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- WELCOME TO A LESSON ON THE POPULATION PARADOX
OF APPORTIONMENT.
THE POPULATION PARADOX IS A COUNTERINTUITIVE RESULT
OF SOME PROCEDURES FOR APPORTIONMENT.
WHEN TWO STATES HAVE POPULATIONS INCREASING AT DIFFERENT RATES,
A SMALL STATE WITH RAPID GROWTH CAN LOSE A LEGISLATIVE SEAT
TO A BIG STATE WITH SLOWER GROWTH.
THE PARADOX ARISES BECAUSE OF ROUNDING IN THE PROCEDURE
FOR DIVIDING THE SEATS.
LET'S TAKE A LOOK AT AN EXAMPLE.
A SMALL COUNTRY HAS 24 SEATS IN A CONGRESS DIVIDED AMONG STATES
ACCORDING TO THEIR RESPECTIVE POPULATIONS.
THE TABLE BELOW SHOWS THE POPULATIONS
AFTER A 2000 AND A 2010 CENSUS.
WE WANT TO USE HAMILTON'S METHOD TO APPORTION THE 24 SEATS.
SO NOTICE HOW THIS COLUMN GIVES US THE POPULATION
AFTER THE 2000 CENSUS.
THIS COLUMN HERE GIVES US THE POPULATION
AFTER THE 2010 CENSUS,
AS WELL AS THE PERCENT GROWTH.
NOTICE HOW DISTRICT "A" IS THE SMALLEST DISTRICT,
BUT ALSO THE DISTRICT WITH THE LARGEST GROWTH RATE
OVER THIS TIME PERIOD.
WE'LL BEGIN BY APPORTIONING THE 24 SEATS
USING THE 2000 POPULATION DATA.
SO REMEMBER THE FIRST STEP IS TO FIND THE DIVISOR
BY TAKING THE SUM OF THE POPULATION
FROM ALL THE STATES OR DISTRICTS,
AND DIVIDING BY THE NUMBER OF SEATS, WHICH IS 24,
GIVING US A STANDARD DIVISOR OF APPROXIMATELY 1,566.667.
AND NOW WE TAKE THE POPULATION FROM EACH DISTRICT
AND DIVIDE BY THE DIVISOR.
SO FOR EXAMPLE, FOR DISTRICT "A" THE QUOTA OF 3.383
CAME FROM TAKING THE POPULATION OF 5,300
AND DIVIDING BY OUR DIVISOR,
WHICH GIVES US APPROXIMATELY 3.383.
WE WOULD DO THE SAME FOR DISTRICT B AND DISTRICT C.
AND NOW TO FIND THE INITIAL ALLOCATION, OR THE LOWER QUOTA,
WE TAKE THE QUOTA VALUES AND REMOVE THE DECIMAL.
NOTICE "A" RECEIVES 3, B RECEIVES 6, AND C RECEIVES 14.
BUT NOTICE HOW THIS SUM IS 23,
AND WE HAVE 24 SEATS TO APPORTION,
SO THE EXTRA SEAT GOES TO THE DISTRICT
WHOSE QUOTA HAS THE LARGEST DECIMAL VALUE.
SO LOOKING AT JUST THE DECIMAL PARTS OF THE QUOTA,
NOTICE HOW DISTRICT "A" HAS THE LARGEST DECIMAL PART,
THEREFORE DISTRICT "A" GETS THE EXTRA SEAT.
SO THE FINAL APPORTIONMENT WOULD BE 4, 6, AND 14,
GIVING US A TOTAL OF 24 SEATS.
AND NOW WE'LL FOLLOW THE SAME PROCEDURE
WITH THE POPULATION FROM 2010.
SO AGAIN, HERE'S THE POPULATION FROM 2010,
AND WE'RE ALSO INCLUDING THE PERCENT OF GROWTH
FROM THE YEAR 2000.
NOTICE THE TOTAL POPULATION FOR THE THREE STATES IS 45,000,
AND THE NUMBER OF SEATS IS STAYING THE SAME.
SO OUR DIVISOR IS 45,000 DIVIDED BY 24, OR 1,875.
SO NOW TO FIND THE QUOTA FOR EACH STATE
WE TAKE EACH STATE POPULATION AND DIVIDE BY THE NEW DIVISOR.
SO FOR EXAMPLE, FOR DISTRICT "A" WE WOULD HAVE
6,800 DIVIDED BY 1,875, GIVING A QUOTA OF APPROXIMATELY 3.627.
AND WE DO THE SAME FOR DISTRICT B AND DISTRICT C.
AND NOW FOR THE INITIAL ALLOCATION, OR THE LOWER QUOTA,
WE TAKE THE QUOTAS AND REMOVE THE DECIMAL PART.
SO NOTICE HOW "A" RECEIVES 3, B RECEIVES 6, AND C RECEIVES 13,
BUT NOTICE HOW THIS SUM IS 22.
WE HAVE TWO EXTRA SEATS TO APPORTION OR ALLOCATE.
AND THOSE WOULD GO TO THE STATES
THAT HAVE THE LARGEST DECIMAL VALUE OF THEIR QUOTA.
SO LOOKING AT JUST THE DECIMAL PARTS OF THE QUOTA,
NOTICE HOW DISTRICT B AND DISTRICT C
HAVE THE TWO LARGEST DECIMAL PARTS,
AND THEREFORE DISTRICT B RECEIVES AN EXTRA SEAT
AND SO DOES DISTRICT C.
SO THE FINAL ALLOCATION WOULD BE 3, 7, 14,
GIVING A TOTAL OF 24 SEATS.
SO NOTICE HOW DISTRICT "A" HAD THE LARGEST PERCENT GROWTH,
BUT NOTICE IN THE YEAR 2000 DISTRICT "A" HAD 4 SEATS,
BUT IN 2010 DISTRICT "A" ONLY RECEIVES 3 SEATS.
THIS IS AN EXAMPLE OF THE POPULATION PARADOX.
I HOPE YOU FOUND THIS HELPFUL.