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

**
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.

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] $

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.

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] $

__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.__

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

$ \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] $

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 = $2$ |

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 | $4.16988417$ | $0.16988417$ |

Student | Lower Quota | Extra Item | Apportion |
---|---|---|---|

Joshua | $10$ | $1$ | $11$ |

Judges | $5$ | $5$ | |

Ruth | $9$ | $1$ | $10$ |

Samuel | $4$ | $4$ | |

$\Sigma Apportion = 30$ $NIA = 30$ $\Sigma Apportion = NIA$ |

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,

Hamilton's method

(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.

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] $

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] $

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] $

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] $

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] $

__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

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

$ \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] $

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] $

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] $

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] $

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] $

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! |

Student | Modified Lower Quota | Apportion |
---|---|---|

Genesis | $5$ | $5$ |

Exodus | $2$ | $2$ |

Leviticus | $3$ | $3$ |

Numbers | $4$ | $4$ |

$\Sigma Apportion = 14$ $NIA = 14$ $\Sigma Apportion = NIA$ |

Looking at the

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,

However, Jefferson's method

*
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.

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] $

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] $

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] $

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] $

__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

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

$ \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] $

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 | $6$ | $6$ |

$\Sigma Apportion = 27$ $NIA = 27$ $\Sigma Apportion = NIA$ |

Looking at the

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,

However, Adams's method

*
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.

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] $

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] $

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] $

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] $

__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

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

$ \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] $

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] $

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] $

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] $

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! |

Student | Modified Normal Quota | Apportion |
---|---|---|

Isaiah | $1$ | $1$ |

Jeremiah | $3$ | $3$ |

Ezekiel | $1$ | $1$ |

Daniel | $2$ | $2$ |

$\Sigma Apportion = 7$ $NIA = 7$ $\Sigma Apportion = NIA$ |

Looking at the

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,

However, Webster's method

*
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.

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] $

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] $

__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

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

$ \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] $

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] $

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! |

Student | Modified Normal Quota | Apportion |
---|---|---|

Hosea | $4$ | $4$ |

Joel | $4$ | $4$ |

Amos | $7$ | $7$ |

Obadiah | $6$ | $6$ |

$\Sigma Apportion = 21$ $NIA = 21$ $\Sigma Apportion = NIA$ |

Looking at the

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,

However, Webster's method

__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.

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] $

__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

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

$ \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] $

Student | Standard Quota | Normal Quota |
---|---|---|

Jonah | $1.31954888$ | $1$ |

Micah | $2.98120302$ | $3$ |

Nahum | $3.61654136$ | $4$ |

Haggai | $5.08270678$ | $5$ |

$\Sigma Apportion = 13$ $NIA = 13$ $\Sigma Apportion = NIA$ |

Looking at the

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,

However, Webster's method

(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] $

$ (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] $

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 $81-85$

$ \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.

$ \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] $

Apportion these seats to five of the counties below using the

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] $

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 $

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.

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] $

Zechariah read the most number of pages per candy.

Assign the remaining candy to Zechariah.

OR

Zechariah has the highest Lowndes's ratio.

Assign the remaining candy to Zechariah.

__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

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

$ \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] $

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 = $1$ |

Student | Pages | Lower Quota of Candy | Pages per Candy |
---|---|---|---|

Malachi | $305$ | $7$ | $43.5714286$ |

Habakkuk | $260$ | $6$ | $43.3333333$ |

Zechariah | $110$ | $2$ | $55$ |

Zephaniah | $85$ | $2$ | $42.5$ |

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.

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

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 | $2.0131579$ | $0.0131579$ | $2$ | $0.00657895$ |

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.

Student | Lower Quota | Extra Item | Apportion |
---|---|---|---|

Malachi | $7$ | $7$ | |

Habakkuk | $6$ | $6$ | |

Zechariah | $2$ | $1$ | $3$ |

Zephaniah | $2$ | $2$ | |

$\Sigma Apportion = 18$ $NIA = 18$ $\Sigma Apportion = NIA$ |

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,

However, Lowndes's method

(14.)

There is no mode.

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] $

$ (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] $

SQ | LQ | UQ | 2 * LQ * UQ | LQ + UQ | $HM = \dfrac{2 * LQ * UQ}{LQ + UQ}$ | Compare SQ with HM | Rounded Number |
---|---|---|---|---|---|---|---|

$9.152$ | $9$ | $10$ | $180$ | $19$ | $9.47368421$ | $9.152 \lt 9.47368421$ | $9$ |

$8.654$ | $8$ | $9$ | $144$ | $17$ | 8.47058824$ | $8.654 \gt 8.47058824$ | $9$ |

$9.7$ | $9$ | $10$ | $180$ | $19$ | $9.47368421$ | $9.7 \gt 9.47368421$ | $10$ |

$9.486$ | $9$ | $10$ | $180$ | $19$ | $9.47368421$ | $9.486 \gt 9.47368421$ | $10$ |

$9.487$ | $9$ | $10$ | $180$ | $19$ | $9.47368421$ | $9.487 \gt 9.47368421$ | $10$ |

$8.484$ | $8$ | $9$ | $144$ | $17$ | $8.47058824$ | $8.484 \gt 8.47058824$ | $9$ |

$8.497$ | $8$ | $9$ | $144$ | $17$ | $8.47058824$ | $8.497 \gt 8.47058824$ | $9$ |

(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 $81-85$

$ \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 $81-85$