
I mean we could have calculated the sum of the modified lower quotas without drawing all these tables.
Also, we could have used the calculator to try different values of the modified divisor.
Teacher: You do not need to draw all the tables.
But, it is important you show your work...the main ones.
I am being very detailed to ensure that all my students understand the solution.

(4.) To encourage her students to study the Holy Bible, Anna intends to distribute $27$ identical pieces of candy among her $4$ students based on the the number of pages they read last month.
The table below lists the total number of pages read by each student.
Student Pages
Ezra $180$
Nehemiah $305$
Esther $280$
Job $230$

Apportion the candy using Adams's Method

$\underline{First\:\:Step} \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 180 + 305 + 280 + 230 = 995 \\[3ex] NIA = 27 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{995}{27} \\[5ex] = 36.8518519 \\[3ex] \underline{Second\:\:Step} \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Ezra \\[3ex] = \dfrac{180}{36.8518519} \\[5ex] = 4.8844221 \\[3ex] SQ\:\:for\:\:Nehemiah \\[3ex] = \dfrac{305}{36.8518519} \\[5ex] = 8.2763819 \\[3ex] SQ\:\:for\:\:Esther \\[3ex] = \dfrac{280}{36.8518519} \\[5ex] = 7.59798994 \\[3ex] SQ\:\:for\:\:Job \\[3ex] = \dfrac{230}{36.8518519} \\[5ex] = 6.24120602 \\[3ex]$
Adams's Method - Part $1$
Student Standard Quota Upper Quota
Ezra $4.8844221$ $5$
Nehemiah $8.2763819$ $9$
Esther $7.59798994$ $8$
Job $6.24120602$ $7$
$\Sigma UQ = 29$
$29 \ne 27$
$27 - 29 = -2$
Balance = $-2$

The balance is negative.
Try a bigger standard divisor
Let the modified divisor = $37$

$MD = 37 \\[3ex] \underline{Third\:\:Step - Part\:\:1} \\[3ex] Modified\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Modified\:\:Divisor} \\[7ex] MQ\:\:for\:\:Ezra \\[3ex] = \dfrac{180}{37} \\[5ex] = 4.86486486 \\[3ex] MQ\:\:for\:\:Nehemiah \\[3ex] = \dfrac{305}{37} \\[5ex] = 8.24324324 \\[3ex] MQ\:\:for\:\:Esther \\[3ex] = \dfrac{280}{37} \\[5ex] = 7.56756757 \\[3ex] MQ\:\:for\:\:Job \\[3ex] = \dfrac{230}{37} \\[5ex] = 6.21621622 \\[3ex]$
Student Modified Quota Modified Upper Quota
Ezra $4.86486486$ $5$
Nehemiah $8.24324324$ $9$
Esther $7.56756757$ $8$
Job $6.21621622$ $7$
$\Sigma UQ = 29$
$29 \ne 27$
$27 - 29 = -2$
Balance = $-2$

The balance is still negative.
Try another bigger standard divisor
Let the modified divisor = $38$

$MD = 38 \\[3ex] \underline{Third\:\:Step - Part\:\:2} \\[3ex] MQ\:\:for\:\:Ezra \\[3ex] = \dfrac{180}{38} \\[5ex] = 4.73684211 \\[3ex] MQ\:\:for\:\:Nehemiah \\[3ex] = \dfrac{305}{38} \\[5ex] = 8.02631579 \\[3ex] MQ\:\:for\:\:Esther \\[3ex] = \dfrac{280}{38} \\[5ex] = 7.36842105 \\[3ex] MQ\:\:for\:\:Job \\[3ex] = \dfrac{230}{38} \\[5ex] = 6.05263158 \\[3ex]$
Student Modified Quota Modified Upper Quota
Ezra $4.73684211$ $5$
Nehemiah $8.02631579$ $9$
Esther $7.36842105$ $8$
Job $6.05263158$ $7$
$\Sigma UQ = 29$
$29 \ne 27$
$27 - 29 = -2$
Balance = $-2$

The balance is still negative.
Try another bigger standard divisor
Let the modified divisor = $39$

$MD = 39 \\[3ex] \underline{Third\:\:Step - Part\:\:3} \\[3ex] MQ\:\:for\:\:Ezra \\[3ex] = \dfrac{180}{39} \\[5ex] = 4.61538462 \\[3ex] MQ\:\:for\:\:Nehemiah \\[3ex] = \dfrac{305}{39} \\[5ex] = 7.82051282 \\[3ex] MQ\:\:for\:\:Esther \\[3ex] = \dfrac{280}{39} \\[5ex] = 7.17948718 \\[3ex] MQ\:\:for\:\:Job \\[3ex] = \dfrac{230}{39} \\[5ex] = 5.8974359 \\[3ex]$
Student Modified Quota Modified Upper Quota
Ezra $4.61538462$ $5$
Nehemiah $7.82051282$ $8$
Esther $7.17948718$ $8$
Job $5.8974359$ $6$
$\Sigma UQ = 27$
$27 = 27$
This works!

Student Modified Upper Quota Apportion
Ezra $5$ $5$
Nehemiah $8$ $8$
Esther $8$ $8$
Job $6$ $6$
$\Sigma Apportion = 27$
$NIA = 27$
$\Sigma Apportion = NIA$

As you can see:
Looking at the standard quota (not the modified quota):
Ezra was apportioned $5$ pieces of candy...the upper quota
Nehemiah was apportioned $8$ pieces of candy...the lower quota
Esther was apportioned $8$ pieces of candy...the upper quota
Job was apportioned $6$ pieces of candy...the lower quota
Each student was apportioned either the lower quota or the upper quota.
Therefore, Quota Rule is met in this case.
However, Adams's method does not always satisfy the Quota Rule as seen in Example $1$ of Adams's Method of Apportionment

Student: Sir, do we need to draw all these tables?...
I mean we could have calculated the sum of the modified upper quotas without drawing all these tables.
Also, we could have used the calculator to try different values of the modified divisor.
Teacher: You do not need to draw all the tables.
But, it is important you show your work...the main ones.
I am being very detailed to ensure that all my students understand the solution.

(5.) To encourage her students to study the Holy Bible, Anna intends to distribute $7$ identical pieces of candy among her $4$ students based on the the number of pages they read last month.
The table below lists the total number of pages read by each student.
Student Pages
Isaiah $230$
Jeremiah $655$
Ezekiel $105$
Daniel $305$

Apportion the candy using Webster's Method

$\underline{First\:\:Step} \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 230 + 655 + 105 + 305 = 1295 \\[3ex] NIA = 7 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{1295}{7} \\[5ex] = 185 \\[3ex] \underline{Second\:\:Step} \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Isaiah \\[3ex] = \dfrac{230}{185} \\[5ex] = 1.24324324 \\[3ex] SQ\:\:for\:\:Jeremiah \\[3ex] = \dfrac{655}{185} \\[5ex] = 3.54054054 \\[3ex] SQ\:\:for\:\:Ezekiel \\[3ex] = \dfrac{105}{185} \\[5ex] = 0.567567568 \\[3ex] SQ\:\:for\:\:Daniel \\[3ex] = \dfrac{305}{185} \\[5ex] = 1.64864865 \\[3ex]$
Webster's Method - Part $1$
Student Standard Quota Normal Quota
Isaiah $1.24324324$ $1$
Jeremiah $3.54054054$ $4$
Ezekiel $0.567567568$ $1$
Daniel $1.64864865$ $2$
$\Sigma NQ = 8$
$8 \ne 7$
$7 - 8 = -1$
Balance = $-1$

The balance is negative.
Try a bigger standard divisor
Let the modified divisor = $186$

$MD = 186 \\[3ex] \underline{Third\:\:Step - Part\:\:1} \\[3ex] Modified\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Modified\:\:Divisor} \\[7ex] MQ\:\:for\:\:Isaiah \\[3ex] = \dfrac{230}{186} \\[5ex] = 1.23655914 \\[3ex] MQ\:\:for\:\:Jeremiah \\[3ex] = \dfrac{655}{186} \\[5ex] = 3.52150538 \\[3ex] MQ\:\:for\:\:Ezekiel \\[3ex] = \dfrac{105}{186} \\[5ex] = 0.564516129 \\[3ex] MQ\:\:for\:\:Daniel \\[3ex] = \dfrac{305}{186} \\[5ex] = 1.63978495 \\[3ex]$
Webster's Method - First Try
Student Modified Quota Modified Normal Quota
Isaiah $1.23655914$ $1$
Jeremiah $3.52150538$ $4$
Ezekiel $0.564516129$ $1$
Daniel $1.63978495$ $2$
$\Sigma NQ = 8$
$8 \ne 7$
$7 - 8 = -1$
Balance = $-1$

The balance is still negative.
Try another bigger standard divisor
Let the modified divisor = $187$

$MD = 187 \\[3ex] \underline{Third\:\:Step - Part\:\:2} \\[3ex] MQ\:\:for\:\:Isaiah \\[3ex] = \dfrac{230}{187} \\[5ex] = 1.22994652 \\[3ex] MQ\:\:for\:\:Jeremiah \\[3ex] = \dfrac{655}{187} \\[5ex] = 3.5026738 \\[3ex] MQ\:\:for\:\:Ezekiel \\[3ex] = \dfrac{105}{187} \\[5ex] = 0.561497326 \\[3ex] MQ\:\:for\:\:Daniel \\[3ex] = \dfrac{305}{187} \\[5ex] = 1.63101604 \\[3ex]$
Webster's Method - Second Try
Student Modified Quota Modified Normal Quota
Isaiah $1.22994652$ $1$
Jeremiah $3.5026738$ $4$
Ezekiel $0.561497326$ $1$
Daniel $1.63101604$ $2$
$\Sigma NQ = 8$
$8 \ne 7$
$7 - 8 = -1$
Balance = $-1$

The balance is still negative.
Try another bigger standard divisor
Let the modified divisor = $188$

$MD = 188 \\[3ex] \underline{Third\:\:Step - Part\:\:3} \\[3ex] MQ\:\:for\:\:Isaiah \\[3ex] = \dfrac{230}{188} \\[5ex] = 1.22340426 \\[3ex] MQ\:\:for\:\:Jeremiah \\[3ex] = \dfrac{655}{188} \\[5ex] = 3.48404255 \\[3ex] MQ\:\:for\:\:Ezekiel \\[3ex] = \dfrac{105}{188} \\[5ex] = 0.558510638 \\[3ex] MQ\:\:for\:\:Daniel \\[3ex] = \dfrac{305}{188} \\[5ex] = 1.62234043 \\[3ex]$
Webster's Method - Third Try
Student Modified Quota Modified Normal Quota
Isaiah $1.22340426$ $1$
Jeremiah $3.48404255$ $3$
Ezekiel $0.558510638$ $1$
Daniel $1.62234043$ $2$
$\Sigma NQ = 7$
$7 = 7$
This works!

Webster's Method of Apportionment
Student Modified Normal Quota Apportion
Isaiah $1$ $1$
Jeremiah $3$ $3$
Ezekiel $1$ $1$
Daniel $2$ $2$
$\Sigma Apportion = 7$
$NIA = 7$
$\Sigma Apportion = NIA$

As you can see:
Looking at the standard quota (not the modified quota):
Isaiah was apportioned $1$ pieces of candy...the lower quota
Jeremiah was apportioned $3$ pieces of candy...the lower quota
Ezekiel was apportioned $1$ pieces of candy...the upper quota
Daniel was apportioned $2$ pieces of candy...the upper quota
Each student was apportioned either the lower quota or the upper quota.
Therefore, Quota Rule is met in this case.
However, Webster's method does not always satisfy the Quota Rule as seen in Example $1$ of Webster's Method of Apportionment

Student: Sir, do we need to draw all these tables?...
I mean we could have calculated the sum of the modified normal quotas without drawing all these tables.
Also, we could have used the calculator to try different values of the modified divisor.
Teacher: You do not need to draw all the tables.
But, it is important you show your work...the main ones.
I am being very detailed to ensure that all my students understand the solution.

(6.) To encourage her students to study the Holy Bible, Anna intends to distribute $21$ identical pieces of candy among her $4$ students based on the the number of pages they read last month.
The table below lists the total number of pages read by each student.
Student Pages
Hosea $430$
Joel $480$
Amos $805$
Obadiah $605$

Apportion the candy using Webster's Method

$\underline{First\:\:Step} \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 430 + 480 + 805 + 605 = 2320 \\[3ex] NIA = 21 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{2320}{21} \\[5ex] = 110.47619 \\[3ex] \underline{Second\:\:Step} \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Hosea \\[3ex] = \dfrac{430}{110.47619} \\[5ex] = 3.8922414 \\[3ex] SQ\:\:for\:\:Joel \\[3ex] = \dfrac{480}{110.47619} \\[5ex] = 4.3448276 \\[3ex] SQ\:\:for\:\:Amos \\[3ex] = \dfrac{805}{110.47619} \\[5ex] = 7.28663796 \\[3ex] SQ\:\:for\:\:Obadiah \\[3ex] = \dfrac{605}{110.47619} \\[5ex] = 5.47629313 \\[3ex]$
Webster's Method - Part $1$
Student Standard Quota Normal Quota
Hosea $3.8922414$ $4$
Joel $4.3448276$ $4$
Amos $7.28663796$ $7$
Obadiah $5.47629313$ $5$
$\Sigma NQ = 20$
$20 \ne 21$
$21 - 20 = 1$
Balance = $1$

The balance is positive.
Try a smaller standard divisor
Let the modified divisor = $109$

$MD = 109 \\[3ex] \underline{Third\:\:Step - Part\:\:1} \\[3ex] Modified\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Modified\:\:Divisor} \\[7ex] MQ\:\:for\:\:Hosea \\[3ex] = \dfrac{430}{109} \\[5ex] = 3.94495413 \\[3ex] MQ\:\:for\:\:Joel \\[3ex] = \dfrac{480}{109} \\[5ex] = 4.40366972 \\[3ex] MQ\:\:for\:\:Amos \\[3ex] = \dfrac{805}{109} \\[5ex] = 7.3853211 \\[3ex] MQ\:\:for\:\:Obadiah \\[3ex] = \dfrac{605}{109} \\[5ex] = 5.55045872 \\[3ex]$
Webster's Method - First Try
Student Modified Quota Modified Normal Quota
Hosea $3.94495413$ $4$
Joel $4.40366972$ $4$
Amos $7.3853211$ $7$
Obadiah $5.55045872$ $6$
$\Sigma NQ = 21$
$21 = 21$
This works!

Webster's Method of Apportionment
Student Modified Normal Quota Apportion
Hosea $4$ $4$
Joel $4$ $4$
Amos $7$ $7$
Obadiah $6$ $6$
$\Sigma Apportion = 21$
$NIA = 21$
$\Sigma Apportion = NIA$

As you can see:
Looking at the standard quota (not the modified quota):
Hosea was apportioned $4$ pieces of candy...the upper quota
Joel was apportioned $4$ pieces of candy...the lower quota
Amos was apportioned $7$ pieces of candy...the lower quota
Obadiah was apportioned $6$ pieces of candy...the upper quota
Each student was apportioned either the lower quota or the upper quota.
Therefore, Quota Rule is met in this case.
However, Webster's method does not always satisfy the Quota Rule as seen in Example $1$ of Webster's Method of Apportionment

Student: Sir, do we need to draw all these tables?...
I mean we could have calculated the sum of the modified normal quotas without drawing all these tables.
Also, we could have used the calculator to try different values of the modified divisor.
Teacher: You do not need to draw all the tables.
But, it is important you show your work...the main ones.
I am being very detailed to ensure that all my students understand the solution.

(7.) To encourage her students to study the Holy Bible, Anna intends to distribute $13$ identical pieces of candy among her $4$ students based on the the number of pages they read last month.
The table below lists the total number of pages read by each student.
Student Pages
Jonah $135$
Micah $305$
Nahum $370$
Haggai $520$

Apportion the candy using Webster's Method

$\underline{First\:\:Step} \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 135 + 305 + 370 + 520 = 1330 \\[3ex] NIA = 13 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{1330}{13} \\[5ex] = 102.307692 \\[3ex] \underline{Second\:\:Step} \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Jonah \\[3ex] = \dfrac{135}{102.307692} \\[5ex] = 1.31954888 \\[3ex] SQ\:\:for\:\:Micah \\[3ex] = \dfrac{305}{102.307692} \\[5ex] = 2.98120302 \\[3ex] SQ\:\:for\:\:Nahum \\[3ex] = \dfrac{370}{102.307692} \\[5ex] = 3.61654136 \\[3ex] SQ\:\:for\:\:Haggai \\[3ex] = \dfrac{520}{102.307692} \\[5ex] = 5.08270678 \\[3ex]$
Webster's Method - Part $1$
Student Standard Quota Normal Quota
Jonah $1.31954888$ $1$
Micah $2.98120302$ $3$
Nahum $3.61654136$ $4$
Haggai $5.08270678$ $5$
$\Sigma Apportion = 13$
$NIA = 13$
$\Sigma Apportion = NIA$

As you can see:
Looking at the standard quota (not the modified quota):
Jonah was apportioned $1$ piece of candy...the lower quota
Micah was apportioned $3$ pieces of candy...the upper quota
Nahum was apportioned $4$ pieces of candy...the upper quota
Haggai was apportioned $5$ pieces of candy...the lower quota
Each student was apportioned either the lower quota or the upper quota.
Therefore, Quota Rule is met in this case.
However, Webster's method does not always satisfy the Quota Rule as seen in Example $1$ of Webster's Method of Apportionment
(8.)

(9.) Round each number using the Huntington-Hill's method rounding rules.

$(a.)\:\: 9.152 \\[3ex] (b.)\:\: 8.654 \\[3ex] (c.)\:\: 9.7 \\[3ex] (d.)\:\: 9.486 \\[3ex] (e.)\:\: 9.487 \\[3ex] (f.)\:\: 8.484 \\[3ex] (g.)\:\: 8.497 \\[3ex]$

Huntingdon-Hill's Method
SQ LQ UQ LQ * UQ $GM = \sqrt{LQ * UQ}$ Compare SQ with GM Rounded Number
$9.152$ $9$ $10$ $90$ $9.48683298$ $9.152 \lt 9.48683298$ $9$
$8.654$ $8$ $9$ $72$ $8.48528137$ $8.654 \gt 8.48528137$ $9$
$9.7$ $9$ $10$ $90$ $9.48683298$ $9.7 \gt 9.48683298$ $10$
$9.486$ $9$ $10$ $90$ $9.48683298$ $9.486 \lt 9.48683298$ $9$
$9.487$ $9$ $10$ $90$ $9.48683298$ $9.487 \gt 9.48683298$ $10$
$8.484$ $8$ $9$ $72$ $8.48528137$ $8.484 \lt 8.48528137$ $8$
$8.497$ $8$ $9$ $72$ $8.48528137$ $8.497 \gt 8.48528137$ $9$
(10.)

$\Sigma f = 3 + 1 + 3 + 4 + 9 = 20 \\[3ex] \dfrac{\Sigma f}{2} = \dfrac{20}{2} = 10 \\[5ex] Begin\:\:from\:\:the\:\:first\:\:class \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:10 \\[3ex] 3 + 1 = 4 \\[3ex] 4 + 3 = 7 \\[3ex] 7 + 4 = 11...stop \\[3ex]$ The class that contains the median is $81-85$
(11.) The legislature in the State of Virtues has $39$ seats.
Apportion these seats to five of the counties below using the Huntingdon-Hill method.

County Population
Faith $278,000$
Hope $467,000$
Peace $300,000$
Love $472,000$
Joy $230,000$

$\underline{First\:\:Step} \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 278000 + 467000 + 300000 + 472000 + 230000 = 1747000 \\[3ex] NIA = 39 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{1747000}{39} \\[5ex] = 44794.8718 \\[3ex] \underline{Second\:\:Step} \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Faith \\[3ex] = \dfrac{278000}{44794.8718} \\[5ex] = 6.20606754 \\[3ex] SQ\:\:for\:\:Hope \\[3ex] = \dfrac{467000}{44794.8718} \\[5ex] = 10.4253005 \\[3ex] SQ\:\:for\:\:Peace \\[3ex] = \dfrac{300000}{44794.8718} \\[5ex] = 6.69719519 \\[3ex] SQ\:\:for\:\:Love \\[3ex] = \dfrac{472000}{44794.8718} \\[5ex] = 10.5369204 \\[3ex] SQ\:\:for\:\:Joy \\[3ex] = \dfrac{230000}{44794.8718} \\[5ex] = 5.13451631 \\[3ex]$
Huntingdon-Hill's Method
County SQ LQ UQ LQ * UQ $GM = \sqrt{LQ * UQ}$ Compare SQ with GM Apportion
Faith $6.20606754$ $6$ $7$ $42$ $6.4807407$ $6.20606754 \lt 6.4807407$ $6$
Hope $10.4253005$ $10$ $11$ $110$ $10.4880885$ $10.4253005 \lt 10.4880885$ $10$
Peace $6.69719519$ $6$ $7$ $42$ $6.4807407$ $6.69719519 \gt 6.4807407$ $7$
Love $10.5369204$ $10$ $11$ $110$ $10.4880885$ $10.5369204 \gt 10.4880885$ $11$
Joy $5.13451631$ $5$ $6$ $30$ $5.47722558$ $5.13451631 \lt 5.47722558$ $5$
$\Sigma Apportion = 39$
$NIA = 39$
$\Sigma Apportion = NIA$
(12.)

Let the unreadable weight = $x$

$For\:\:20\:\:patients \\[3ex] \bar{x} = 82 \\[3ex] n = 20 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 82(20) = 1640 \\[3ex] For\:\:19\:\:patients \\[3ex] \bar{x} = 86 \\[3ex] n = 19 \\[3ex] \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 86(19) = 1634 \\[3ex] \Sigma x_{19} + x = \Sigma x_{20} \\[3ex] 1634 + x = 1640 \\[3ex] x = 1640 - 1634 \\[3ex] x = 6$
(13.) To encourage her students to study the Holy Bible, Anna intends to distribute $18$ identical pieces of candy among her $4$ students based on the the number of pages they read last month.
The table below lists the total number of pages read by each student.
Student Pages
Malachi $305$
Habakkuk $260$
Zechariah $110$
Zephaniah $85$

Apportion the candy using Lowndes's Method

$\underline{First\:\:Step} \\[3ex] Standard\:\:Divisor = \dfrac{Population\:\:Size}{Number\:\:of\:\:Items\:\:to\:\:Apportion} \\[7ex] N = 305 + 260 + 110 + 85 = 760 \\[3ex] NIA = 18 \\[3ex] SD = \dfrac{N}{NIA} \\[5ex] = \dfrac{760}{18} \\[5ex] = 42.2222222 \\[3ex] \underline{Second\:\:Step} \\[3ex] Standard\:\:Quota\:\:for\:\:Each\:\:Sample = \dfrac{Sample\:\:Size}{Standard\:\:Divisor} \\[7ex] SQ\:\:for\:\:Malachi \\[3ex] = \dfrac{305}{42.2222222} \\[5ex] = 7.22368421 \\[3ex] SQ\:\:for\:\:Habakkuk \\[3ex] = \dfrac{260}{42.2222222} \\[5ex] = 6.15789474 \\[3ex] SQ\:\:for\:\:Zechariah \\[3ex] = \dfrac{110}{42.2222222} \\[5ex] = 2.60526316 \\[3ex] SQ\:\:for\:\:Zephaniah \\[3ex] = \dfrac{85}{42.2222222} \\[5ex] = 2.0131579 \\[3ex]$
Lowndes's Method - Part $1$
Student Standard Quota Lower Quota
Malachi $7.22368421$ $7$
Habakkuk $6.15789474$ $6$
Zechariah $2.60526316$ $2$
Zephaniah $2.0131579$ $2$
$\Sigma LQ = 17$
$17 \ne 18$
$18 - 17 = 1$
Balance = $1$

Lowndes's Method - Part $2$
First Approach: Persons per Representative
(Pages per Candy)
Student Pages Lower Quota of Candy Pages per Candy
Malachi $305$ $7$ $43.5714286$
Habakkuk $260$ $6$ $43.3333333$
Zechariah $110$ $2$ $55$
Zephaniah $85$ $2$ $42.5$

Lowndes's Method - Part $2$
First Approach: Pages per Candy (Sorted)
Student Pages per Candy Pages per Candy (Descending Order)
Zechariah $55$ First
Malachi $43.5714286$ Second
Habakkuk $43.3333333$ Third
Zephaniah $42.5$ Fourth

Zechariah read the most number of pages per candy.
Assign the remaining candy to Zechariah.

Lowndes's Method of Apportionment
Student Lower Quota Extra Item Apportion
Malachi $7$ $7$
Habakkuk $6$ $6$
Zechariah $2$ $1$ $3$
Zephaniah $2$ $2$
$\Sigma Apportion = 18$
$NIA = 18$
$\Sigma Apportion = NIA$

OR

Lowndes's Method - Part $2$
Second Approach: Lowndes's Ratio
Student Standard Quota Decimal Part of Standard Quota Lower Quota Lowndes's Ratio
Malachi $7.22368421$ $0.22368421$ $7$ $0.0319548871$
Habakkuk $6.15789474$ $0.15789474$ $6$ $0.02631579$
Zechariah $2.60526316$ $0.60526316$ $2$ $0.30263158$
Zephaniah $2.0131579$ $0.0131579$ $2$ $0.00657895$

Lowndes's Method - Part $2$
Second Approach: Lowndes's Ratio (Sorted)
Student Lowndes's Ratio Lowndes's Ratio (Descending Order)
Zechariah $0.30263158$ First
Malachi $0.0319548871$ Second
Habakkuk $0.02631579$ Third
Zephaniah $0.00657895$ Fourth

Zechariah has the highest Lowndes's ratio.
Assign the remaining candy to Zechariah.

Lowndes's Method of Apportionment
Student Lower Quota Extra Item Apportion
Malachi $7$ $7$
Habakkuk $6$ $6$
Zechariah $2$ $1$ $3$
Zephaniah $2$ $2$
$\Sigma Apportion = 18$
$NIA = 18$
$\Sigma Apportion = NIA$

As you can see:
Malachi was apportioned $7$ pieces of candy...the lower quota
Habakkuk was apportioned $6$ pieces of candy...the lower quota
Zechariah was apportioned $3$ pieces of candy...the upper quota
Zephaniah was apportioned $2$ pieces of candy...the lower quota
Each student was apportioned either the lower quota or the upper quota.
Therefore, Quota Rule is met.
However, Lowndes's method does not always satisfy the Quota Rule as seen in Example $1$ of Lowndes's Method of Apportionment
(14.)

 Data Value Frequency $51$ $1$ $61$ $1$ $62$ $1$ $57$ $1$ $50$ $1$ $67$ $1$ $68$ $1$ $58$ $1$ $53$ $1$

There is no mode.
(15.) Round each number using the Dean's method rounding rules.

$(a.)\:\: 9.152 \\[3ex] (b.)\:\: 8.654 \\[3ex] (c.)\:\: 9.7 \\[3ex] (d.)\:\: 9.486 \\[3ex] (e.)\:\: 9.487 \\[3ex] (f.)\:\: 8.484 \\[3ex] (g.)\:\: 8.497 \\[3ex]$

Dean's Method
SQ LQ UQ 2 * LQ * UQ LQ + UQ $HM = \dfrac{2 * LQ * UQ}{LQ + UQ}$ Compare SQ with HM Rounded Number
$9.152$ $9$ $10$ $180$ $19$ $9.47368421$ $9.152 \lt 9.47368421$ $9$
$8.654$ $8$ $9$ $144$ $17$ 8.470588248.654 \gt 8.4705882499.7910180199.473684219.7 \gt 9.47368421109.486910180199.473684219.486 \gt 9.47368421109.487910180199.473684219.487 \gt 9.47368421108.48489144178.470588248.484 \gt 8.4705882498.49789144178.470588248.497 \gt 8.470588249$(16.)$ \Sigma f = 3 + 1 + 3 + 4 + 9 = 20 \\[3ex] \dfrac{\Sigma f}{2} = \dfrac{20}{2} = 10 \\[5ex] Begin\:\:from\:\:the\:\:first\:\:class \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:10 \\[3ex] 3 + 1 = 4 \\[3ex] 4 + 3 = 7 \\[3ex] 7 + 4 = 11...stop \\[3ex] $The class that contains the median is$81-85\$