Dice and Data A comic about statistics with a 95% chance of being one Dice and Data A comic about statistics with a 95% chance of being one

GOVERN DE LES ILLES BALEARS Vicepresidència i Conselleria d’Economia, Comerç i Indústria Direcció General d’Economia

CCIX © Edition: Directorate General of Economy Project Management: Antoni Monserrat I Moll. General Director of Economy General Coordination: Jose Antonio Pipo Jaldo. Director of IBAE Produced by: Balearic Institute of Statistics Palau Reial, 19 07001 – Palma () - Telephone: (34) 971-176-706 http://ibae.caib.es E-mail: [email protected]

Author: Javier Cubero

Management and Production: Inrevés SLL Illustrations: Alex Fito and Linhart Color: Pau Genestra Layout: Xisco Alario and Margalida Capó Script adapted by: Felipe Hernández Translation to English: T&i-Traducción e interpretación Revision: Roser Belmonte Juan Coordination: Sebastià Marí and Pere Joan

Collection: Estadística al carrer. Volume 1 Title: Dice and Data. A comic about statistics with a 95% chance of being one No. IBAE: CCIX Legal Deposit: 2PM-2462-2002 ISBN: 84-89745-53-6

Printed: Planografica Balear SL Publishing date of the English version: November 2002

© Copyright: Directorate General of Economy. Regional Ministry of Economy, Trade and Industry. PRESENTATION

The study of mathematics and the concepts of statistics have always been known to be difficult disciplines that are not very attractive to students. Therefore, the Government of the Balearic Islands wanted to contribute to the Worldwide Year of Mathematics in the area of the popularization of these areas with the publishing of the comic Dice and Data.

The present edition, elaborated by the Balearic Institute of Statistics (IBAE) of the Regional Ministry of Economy, Trade, and Industry, is an effective instrument that meets the teaching requirements of the Spanish High School study plan (ESO) and can be used to educate adults. This document’s purpose is to enable the aforementioned people to learn the information it contains in an easier and more attractive way.

This publication is included in the teaching plan that was initiated by the IBAE with the purpose of communicating to the society the different studies and analyses that have been carried out by the IBAE. However, the objective of the institute is not only to make known this statistical data that shows the socio-economic reality of the Balearic Islands, but also to present a work methodology that is important in planning decisions regarding the future of our country, building on a good solid base.

Finally, we would like to thank the team that worked on this publication and their par- ticipation in an experience that we consider to be innovative. We would also like to make special mention of the group of creative and graphic designers that participated in the making of this publication, having demonstrated the high quality of this sector in the Balearic Islands.

Pere Sampol i Mas

Vice-president of the Government of the Balearic Islands and Regional Ministry of Economy, Trade and Industry PROLOGUE

It is a true pleasure to preface the work that you presently have in your hands for many reasons. First of all, undoubtedly not the most important reason, but nevertheless important, would be the longstanding friendship that I have with the author, since just 35 years ago we were in the same class in the University. In those days, we would have never thought that we would meet up again for the love of statistics.

The second is the very work itself, Dice and Data that, as a sharp reader, you will observe that it is much more than a comic. From the choice of names of the characters, that is definitely not random but has its story, like 55’s birthday at the end or the rock group called quartile.

To point out some parts that I particularly enjoyed and that will make you think, I will begin with the historical brushstrokes that each chapter begins with, followed by an ele- gant way of explaining the difference between a continuous variable (snail trails) and a discrete variable (grasshopper leaps), and finally the method of teaching that the data contains more information than it appears to contain at first (the age problem of the four siblings) is, in the least, original.

The way in which it avoids reasoning by way of charts, since they can lead you to evi- dent errors (areas of the square and rectangle), reminded me of a professor that we once had in our studies of Mathematical Science.

The introduction of the concepts of population density and the age profile are very illustrative with their applications to the different cities of the Balearic Islands along with the motor vehicle accident rates, which is something that effects our youth today and that show why there are irregularities in the age profile.

It is probably in chapter 8 that the author shows his genius with cartoons used to introduce index numbers depending on the old (unlabeled corked bottles) and new quan- tities (tetra briks of milk), in conjunction with the old prices (spiral notebook) and the new prices (computer screen).

May these last words encourage Javier to continue with the work at hand and to delight us in the near future with a sequel.

Granada, April 2000

Rafael Herrerías Pleguezuelo Professor of Applied Economics INDEX

Chapter 1 – PIERRE DE FERMAT pg. 8

Chapter 2 – THOMAS BAYES pg. 13

Chapter 3 – BLAISE PASCAL pg. 20

Chapter 4 – ADOLPHE QUETELET pg. 26

Chapter 5 – JAKOB BERNOUILLI pg. 34

Chapter 6 – CHARLES DODGSON pg. 43

Chapter 7 – WILLIAM SEALEY GOSSET pg. 50

Chapter 8 – LASPEYRES AND PAASCHE pg. 59

Chapter 9 – DE MOIVRE AND GAUSS pg. 68

ANNEXES pg. 87 THE SUPERHERO

The Superhero

…THE MOST IMPORTANT CHARACTER IN THIS COMIC: YOU!

6 THE CHARACTERS

55

Riddle

Chance Binomial

GAUSS

Graph

7 CHAPTER 1

PIERRE DE FERMAT

French mathematician (1601-1665) His knowledge earned him the nickname “prince of amateurs” He was one of the pioneers of the theory of probabilities. Chapter 1 – PIERRE DE FERMAT

THE GRASSHOPPER’S GONNA WIN!

WHAT A RACE!

I’M GOING TO MAKE A HYPOTHE- OF COURSE THE GRASSHOPPER IS LUCKILY FOR THE SIS: IT’S MOST LIKELY THAT THE GOING TO WIN, HE’S FASTER. SNAIL, THE SNAIL WILL LOSE. GRASSHOPPER STOPS BETWEEN JUMPS.

IT’S AS CLEAR AS DAY! THE GIVE ME A PENCIL AND SOME WAIT A MINUTE. THE GRASSHOPPER’S GONNA WIN. PAPER AND I WILL KEEP TRACK GRASSHOPPER ADVANCES BY OF HOW THE RACE IS GOING. JUMPING AND LEAVES TRACKS…

…IN CERTAIN PLACES ALONG THE WAY. THIS WAY WE CAN TRACE THE MEANWHILE, THE SNAIL ADVANCES PATH OF THE COMPETITORS. LEAVING A CONTINUOUS TRAIL BEHIND HIM.

9 Chapter 1 – PIERRE DE FERMAT

YEAH, BUT THERE’S ONE PROBLEM. WE’LL CALL HIM HA, HA, HA! …WE’LL CALL “MR. SEE HOW THE GRASSHOPPER “MR. DISCRETE”! THEN THE SNAIL… CONTINUOUS”. THINKS BEFORE HE LEAPS…

THEY ARE CALLED EVENTS THAT REMINDS ME THAT PERHAPS THE STUDY OF BECAUSE THEY HAPPEN, NOT THEY STUDY EVENTS WITH STATISTICS BEGAN WITH BECAUSE THEY ARE DISAS- “DISCRETE” VARIABLES AND THE STUDY OF GAMES… TROUS, BLOCKHEAD! “CONTINUOUS” VARIABLES.

WE COULD MAKE A LAB AND DO SOME EXPERIMENTS!

I DON’T KNOW IF THIS HAS GREAT! ANYTHING TO DO WITH IT COOL! BUT I HEARD THAT IT IS HARDER TO GUESS WHAT ONE PERSON WOULD WISH FOR THAN WHAT A MILLION PEOPLE WOULD WISH FOR.

THAT JUST GAVE ME AN IDEA! LISTEN. I SAY WE PLAY A GAME… Experiments

EXERCISE WE FLIP A COIN IN THE AIR 8 TIMES AND WRITE DOWN THE RESULTS. WE DO IT THREE TIMES AND WRITE DOWN ALL THE TIMES THAT HEADS COMES UP.

THEN WE WILL FLIP A COIN 50 TIMES IN A ROW. I BET YOU THAT THE NUMBER OF TIMES THAT THE COIN COMES UP HEADS THIS TIME IS CLOSER TO 25 THAN IT WAS TO 4 WHEN WE FLIPPED IT THE OTHER TIMES.

SO WHAT YOUR SAYING IS THAT THE MORE WE REPEAT THE EXPERIMENT, THE MORE SURE WE WILL BE TO GET IT TO COME UP HEADS HALF THE TIME WE FLIP IT.

IF WE DID IT A MILLION TIMES, WE WOULD GET IT EVEN CLOSER TO BEING HALF OF THE TIMES. AND IF WE DID IT 10 MILLION TIMES, IT’D BE EVEN CLOSER. THEREFORE, THE PROBABILITY THE COIN WILL COME 1 UP HEADS OR TAILS IS CLOSER TO 2 THE MORE TIMES WE FLIP IT.

10 Chapter 1 – PIERRE DE FERMAT

Experiments

OK, BUT THAT’S EASY BECAUSE A COIN ONLY HAS TWO SIDES: HEAD AND TAILS, YOU WIN OR YOU LOSE. IF WE TRY THE EXPERIMENT OUT WITH A DIE, WE’LL SEE THAT THERE ARE 6 POSSIBILITIES. LET’S SEE HOW MANY TIMES 5 COMES UP. WE WILL ROLL THE DIE 90 TIMES AND WRITE DOWN THE RESULTS. LET’S SEE HOW IT TURNS OUT.

SO, IF WE HAVEN’T CHEATED AND THE DIE IS NOT RIGGED, WE CAN SAY THAT WE HAVE WON 15 TIMES AND HAVE LOST 75 TIMES.

15 1 THE PROBABILITY THAT WE ROLL A 5 IS 90 OR 6 , WHILE THE PROBABILITY OF NOT 75 5 ROLLING A 5 IS 90 OR 6 . SO WE CAN SEE THAT THE PROBABILITY OF NOT ROLLING A 5 IS EQUAL TO 1 MINUS THE PROBABILITY OF SUCCEEDING.

SO, THE PROBABILITY OF GUESSING THE OUTCOME OF A SOCCER MATCH BETWEEN TWO EVEN TEAMS WOULD BE 1 (SINCE THERE ARE ONLY THREE POSSIBLE OUTCOMES: 3 2 WIN, LOSS, OR TIE) AND THE PROBABILITY OF NOT GUESSING WOULD BE 3 . WHICH BRINGS US TO SEEING THAT GUESSING IS NOT ALWAYS EASY, JUST AS THE QUESTIONS IN CLASS AREN’T ALWAYS EASY, UNLESS I AM THE ONE ANSWERING.

WE’LL GET IT RIGHT WHAT IF YOU HAVE TO ROLL A 7 IF WE SAY THAT THE WHEN YOU ROLL A DIE? OUTCOME WILL BE “1”, “X”, OR “2”. THAT THEN I WOULD HAVE 0 CHANCE WAY, OF COURSE, OF SUCCEEDING. IN THAT CASE, THE CHANCES OF I WOULD CHOOSE TO BET THAT I SUCCEEDING ARE WOULDN’T ROLL A 7. GUARANTEED!

First, we must agree about how to record the data. We will mark them off with vertical lines until the number 4. The number 5 will be marked by diagonally crossing the previous four marks made. That way they will be divided in groups of 5 and it will be easy for us to count and add the ones from the last group which will be a maximum of 4. Let’s look at an example:

This count totals 33. Another way to do it is:

1 10 20 30

11 Chapter 1 - PIERRE DE FERMAT

Experiments

To save the results from the experiment, which could be useful to us on other researches, we will fill out the boxes below. These games are serious work!

COIN TOSS 8 TIMES Heads: _____ Tails: _____

Heads Tails (in red) (in blue)

COIN TOSS 50 TIMES

Heads (in red) Tails (in black)

Number of times heads: Number of times tails: 50 - _____ Number of times heads: _____

PROBABILITY:

Coin toss: Success=heads Failure=Tails

Chance of success in just one toss: 1 Total chances: PROBABILITY= 2 Roll of the die: Success=”Rolling a 5” Failure = "1 or 2 or 3 or 4 or 6"

Chance of success in just one roll: 1 Total chances (success + failure): PROBABILITY= 6 Roll of the die: Success = “Rolling a 3” Failure = "1 or 2 or 4 or 5 or 6"

Chance of success in just one roll: 1 Total chances (success + failure): PROBABILITY= 6 Roll of the die: Success = “Rolling a 3 or 5” Failure = "1 or 2 or 4 or 6"

Chance of success in just one roll: 2 Total chances (success + failure): PROBABILITY= 6 CHANCE OF ROLLING A 3 OR A 5 1 1 Chance of rolling a 3 + chance of rolling a 5 = + = 2 6 6 6

12 CHAPTER 2

THOMAS BAYES

(1702?-1761) English clergyman of the first half of the XVIII century, father of Bayes’ statistics. Chapter 2 - THOMAS BAYES

WOW! I’M SO LUCKY! I FOUND THREE COINS! WELL, IT DOESN’T LOOK LIKE A FORTUNE. BESIDES, THEY ARE FOREIGN COINS… LOOK HOW STRANGE THEY ARE. THEY’RE PROBABLY WORTHLESS.

NO, THAT’S NOT WHAT I MEAN. LOOK HOW THEY JUST COINCIDENTALLY YEAH. BUT I DON’T LANDED HEADS UP. THINK THAT IS SO STRANGE.

IT’S SOMETHING THAT WOULD BE WORTH IMPOSSIBLE! THAT TESTING OUT. FOR WELL… SINCE WE SAW THAT CAN’T BE RIGHT EXAMPLE, IF I FLIP A COIN THE PROBABILITY OF A COIN BECAUSE IT WOULD 3 TIMES, WHAT ARE THE COMING UP HEADS IS 1 , 2 BE A CHANCES OF IT COMING UP THEN WITH THREE COINS IT PROBABILITY OF HEADS 3 TIMES? WHAT WOULD BE: 3 ABOUT TWICE TAILS AND 2 AND THE ONCE HEADS? PROBABILITY OF AN 1 + 1 + 1 = 3 EVENT 2 2 2 2 CANNOT BE MORE THAN 1!

I THINK IT WOULD BE BETTER TO SEE IT ON A CHART. 1 PROBABILITY H = 2 1 PROBABILITY T = 2

TOTAL = 1

14 Chapter 2 - THOMAS BAYES

Experiments LIKE THIS.

OR LIKE THIS. H HHH H T HHT H H HTH T T HTT H THH H T THT H TTH T T TTT

15 Chapter 2 - THOMAS BAYES

Experiments

WE’LL LET THE SUPERHERO DO THE DIAGRAM FOR 4 TOSSES.

Total number: 8 chances.

Number of 3 heads = 1 Probability= 1 8 Number of 2 heads and 1 tail = 3 Probability= 3 8 Number of 1 head and 2 tails = 3 Probability= 3 8 Number of 3 tails = 1 Probability= 1 8

Total number: 16 chances

Number of HHHH = 1 Probability of HHHH = 16 4 1 Number of HHHT = 4 Probability of HHHT = = 16 4 Number of HHTT = Probability of HHTT = = 16 8

Number of HTTT = Probability of HTTT = = 16 Number of TTTT = Probability of TTTT =

Check

+ + + + =1

16 Chapter 2 - THOMAS BAYES

SO, AFTER DOING THESE EXPERIMENTS, WE CAN SAY THAT:

1. THE PROBABILITY OF A SURE EVENT IS 1.

2. THE PROBABILITY OF SUCCESS + THE PROBABILITY OF FAILURE ARE EQUAL TO 1.

3. THE PROBABILITY OF FAILURE = 1 – THE PROBABILITY OF SUCCESS.

4. THE PROBABILITY OF AN EVENT = 1 – THE PROBABILITY OF THE OPPOSITE.

OH! DID YOU GUYS GAUSS, WHAT ARE THE IN THE SNOW AND BEING KNOW THAT I LOST CHANCES THAT SO RARE? I SUPPOSE IN THAT THREE VERY VALUABLE THESE ARE HERS? CASE THE PROBABILITY OF COINS? THEY WERE THE EVENT IS SURE, THAT IS SOME REAL COLLECTOR’S TO SAY, 1. ITEMS THAT MY DAD FORGOT IN THE POCKET OF HIS JACKET!

LET’S GO BACK TO THE LODGE. IT’S WHAT LUCK! IT’S THEM! STARTING TO SNOW AND I WANT THANKS! YOU REALLY GOT TO SEE THE SOCCER SCORES. ME OUT OF A BIND!

17 Chapter 2 - THOMAS BAYES

I FOUND ONE IN MY DAD’S SO, YOU WANT TO FIND FANTASTIC! IT’S OUR CHANCE TO JACKET! THE DIE HAS SIX SIDES OUT THE SOCCER SCORES? DO ANOTHER EXPERIMENT! ANYBODY BUT TWO HAVE A “1”, TWO AN “X”, WELL IT LOOKS LIKE THE AND TWO A “2”. TV IS BROKEN AND YOU HAS A SOCCER POOL DIE? WILL HAVE TO IMAGINE THE SCORES…

Experiments Experiences

1 X 2

1 X 2 1 X 2 1 X 2

Deliberations

According to the current standings

3 if they win 3 1 0 1 if they tie 0 if they lose 3 1 0 3 1 0 3 1 0 6 4 34 2 1 3 1 0

According to what we made up

1 if they win 1 0 -1 0 if they tie -1 if they lose 1 0 -1 1 0 -1 1 0 -1 2 1 0 1 0 -1 0-1-2

WE’LL LET OUR SUPERHERO TAKE A HMM… THERE MUST BE A HEY! NOW CRACK AT FINDING OUT THE 15 DIF- FORMULA. ONE THAT IS IT’S OUR FERENT OUTCOMES, ALTHOUGH I FASTER AND NOT SO MUCH TURN… THINK HE’LL NEED A LOT OF PAPER… WORK… I HOPE GAUSS FIGURES ONE OUT. OK?

18 Chapter 2 - THOMAS BAYES

EUREKA! I’VE GOT IT! IF I WANT TO FIND OUT THE OUTCOME OF JUST ONE MATCH, THERE ARE 3 POSSIBILITIES. IF I WANTED TO FIND OUT TWO MATCHES… IT WOULD BE A TOTAL OF 9. GRAPH WILL MAKE US A CHART FOR 3 MATCHES. LET’S SEE WHAT THE RESULT IS!

THAT MAKES FOR A TOTAL OF 27 POSSIBILITIES WHICH WOULD BE: FOR ONE MATCH: 3 POSSIBILITIES = 31 FOR 2 MATCHES: 9 POSSIBILITIES = 3 X 3 = 32 FOR 3 MATCHES: 27 POSSIBILITIES = 3 X 3 X 3 = 33 HYPOTHETICALLY FOR 15 MATCHES THE NUMBER OF POSSIBILITIES WOULD BE 3 X 3 X 3 … (15 TIMES) = 315

OUR SUPERHERO WOULD HAVE TO DRAW OUT ALL THE DIAGRAMS IN ORDER TO KNOW HOW THE SOCCER POOL WOULD COME OUT IF WE WANTED TO FIND OUT 15 MATCHES:

5 FOR SURE, 6 WITH A DOUBLE OUTCOME, AND 4 WITH A TRIPLE OUTCOME…

THAT WOULD MAKE THE ANSWER:

1 x 1 x 1 x 1 x 1 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 = 15 x 26 x 34

EVEN SO, THE ONE WHO GUESSES IT WOULD WHILE YOU WERE ALL TALKING, I FIXED HAVE TO BE A LUCKY THE ANTENNA. IT WAS BURIED UNDER A DOG. BUNCH OF SNOW! YOU’D BETTER WRITE DOWN THE OUTCOME OF THE MATCHES BEFORE IT STARTS SNOWING AGAIN!

19 CHAPTER 3

BLAISE PASCAL French scientist (1623 – 1662) Probably one of the most important pioneers of the theory of probabilities and the study of the combinatories. His scientific and writing relationship with Fermat is fascinating. Chapter 3 - BLAISE PASCAL

I AM STILL SORE FROM THE THERE’S NO BASKETBALL HOW IS IT GOING BINOMIAL? WEEKEND… SO, I AM NOT GAME TODAY. THEY ARE GOING TO PLAY BASKET BALL GOING TO USE THE COURT THIS BREAK. FOR FIGURE SKATING.

LOOK, HERE COME OUR FRIENDS. MAYBE YOU KNOW, MAYBE WE COULD THEY COULD HELP US. CONTINUE WITH OUR DICE EXPERIMENTS… I ONLY HAVE ONE DIE BUT WE COULD TRY.

Experiments

NOW TRY ROLLING IT TWICE AND RIGHT DOWN THE RESULTS TWO AT A TIME. LET’S SEE WHAT HAPPENS… DID YOU KNOW THAT I SAW THAT THIS DIAGRAM IS CALLED “SAMPLE SPACE”? SO FROM NOW ON WE WILL CALL IT THAT ALSO. IT’S MORE TECHNICAL THAT WAY.

21 Chapter 3 - BLAISE PASCAL

HI GUYS! HOW INTERESTING! WHAT IF WE MAKE IT MORE COMPLICATED? WHAT WOULD HAPPEN IF ONLY THE SUM OF THE TWO DICE ROLLED WAS THE ONLY VALID ANSWER?

Experiments

I DON’T LIKE THIS. YOU’RE RIGHT. IF WE PLAY “ADD UP TO 7” SOMETHING’S WEIRD HERE. AND WE LET OUR SUPERHERO PLAY “ADD UP TO 12”, WE’LL BLOW HIM OUT OF THE WATER.

TOTAL NUMBER OF POSSIBILITIES: 6 x 6 = 36

1 CHANCE OF ADDING UP TO “2” ...... = WOW! IT’S TRUE THAT 36 THERE ARE MORE “3” ...... = 2 CHANCES THAT THE 36 “ “ “ “ “ 3 SUM OF THE DICE WILL “4” ...... = 36 COME UP 7 THAN 11 OR “ “ “ “ “ “5” ...... = 4 12. LET’S WORK IT “ “ “ “ “ 36 OUT. LOOKING AT THE “6” ...... = 5 DIAGRAM WILL BE 36 “ “ “ “ “ 6 EASIER FOR US. “7” ...... = “ “ “ “ “ 36 “8” ...... = 5 “ “ “ “ “ 36 “9” ...... = 4 “ “ “ “ “ 36 “10” ...... = 3 “ “ “ “ “ 36 “11” ...... = 2 “ “ “ “ “ 36 “12” ...... = 1 “ “ “ “ “ 36

22 Chapter 3 - BLAISE PASCAL

YOU SEE. I THOUGHT GAUSS, COULDN’T YOU MAKE THERE WAS UP SOMETHING THAT ALL OF I MIGHT. IF WE ASSIGN A SOMETHING WEIRD ABOUT US COULD PLAY IN WHICH WE DIFFERENT PRIZE TO EACH SUM, IT. WE HAVE 6 CHANCES COULD CHOOSE THE SUM OF I THINK THAT WE COULD OF WINNING AND OUR THE DICE WE WANTED AND BALANCE THE GAME OUT. BUT, SUPERHERO JUST 1. WINNING WOULD ONLY DEPEND ON LUCK AND NOT ON I WILL NEED GRAPH’S HELP IN KNOWING HOW TO DO A ORDER TO DRAW OUT THE RIGHT BUNCH OF MATH? PRICE TABLES.

GRAPH! CAN YOU COME OVER HERE AND HELP US?

OF COURSE! I’M ON MY WAY!

Experiments

HERE YOU CAN SEE THAT THE PROBABILITIES ARE PROPORTIONAL TO 1, 2, 3, 4, 5, AND 6. RIGHT? WELL, THE PRIZES WOULD HAVE TO BE JUST THE OPPOSITE.

LET’S GET THIS STRAIGHT. IF I CHOOSE THE SUM “7”, I HAVE 6 TIMES BETTER CHANCE OF ROLLING IT THAN SOMEONE WHO CHOOSES THE SUM “12”. SO THEN, MY PRIZE WOULD HAVE TO BE INVERSELY PROPOR- INCREDIBLE! I DON’T TIONAL TO HIS. IT’S KNOW HOW YOU DO IT? FAIR AND LOGICAL! WHERE DID YOU GET THAT COOL CHART?

YOU LIKE IT? HERE, IT’S YOURS… EXACTLY.

23 Chapter 3 - BLAISE PASCAL

Experiments LET’S SEE IF I’VE GOT THIS STRAIGHT… I WANT THE GAME TO DEPEND ONLY UPON THAT DOESN’T COINCIDE CHANCE, LIKE MY WITH YOUR TABLE! NAME… SO I WILL MAKE A LIST: HOLD ON! THE RESULTING PRIZES WERE:

1 36 Probability of rolling “2” =36 Prize = 1 2 36 Probability of rolling “3” =36 Prize = 2 3 36 Probability of rolling “4” =36 Prize = 3 4 36 Probability of rolling “5” =36 Prize = 4 5 36 Probability of rolling “6” =36 Prize = 5 6 36 Probability of rolling “7” =36 Prize = 6 5 36 Probability of rolling “8” =36 Prize = 5 4 36 Probability of rolling “9” =36 Prize = 4 3 36 Probability of rolling “10” = 36 Prize = 3 2 36 Probability of rolling “11” = 36 Prize = 2 1 36 Probability of rolling “12” = 36 Prize = 1

...

36 YEAH, LIKE THAT 5 … THAT MESSES THE WHOLE THING UP! 36 36 36 36 36 36 1 , 2 , 3 , 4 , 5 , 6 WOULD WORK FINE FOR PLAYING THE GAME FAIRLY. THAT IS WHY WE HAVE TO FIND SOME FRACTIONS THAT GIVE US WHOLE NUMBERS. BUT, I THINK THEY ARE HARD NUMBERS IF WE LEFT AS IT IS, WE WOULD GET: 36, 18, 12, 9, 36 , AND 6. BUT, IF WE MULTIPLY 5 TO WORK WITH SINCE THE RESULTS THEM ALL BY 5, THEY WOULD ALL BE WHOLE NUMBERS: 180, 90, 60, 45, 36, AND 30. WOULD BE NUMBERS WITH DECIMALS… BUT IT WOULDN’T BE THE SAME RESULTS…

24 Chapter 3 - BLAISE PASCAL

AS YOU CAN SEE, THE FIGURES ARE CORRECT AS LONG AS WE MAINTAIN THE SAME PROPORTIONS. OUR SUPERHERO WILL CHECK IT FOR US. PATIENCE MY FRIENDS… SO THAT WE CAN PLAY WITH SMALLER NUMBERS, I HAVE DIVIDED THEM BY THREE. WHAT DO YA THINK? THE RESULTS WOULD WE’LL ALL PLAY A FEW GAMES. FIRST WE’LL BE: 60, 30, 20, 15, 12, AND 10. PLAY WITHOUT EVENING OUT THE PRIZES.

ALRIGHT! LET’S CONTINUE!

OF COURSE! BUT SO THAT WE CAN SEE HOLD ON, BECAUSE AFTER WE WILL TRY THE THE RESULTS OF OUR EXPERIMENT, I GAME WE MADE UP AND WILL USE THE PRIZES ON PROPOSE WE ROLL THE DICE 50 TIMES THE TABLE GAUSS INVENTED. THEN WE WILL SEE FOR EACH, BOTH WAYS. WHO IS THE LUCKIEST!

I CHOOSE THE SUM “12”. THAT WAY I WILL SHOW YOU THAT CHOOSING THIS NUMBER WILL WHAT ARE YOU DOING? MAKE YOU LIKELY TO LOSE NOT SO HIGH! ALMOST EVERY GAME.

25 CHAPTER 4

ADOLPHE QUÉTELET Belgian statistician and astronomer (1796 – 1874). Worthy to be mentioned for his discovery of normal distribution. Chapter 4 - ADOLPHE QUÉTELET

YOU KNOW, I’VE BEEN THINKING ANSWER THIS. WHICH IS GREATER? ABOUT FOUR INTERESTING PROBLEMS A HALF METER SQUARED OR ONE HALF THAT COULD BE USEFUL TO US IN OF A SQUARE METER? OUR RESEARCH.

WE’LL HAVE TO PAY CARE- FUL ATTENTION TO THE CONCEPTS AND THINK THEM OVER WELL SO THAT WE DON’T MESS UP.

THEY’RE THE SAME. I DOUBT IT. LET’S DRAW IT OUT AND WE CAN SEE IT BETTER.

1 Half of 1 sq. m. m. squared 2

OH! NOW I SEE IT. I WELL NOW, YOU HAVE TO PAY BELIEVE IT. HALF OF A CAREFUL ATTENTION TO THE METER SQUARED IS DOUBLE CONCEPTS… SINCE THE PROBLEMS A HALF METER SQUARED. ARE DIFFERENT…

…THE OTHER DAY I WAS WALKING DOWN THE STREET WHEN I WAS PRESENTED WITH THE FOLLOWING PROBLEM: THE PRODUCT OF THE AH! I FORGOT! THEY ALSO TOLD AGES OF 4 SIBLINGS IS 36, AND THE SUM OF THEM IS A NUMBER ON ME THAT THE OLDEST SISTER IS THE OTHER SIDE OF THE STREET. WHAT ARE THEIR AGES? GETTING THROUGH HER PRIMARY EDUCATION COURSES JUST FINE.

HM! ONE PIECE OF INFORMATION IS MISSING.

27 Chapter 4 - ADOLPHE QUÉTELET

WE’LL JUST DO WHAT WE DID IN THE OTHER GAMES. WE HAVE TO DEFINE THEM WELL, BECAUSE IF WE DON’T, WE COULD COME UP WITH SOME WRONG ANSWERS.

IT’S BETTER THAT WAY. DO YOU GUYS SEE HOW IMPORTANT IT IS TO DEFINE THE DATA IN ORDER TO SOLVE A PROBLEM?

IMAGINE FOR EXAMPLE THE STANDINGS OF THE SOCCER TEAMS. WE ALL KNOW THAT A WIN REPRESENTS 3 POINTS, A TIE 1, AND A LOSS 0. WELL, LET’S MAKE A SIMILAR LEADER BOARD WITH THE SAME NUMBER OF WINS, TIES, AND LOSSES. THE ONLY DIFFERENCE WOULD BE THAT WINS WOULD BE WORTH 1 POINT, TIES 0, AND LOSSES -1. THE RESULT WOULD BE:

Experiments J G E P GF GC P G P TOTAL

1.Deportivo 23 13 4 6 42 30 43 13 -6 7 2.Saragossa 23 10 9 4 40 24 39 10 -4 6 3.Barcelona 23 11 5 7 43 29 38 11 -7 5 4.Celta 23 11 5 10 31 29 35 11 -10 1 5.Alabés 23 10 5 8 26 25 35 10 -8 2 6.Ath.Bilbao 23 9 8 6 34 34 35 9 -6 3 7.València 23 9 6 8 32 25 33 9 -8 1 8.Reial Madrid 22 8 9 5 38 37 33 8 -5 3 9.Rayo Vallecano 23 10 3 10 32 32 33 10 -10 0 10.REIAL MALLORCA 23 9 5 9 31 31 32 9 -9 0 11.Numància 23 8 7 8 32 36 31 8 -8 0 12.Màlaga 23 7 8 8 33 32 29 7 -8 -1 13.At.Madrid 23 8 5 10 36 37 29 8 -10 -2 14.Espanyol 23 7 7 9 33 34 28 7 -9 -2 15.Valladolid 22 7 7 8 22 25 28 7 -8 -1 16.Betis 23 8 3 12 21 37 27 8 -12 -4 17.Racing 23 6 8 9 35 34 26 6 -9 -3 18.Reial Societat 23 5 10 8 25 29 25 5 -8 -3 19.Reial Oviedo 23 6 7 10 24 38 25 6 -10 -4 20.Sevilla 23 4 8 11 24 36 20 4 -11 -7

CURRENT STANDINGS PROPOSAL

28 Chapter 4 - ADOLPHE QUÉTELET

WELL IT DOES YOU CAN SEE HOW OUR SYSTEM WOULD BE CHANGE USING THE THE ORDER OF THE ARGUABLE BUT VALID IF WE SAME DATA… STANDINGS HAS CHANGED COULD WORK OUT A FEW DUE TO THE CHANGE IN LITTLE PROBLEMS LIKE HOW POINTS AWARDED! TO ORDER THOSE WHO ARE TIED…

BUT AS A MATHEMATICAL THAT’S EASY! THE ONES WHO ARE TIED OK, EVEN THOUGH I DON’T SYSTEM IT WOULD BE WILL BE ORDERED BY A DRAWING. THINK THEY’D LIKE THAT CORRECT. IT IS A SPECIFIC VERY MUCH. WAY OF ORDERING THEM WITH ONLY ONE INTERPRETATION.

SO, TO EACH ONE WHO IS TIED IN THE STANDINGS WE WOULD GIVE HM! I THINK WE ARE THEM A CORRELATIVE NUMBER AND TRYING TO CREATE A PUT IT INTO THE DRUM AS MANY TABLE OF RANDOM NUMBERED BALLS NECESSARY AND NUMBERS… WE COULD DO A DRAWING.

HEY! OUR SUPERHERO HAS WE’LL LET OUR SUPERHERO ALREADY WORKED THE CHECK IT WITH THE PROBLEM OF THE SIBLINGS OUT STANDINGS FROM LAST AND I THINK IT COINCIDES WEEK. WITH MY ANSWER…

WE WOULD HAVE TO KEEP THAT IN MIND WHEN TAKING ABOUT STATISTICS BECAUSE, IN USING THE SAME DATA, THE RESULTS WOULD VARY ACCORDING TO THE RULES WE APPLY.

29 Chapter 4 - ADOLPHE QUÉTELET

WHAT IS THE ANSWER? THE PRODUCT MUST BE 36. THE SUM MUST BE THE AGES ARE 9, 2, 2, 1. AN EVEN NUMBER SINCE THEY ALREADY TOLD YOU IT ON THE SIDE OF THE STREET WITH THE ODD NUMBERS. THERE IS SISTER THAT IS OLDER THAN THE OTHER THREE AND SHE MUST BE EITHER 8, 9, 10, OR 11 YEARS OLD BECAUSE SHE IS IN ELEMENTARY SCHOOL AND HAS ALREADY STUDIED A FEW YEARS. HOW DID YOU GET THAT?

BOY THERE SURE WAS A LOT OF Various Even DATA… AND I THOUGHT THAT LET’S LOOK AT THE TOTAL Ages of the MUCH OF THE DATA DIDN’T HAVE siblings elementary Sum POSSIBILITIES AND APPLY school years ANYTHING TO DO WITH THE PROBLEM! THE PREMISES OF THE PROBLEM TO THEM. (DID 36 1 1 1 YOU GUYS NOTICE WHAT “BIG WORDS” I USE?) I 18 2 1 1 THINK I HAVE SOLVED THE PROBLEM RATHER WELL. 12 3 1 1 WHAT A GREAT LITTLE INVENTION THIS PAPER 9 4 1 1 x ALONG WITH THINKING, 9 2 2 1 x x THE DESIRE TO FIND OUT THE ANSWER, AND CARE- 6 6 1 1 FUL AND DETAILED STUDY! 6 3 2 1 4 3 3 1 3 3 2 2

THAT’S EASY. JUST SEE IF THEIR NAME IS MASCULINE OR FEMININE… HERE I HAVE ANOTHER FRIGHTFUL PROBLEM: “A FAMILY HAS 4 KIDS: PRAXEDES, 10 YEARS OLD, AMOR, 8 YEARS OLD, MONSERRAT, 5 YEARS OLD, AND REYES, 2 YEARS OLD. WHICH ARE BOYS AND WHICH ARE GIRLS?

THAT JUST GOES TO SHOW THAT THE PARENTS DECIDED ON A SINGLE THAT WAS WHAT I THOUGHT THEN WE CAN NAME AND THEN LOOKED AT THE AT FIRST. BUT IT TURNS OUT ONLY GET THE ULTRASOUND RESULTS LATER. THAT THAT THE FOUR NAMES ARE ANSWER BY USING WAY, THEY CHOSE A NAME THAT USED FOR BOTH BOYS AND PROBABILITIES. WOULD WORK FOR BOTH BOY AND GIRL. GIRLS.

30 Chapter 4 - ADOLPHE QUÉTELET

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THE FUNNIEST THING OF ALL IS THAT I TOLD IT TO THE MATH PROFESSOR AND HE TOLD ME THAT IT WAS JUST WHAT HE WAS EXPLAINING RIGHT NOW, BINOMIAL POWERS, WOULD WORK TO SOLVE THE PROBLEM.

LET’S TRY TO SOLVE THE PROBLEM JUST AS WE DID IN THE PREVIOUS EXPERIMENTS AND THEN WE WILL SEE THE POWER OF A IT’S ALREADY HARD ENOUGH TO PUT UP BINOMIAL. LET’S SEE, IF 55 GETS A GOOD WITH BINOMIAL AT TIMES. I DON’T WANT GRADE, THEY’LL LET HER HANG OUT WITH TO THINK WHAT IT WOULD BE LIKE TO PUT US MORE. THAT’S, IF SHE WANTS TO UP WITH HIM TO THE THIRD POWER! OF COURSE. THAT’S CUBED.

LET’S MAKE A TABLE. IN STATISTICS, TABLES ARE JUST AS ESSENTIAL AS IN A CARPENTER’S SHOP.

POSSIBLE CASES

PRAXEDES 10 ? M M M M M M M M F F F F F F F AMOR 8 ? M M M M F F F F M M M M F F F MONSERRAT 5 ? M M F F M M F F M M F F M F F REYES 2 ? M F M F M F M F M F M F F M F

HOW DID YOU KNOW THERE WERE 16 CASES? BECAUSE THE FIRST ONE HAS 2 POSSI- BILITIES THAT MUST BE COMBINED WITH THE SECOND WHICH MAKE 4. THESE FOUR HAVE TO BE COMBINED WITH THE 2 OF THE THIRD. SO WE HAVE 8 TO COMBINE WITH THE 2 OF THE FOURTH AND THAT MAKES 16.

2 X 2 X 2 X 2 = 24 = 16

REMEMBER WHEN WE DID THE EXPERIMENTS WITH HEADS AND TAILS?

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LOOKING AT THE TABLE WE WOULD HAVE:

A. 1 CHANCE IN 16 OF BEING 4 BOYS B. 4 CHANCES IN 16 OF BEING 3 BOYS AND 1 GIRL C. 6 CHANCES IN 16 OF BEING 2 BOYS AND 2 GIRLS D. 4 CHANCES IN 16 OF BEING 1 BOY AND 3 GIRLS E. 1 CHANCE IN 16 OF BEING 4 GIRLS

WE’LL JUST DO WHAT WE DID IN THE OTHER GAMES. WE HAVE TO DEFINE THEM WELL, BECAUSE IF WE DON’T, WE COULD COME UP WITH SOME WRONG ANSWERS.

SO, I WILL SAY THAT THERE ARE TWO BOYS AND TWO GIRLS, THAT WAY I HAVE MORE CHANCES OF GETTING IT RIGHT.

I’M THINKING THIS WOULD MAKE FOR SOME INTERESTING CHARTS.

6 / 16 5 / 16 4 / 16

3 / 16 2 / 16

1 / 16

32 Chapter 4 - ADOLPHE QUÉTELET

MAN! THIS CHART REMINDS ME OF ANOTHER ONE I SAW IN A BOOK: IT WAS CALLED THE BINOMIAL DISTRIBUTION CHART.

I THINK I HAVE A STARTING POINT FOR WHAT 55 SAID ABOUT BINOMIAL POWERS. I AM GOING TO SPEND SOME TIME ON IT, ALTHOUGH I THINK IT WOULD BE IMPORTANT TO TAKE A MORE IN-DEPTH LOOK AT IT AFTERWARDS WITH SOME CHARTS,

I KNOW WHO’S GOING TO HAVE TO WORK HARD ON THAT. I CAN SLEEP IN TOMORROW. SO, I AM GOING TO STAY UP TONIGHT STUDYING BINOMIAL POWERS AND HOW THEY RELATE TO PROBABILITY PROBLEMS. WHERE ARE YOU GOING? HEY! WAIT FOR US!

33 CHAPTER 5

JAKOB BERNOUILLI

French mathematician (1601-1665) Coming from a family of great scientists (Jakob, Daniel, Nicholas) Some of his highlighted works are the art of predicting (posthumous) and the law of large numbers. Chapter 5 - JAKOB BERNOIULLI

SURE A WEIRD COUNTRY TO TAKE WHO SUGGESTED THIS DESTINATION? A CLASS TRIP TO! HANOI! AT LAST! DIDN’T WE TALK ABOUT GOING TO AN AMUSEMENT PARK?

WHAT COUNTRY IS THIS? THIS IS DEFINITELY NOT THE HANOI OF ASIA.

IT’S HANOI FROM THE REGION OF TARTAGLIA IN THE COUNTRY OF AVERAGES.

IT’S THE LAND OF PROBABILITIES.

WHAT PLACES HOW EXCITING! LET’S GO THERE! HAVE YOU SEEN THE SOMETHING LIKE THAT. WILL WE SEE? IS IT AN AMUSEMENT PARK? TOWER OF HANOI? BUT I’M AFRAID IT IS RIGHT BEFORE OUR EYES.

35 Chapter 5 - JAKOB BERNOUILLI

BUT, IT’S NOT ONLY A MATHEMATICAL STRUCTURE BUT ALSO A GAME. HOW DOES IT WORK?

WELL. WE USED TO PLAY IT LAST YEAR. WE BUILT IT WITH THREE STAKES ON WHICH WE WERE TO PLACE DIFFERENT SIZED DISCS. THE OBJECT OF THE GAME IS TO MOVE THE DISCS FROM ONE STAKE TO ANY ONE OF THE OTHER STAKES FOLLOWING THE RULES BELOW:

1. YOU CAN ONLY MOVE ONE DISC AT A TIME. 2. A LARGER DISC CAN NEVER BE ON TOP OF A SMALLER DISC. SO AS YOU CAN SEE WITH ONLY THREE DISCS YOU NEED TO MAKE 7 MOVES.

THAT IS TO SAY, 23 - 1 = 8 - 1 = 7 5 MOVES WOULD NEED: 25 - 1 = 32 - 1 = 31 6 MOVES WOULD NEED: 26 - 1 = 64 - 1 = 63

GET TO THE POINT. WITH “N” DISCS, LET’S BUILD ONE AND PRACTICE WITH A FEW YOU WOULD NEED 2n - 1. DISCS. I THINK THAT IF WE INCREASE THE NUMBER OF DISCS BY TOO MANY WE’LL GROW OLD TRYING TO FINISH THE GAME.

HEY GAUSS! I BET YOU KNOW A LITTLE, I THINK… WE ALL HERE WE GO AGAIN… EVERYTHING ABOUT BINOMIALS NOW! KNOW THAT A BINOMIAL IS THE ALGEBRAIC SUM OF TWO MONOMIALS.

36 Chapter 5 - JAKOB BERNOUILLI

Experiments

TO PUT IT ANOTHER WAY, IT IS THE SUM OF TWO TERMS. THE MOST SIMPLE CASE WOULD BE (a + b). SINCE WE ARE DEALING WITH BINOMIAL POWERS, WE WOULD HAVE TO CALCULATE:

(a + b) X (a + b) = (a + b)2 (a + b) X (a + b) X (a + b) = (a + b)3

BUT MULTIPLYING IT IS SOMETHING WE ALREADY DID IN CLASS. WE COULD MAKE SOME DRAWINGS IN ORDER TO SEE WHAT IT LOOKS LIKE. DO YOU KNOW HOW TO DRAW IT GRAPHICALLY? (a + b)2 ab a b b b a b b2 b a

2 a a a b a a

b a b a b

IN THE FIRST ONE, I TOOK THE BOX OF THE SIDE (a + b). THAT WAY, THAT SIDE SQUARED IS = (a + b)2. THE RESULT IS: a2 + a X b+ aX b+ b2 = a2 + 2ab + b2 FOR DOING IT TO THE POWER OF 3 I DREW A HEXAHEDRON…

WHAT? A CUBE ON SIDE a + b, THAT BROKEN DOWN IS: 2 CUBES a3 and b3, 3 PARALLELEPIPEDS a x a x b, AND ANOTHER 3 a x b x b.

BUT SINCE 3 DIMENSIONS ARE HARDER TO DRAW, I ALSO DID IT ON A FLAT CHART LIKE THIS:

SLOW DOWN, I’M STILL PROCESSING I TOOK A RECTANGLE SO THAT ONE SIDE “PARALLELEPIPEDS”… WAS a + b, AND SO THAT ANOTHER (a + b)2, WHOSE RESULT WE SAW BEFORE…

37 Chapter 5 - JAKOB BERNOUILLI

Experiments

(a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3a2b + b3 ba a2 2ab b2

2 3 2 a b a a a a 2a b ab2

2 2 3 b a b b b a x b 2ab2 b

LET ME DO THOSE SAME DRAWINGS ON GRAPHS WITH THE FOLLOWING DIMENSIONS: (a + b) = (3 + 2) = 5. THAT WAY WE CAN COUNT, UNIT BY UNIT.

2 2 =12 ab a =9 2ab b =4 3 2 2 a 96 a=3 a 2a b ab 2 2 3 b 64 b=2 a b 2ab b a3 =27 a2 =9 3a2b= 54 2ab= 12 3ab2 =36 b2 =4 b3 =8 (2 + 3)2 = 52 = 5 x 5 = 25 (2 + 3)3 = 53 = 5 x 5 x 5 = 125

I SEE. THAT’S BETTER. THEN LET’S TRY AND DRAW (a + b)4, WHICH WOULD BE THE SQUARE OF A SIDE THAT IS (a + b)2.

38 Chapter 5 - JAKOB BERNOUILLI

Experiments a2 2ab b2

2 4 3 2 2 a a 2a b a b a4 + 4a3b + 6a2b2 + 4ab3 + b4

3 2 2 3 2ab 2a b 4a b 2ab

2 2 2 3 b a b 2ab b4

I THINK I DISCOVERED SOMETHING HERE… LET ME SEE… ACCORDING TO YOUR LAST DRAWING, THE RESULT OF (a + b)4 IS: (a4 + 4a3b + 6 a2b3 + b4), THAT COULD ALSO BE WRITTEN OUT LIKE THIS: (aaaa + 4aaab + 6aabb + 4abbb + bbbb)

YEAH, SO WHAT?

WAIT, YOU’LL SEE. IF YOU REMEMBER THE ANSWER TO THE FOUR GUYS AND GIRLS THERE WERE: 1 CHANCE FOR THE CASE MMMM (4 BOYS) 4 CHANCES FOR THE CASE MMMF (3 BOYS AND 1 GIRL) 6 CHANCES FOR THE CASE MMFF (2 BOYS AND 2 GIRLS) 4 CHANCES FOR THE CASE MFFF (1 BOY AND 3 GIRLS) 1 CHANCE FOR THE CASE FFFF (FOUR GIRLS) EUREKA! THE COEFFICIENTS COINCIDE! THE ANSWER GIVES ME THE BOY/GIRL, HEADS/TAILS, SUCCESS/FAILURE DISTRIBUTION…

39 Chapter 5 - JAKOB BERNOUILLI

BUT WHAT IF I WANT TO KNOW THE PROBABILITY THAT IN A GANG OF 9 KIDS, 5 WOULD BE I WAS HOPING YOU WOULD SAY THAT. GUYS AND 4 GIRLS? I WOULD SO, WHILE WE WERE LEARNING A HAVE TO MAKE A TABLE WITH GENERIC FORMULA FOR BINOMIAL ALL OF THE POSSIBILITIES POWERS, I FOUND A NUMERIC PYRAMID. WHICH WOULD TAKE FOREVER IT’S EASY TO BUILD AND WILL SOLVE OR I WOULD HAVE TO DRAW OUR PROBLEM: OUT A COMPLICATED GRAPH IN ORDER TO FIND OUT THE ANSWER TO (a + b)9.

1 11 121 1331 1 4641 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1

I AM SURE YOU GUYS ALREADY KNEW THAT! EXCEPT FOR THE ONES THAT ARE ALWAYS ADDED TO THE ENDS, YOU GET THE NUMBERS BY ADDING THE TWO FIGURES THAT ARE DIRECTLY ABOVE IT.

GAUSS, I THINK I CAN DRAW A FIGURES ARE NUMBERS. FEW USEFUL CONCLUSIONS:

THE PROBLEM WOULD BE TO FIND OUT IN A GROUP OF 9 THE PROBABILITY OF: HMMMMMM.

40 Chapter 5 - JAKOB BERNOUILLI

TO FIND OUT THE CHANCES, I WILL COMPARE IT TO WHAT WE DID WITH THE GROUP OF 4 AND NOTE THAT FOR 0 GIRLS WAS THE FIRST COEFFICIENT, FOR ONE, THE SECOND. WE ALSO SAW THAT THE GROUP OF 4 WERE IN A ROW WHOSE SECOND NUMBER WAS 4 (THE FIRST ONES ARE ALWAYS ONES). SINCE THE GROUP IS MADE UP OF 9, IT IS FOUND IN THE SECOND NUMBER OF THE ROW:

1 9 36 84 126 126 84 36 9 1

SINCE HERE I HAVE 4 GIRLS, I MUST FIND OUT THE FIFTH FIGURE, THAT IS TO SAY, 126 CHANCES. THEN I CALCULATE THE TOTAL CASES, FOLLOWING YOUR RULE… OOPS! FOR TWO THROWS IT WAS 22, FOR 3 IT WAS 23, FOR 4 IT WAS 24, 16. SO, FOR 9 IT WOULD BE 29…

55, DID YOU KNOW THAT YOU THAT MEANS THE LOOK LIKE YOU’RE PART OF LET MY CALCULATOR CHANCES WOULD THE BERNOUILLI FAMILY? AND I FIGURE IT OUT… BE 126 . YOU ALMOST DESCRIBED THE 512! 512 PRINCIPLE OF INDUCTION. BUT WE’LL TALK ABOUT THAT SOME OTHER DAY.

I HAVE TO SAY THAT IN ORDER TO CALCULATE THE ENTIRE NUMBER OF POSSIBILITIES, YOU CAN JUST ADD UP THE LINE YOU WORKED WITH:

1 + 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1= 512

WE’LL LET OUR SUPERHERO FIGURE OUT THE FOLLOWING PROBLEM: FIND THE PROBABILITY THAT IN A CLASS OF 12 STUDENTS, 3 OF THEM ARE BOYS AND THE REST GIRLS, ASSUMING THAT THE CHANCES OF BOYS AND GIRLS BEING IN THE SAME CLASS ARE THE SAME.

41 Chapter 5 - JAKOB BERNOUILLI

NOTE THAT WE EVEN THOUGH WE ASSUMED THAT WERE TRYING TO FIND THE THE CHANCES OF A GIRL OR BOY PROBABILITY OF WERE THE SAME, THERE WILL BE MMMMMFFFF. OUR FORMULA SOME CASES IN WHICH THEY WON’T. SHOULD LOOK LIKE THIS: THAT IS WHY WE WILL TAKE A (a5 + b4) 126 a5 b4 LOOK AT A GENERAL METHOD FOR SOLVING THE PROBLEM. LET’S DO THE FOLLOWING: 1 1 126 126 X ( )5 X ( )4 = 2 2 512 THE ANSWER.

THEN I AM GOING TO TRY ONE I’LL READ IT TO YOU. I AM WHERE THE CHANCES ARE NOT GOING TO DO IT WITH A VERY THE SAME. IMPRESSIVE, DEEP VOICE.

"FIND THE PROBABILITY OF ROLLING A DIE 9 TIMES AND ROLLING A 3 FIVE TIMES AND THE OTHER ROLLS ANY OTHER NUMBER”.

1 SO I GET: LET’S SEE… THE PROBABILITY OF ROLLING A 3 IS 1 5 1 625 5 6 126 X ( )5 X ( )4 = 126 X ( ) X ( ) = THE PROBABILITY OF NOT ROLLING A 3 IS 6 6 7771 1296 6 78750 = 0, 0078 10077696 1 5 ANSWER: 126 X ( )5 X ( )4 6 6

I’D NEVER BET ON THAT!

WE’VE JUST FINISHED ANOTHER GREAT EXPERIMENT. OK, BUT NOW LET’S GO BACK TO THE HOTEL. I NOW YOU GUYS SEE THE WANTED TO TAKE A DIVE IN THE POOL! MEANING OF OUR VISIT TO THIS LAND OF TOWERS AND PYRAMIDS.

42 CHAPTER 6

CHARLES DODGSON Known by his fans as Lewis Carroll, English mathematician (1832-1898), author of “Alice in Wonderland”, “Through the Looking Glass”. His stories are related with the games theory and in certain cases can be used in statistical applications. Chapter 6 - CHARLES DODGSON

IT WAS SURE WORTH GOING ON A LITTLE HIKE. WHAT A GREAT VIEW!

WE’RE SO LUCKY IT’S SUCH A CLEAR DAY! YOU CAN SEE ALMOST THE ENTIRE ISLAND, THE COAST, THE TOWNS, THE WOODS.

OF COURSE, IN FACT, I GOT ONE IN CLASS YESTERDAY OF THE BINOMIAL, DID YOU BRING A MAP ENTIRE BALEARIC ISLANDS WITH THE km2 OF THE CITIES AND SO THAT WE CAN FIND OUT WHAT TOWNS, AND THEIR POPULATION. THE TEACHER WANTS US TO WE’RE LOOKING AT? FIND OUT THE POPULATION DENSITY.

VERY INTERESTING. SO, YOU I THINK THE ANSWER IS EASY: IN THE HAVE THE DATA REGARDING SUMMERTIME THERE ARE A LOT OF PEOPLE AT THE THE AREA AND POPULATION. BEACH, BUT IN THE WINTER THERE ARE FEW.

THAT’S NOT BAD TO BEGIN WITH!

THAT’S MORE OR LESS WHAT IT’S ABOUT… BY THE WAY, I HAVE AN EXAMPLE THAT IS RELATED… YOU’LL LIKE THIS ONE RIDDLE, IT HAS TO DO WITH MUSIC. SO, I WAS TALKING TO MY FRIENDS WHO ARE ROCKERS. THEY HAD GOTTEN A GROUP TOGETHER AND NAMED IT “QUARTILE” AND RENTED OUT A PLACE TO PRACTICE. THEY COM- PLAINED THAT THE PLACE WAS TOO SMALL BECAUSE IT WAS 12 m2, WHILE ANOTHER GROUP HAD A PLACE THAT WAS 25 m2.

44 Chapter 6 - CHARLES DODGSON

THEY MUST BE RIGHT… THAT’S TRUE. THE OTHER GROUP IS CALLED “DECILE”, I KNOW, BUT SOME DATA AND OF COURSE, IT IS MADE IS MISSING. UP OF 10 MEMBERS.

IT’S MUCH CLEARER NOW. THE BAND MEMBERS OF “QUARTILE” ARE NOT RIGHT SINCE 12 THEY HAVE A PROPORTION OF PER MEMBER WHILE THE GROUP “DECILE” HAS A 4 25 PROPORTION OF PER MEMBER. 10

EXCELLENT! THAT’S IT. NOW WE CAN FIND OUT THE POPULATION DENSITY WHICH WOULD BE THE COEFFICIENT BETWEEN THE NUMBER OF INHABITANTS OF THE CITY DIVIDED BY THE AREA OR THE NUMBER OF km2.

YES. IT IS BETTER TO CALCULATE THE AREA IN km2, BECAUSE IN m2 IT WOULD BE TOO SMALL OF A MEASUREMENT FOR AN ENTIRE CITY.

LET’S SEE IF I VIRTUALLY SPEAKING, WE SHOULD CHOOSE A CITY, DRAW A RATHER 2 GOT IT RIGHT. LARGE MAP TO SCALE, DIVIDE IT INTO SQUARE PLOTS OF LAND 1 km EACH AND DISTRIBUTE THE PEOPLE SO THAT EACH PLOT HAS THE SAME AMOUNT OF PEOPLE.

THEN WHAT WOULD HAPPEN IF WE CAME UP WITH A NUMBER WITH DECIMALS…

45 Chapter 6 - CHARLES DODGSON

YEAH BUT WE’RE TALKING ABOUT A MATHEMATICAL MEASUREMENT, WITH A RATIO, A COEFFICIENT, A RATE, NOT A CARPENTER’S MEASUREMENT… I MEAN WE DON’T HAVE TO SAW ANYBODY IN HALF…

DUDE YOU SURE USE A LOT OF “25 CENT” WORDS!

DON’T YOU WORRY, THOSE WORDS ALL MEAN THE SAME THING BUT COME ACROSS WELL IF YOU DON’T WANT TO REPEAT YOURSELF.

IN OTHER WORDS, WE MUST TAKE INTO ACCOUNT THE NUMBER OF INHABITANTS OF MY TOWN (THE ENTIRE CITY) AND DIVIDE IT BY ITS AREA.

I’M THINKING THAT IT’S A GOOD THING WE HAVE TO CALCULATE THE POPULATION DENSITY OF THE BALEARIC IT GETS EVEN BETTER, RIDDLE! ISLANDS AND NOT HONG KONG. THE AREA IS JUST A FEW km2 AND WE WOULD HAVE TO STACK THE PEOPLE UP IN PYRAMIDS TO MAKE THEM FIT.

UMM…. WELL….

46 Chapter 6 - CHARLES DODGSON

OK, LET’S GET GOING. WE’VE GOT THE MAP AND WILL HAVE TO STICK THE CORRESPONDING STICKER OF ON EACH TOWN.

LET’S START CALCULATING:

THE POPULATION DENSITY OF THE BALEARIC ISLANDS:

796.483 = 158,896 PERSONS/km2 5.012,6

THE POPULATION DENSITY OF :

5.859 = ...... 83,20

THE POPULATION DENSITY OF :

8.444 = ...... 572,6

THE POPULATION DENSITY OF MALLORCA:

...... = ...... 3.640

THE POPULATION DENSITY OF :

...... = 96,466 PERSONS/km2 ......

47 Chapter 6 - CHARLES DODGSON City Inhabitants Area Population Density

Balearic Islands 796483 5012,60 158,896 Per./km2

Alaró 3834 45,70 Per./km2 10581 60,00 Per./km2 3542 89,80 Per./km2 8333 81,50 Per./km2 Artà 5936 139,80 Per./km2 503 18,10 Per./km2 Binissalem 5019 29,80 Per./km2 Búger 951 8,30 Per./km2 4338 84,70 Per./km2 Calvià 32587 145,00 Per./km2 2277 34,70 Per./km2 Campos 6944 149,70 Per./km2 6752 54,90 Per./km2 2210 13,70 Per./km2 849 15,40 Per./km2 Deià 625 15,20 Per./km2 275 139,40 Per./km2 3811 35,30 Per./km2 338 13,40 Per./km2 14600 169,80 Per./km2 580 19,50 Per./km2 Inca 21103 58,30 Per./km2 837 17,40 Per./km2 4529 12,10 Per./km2 Llubí 1893 34,90 Per./km2 21771 327,30 Per./km2 30177 260,30 Per./km2 936 19,90 Per./km2 1733 30,50 Per./km2 Marratxí 18084 54,20 Per./km2 Montuïri 2235 41,10 Per./km2 Muro 6028 58,60 Per./km2 Palma 319181 208,60 Per./km2 Petra 2571 69,90 Per./km2 Pollença 13450 151,70 Per./km2 4226 86,90 Per./km2 10064 48,60 Per./km2 1163 42,30 Per./km2 Chapter 6 - CHARLES DODGSON

Sencelles 1969 52,90 Per./km2 1662 38,50 Per./km2 Sant Llorenç 5594 82,10 Per./km2 Santa Eugènia 1114 20,30 Per./km2 7107 86,50 Per./km2 Santa Maria del Camí 4558 37,60 Per./km2 Santanyí 7974 124,90 Per./km2 Selva 2918 48,70 Per./km2 3240 39,10 Per./km2 2616 47,70 Per./km2 Sóller 11207 42,80 Per./km2 8065 42,60 Per./km2 1599 42,90 Per./km2 2249 24,00 Per./km2 Ariany 772 23,90 Per./km2

MAJORCA 637510 3640,80 Per./km2

Alaior 7046 109,90 Per./km2 Ciutadella 21785 186,30 Per./km2 3921 66,10 Per./km2 Maó 22358 117,20 Per./km2 2723 158,00 Per./km2 Sant Lluís 4106 34,80 Per./km2 6005 11,70 Per./km2 1126 32,00 Per./km2

MINORCA 69070 716,00 96,466 Per./km2

Formentera 5859 83,20 Per./km2

Eivissa 31582 11,10 Per./km2 S. Antoni 14849 126,80 Per./km2 Sant Josep 13364 159,40 Per./km2 S. Joan 3943 121,70 Per./km2 Santa Eularia 20306 153,60 Per./km2

IBIZA 84044 572,60 Per./km2 CHAPTER 7

WILLIAM SEALEY GOSSET British Statistician (1876-1937) Chemist for the Guinness factory with some extraordinary statistical work done on small samples, he is not known by his name but by his nickname of Student and that is how t-distribution came about. Chapter 7 - WILLIAM SEALEY GOSSET

TODAY WE ARE GOING TO VISIT THE BALEARIC INSTITUTE OF STATISTICS. I ARRANGED FOR US TO BE ABLE TO MEET THERE, THAT WAY WE WILL HAVE MATERIAL TO WORK WITH.

I’D LIKE TO SEE HOW THAT DON’T SHOW OFF TOO NOW WE’LL REALLY WORKS. I’M BEGINNING TO LIKE MUCH UNLESS YOU WANT HAVE SOME DATA FOR STATISTICS AND THEY WILL HAVE THEM TO PUT YOU IN CHARGE OUR EXPERIMENTS. TO GET TO KNOW ME THERE SO THAT OF DOING A RESEARCH. ONE DAY, WHEN I KNOW IT ALL, THEY CAN APPOINT ME DIRECTOR.

LOOK AT THIS CHART. IT EXPLAINS ALL THE POS- SIBILITIES OF THE VARIOUS TOSSES OF A COIN OR THE PROBABILITIES OF DIFFERENT SUCCESS- 1 FAILURE GAMES WITH A PROBABILITY OF . 2

51 Chapter 7 - WILLIAM SEALEY GOSSET

Experiments

1

1/2 1/2

1/4 3/4 1/4

1/8 3/8 3/8 1/8

1 4 6 4 1 16 16 16 16 16

WE CAN COPY IT AND COMPARE IT TO THE DIAGRAM I MADE.

H T

HH HT TT

HHH HHT HTT TTT

HHHH HHHT HHTT HTTT TTTT

52 Chapter 7 - WILLIAM SEALEY GOSSET

Experiments

THIS IS GETTING INTERESTING. LOOK AT THIS GRAPH. IT MUST HAVE SOMETHING TO DO WITH WHAT WE WERE WORKING ON, THEY CALL IT THE POPULATION PYRAMID.

100 - .... 95 - 99 90 - 94 85 - 89 80 - 84 75 - 79 70 - 74 65 - 69 60 - 64 55 - 59 50 - 54 45 - 49 40 - 44 35 - 39 30 - 34 25 - 29 20 - 24 15 - 19 10 - 14 5 -9 0 - 4

40.00030.000 20.000 10.000 10.000 20.000 30.000 40.000 50.000

53 Chapter 7 - WILLIAM SEALEY GOSSET

YES, THIS IS A SUBJECT THAT COMPLIMENTS OUR PREVIOUS STUDY ON THE POPULATION OF INHABITANTS…

WHY DID YOU SAY POPULATION OF I COULD HAVE SAID INHABITANTS? POPULATION OF PERSONS, BUT AS YOU MIGHT KNOW, IN STATISTICS POPULATION REFERS TO EVERYTHING WE STUDY, WHETHER THEY ARE PEOPLE, PLANTS, PRICES, EXPENDITURES, ETC…

ACTUALLY, THEY ARE CHARTS THAT REPRESENT TWO SIDES OF GET TO THE POINT GAUSS. PEOPLE ALIVE BY AGE, ON ONE SIDE ARE THE WOMEN AND THE WHAT IS A POPULATION OTHER SIDE THE MEN. IT’S EASY TO UNDERSTAND. PYRAMID?

DON’T SAY IT’S EASY. IT’S RATHER HARD. LOOK. EACH BAR INDICATES THE AMOUNT OF BUT SINCE WE HAVE PROGRESSED A LOT, PEOPLE ALIVE WITH THE AGE SHOWN IN THE MIDDLE WE GET IT. OF THE CHART: THE GUYS ON ONE SIDE AND THE GIRLS ON THE OTHER. IT’S CALLED A PYRAMID BECAUSE AS YOU GO UP IN YEARS, THE NUMBER BEGINS TO DECREASE.

I MADE ANOTHER ONE AND YOU WHY’S GUYS SHOULD BE CAREFUL… AND THAT? YOU GIRLS TOO…

54 Chapter 7 - WILLIAM SEALEY GOSSET

JUST LOOK AT THE PERIOD FROM 15 TO THAT’S BECAUSE OF CARS AND 25 YEARS OLD. MOTORCYCLES. I MEAN, WHAT THE PYRAMID PEOPLE SUFFER FROM TODAY: DECREASES MORE ACCIDENTS. THAN IT SHOULD.

YOU SEE HOW SOME GOOD WE SHOULD HAVE FUN, BUT UNDER THE DATA CAN HELP YOU CONDITION THAT WHEN WE GET TO BE 90 YEARS DRAW MANY RELIABLE OLD, WE CAN STILL BE PART OF THE STATISTICS CONCLUSIONS. IF WE WANT.

YES. BECAUSE YOU GUYS WILL ALL HAVE TO COME WHEN THEY GIVE ME THE NOBEL PRIZE.

55 Chapter 7 - WILLIAM SEALEY GOSSET

BRRRR….!

HA, HA, HA! HA, HA, HA!

YOU ALWAYS HAVE TO PAY ORDER IN THE COURT! I SO THAT YOU CAN ATTENTION TO THE GRAPHS, HAVE TO WARN YOU ABOUT CONFIRM THEM OBSERVE AT WHAT SCALE WITH NUMERICAL A PROBLEM AND I WILL DO THEY ARE DONE AND CHECK DATA. IT THE BEST WAY I CAN. ALL OF THE DETAILS. WHY DO YOU SAY THAT?

Experiments I’LL SHOW YOU AND EXAMPLE THAT YOU WILL FIND IN PUZZLE BOOKS THAT SHOW US WHY.

LOOK AT THESE TWO DIAGRAMS.

56 Chapter 7 - WILLIAM SEALEY GOSSET

Experiments THEY ARE MADE UP OF THE SAME SHAPES BUT POSITIONED IN DIFFERENT WAYS:

TWO TRIANGLES THAT ARE 16 X 6 BOXES TWO RECTANGULAR TRAPEZOIDS WHOSE BASES ARE 6 AND 10 AND WHOSE HEIGHT IS 10.

THEREFORE, THE AREAS OF THE SQUARE FIGURE AND THE RECTANGULAR FIGURE WOULD HAVE TO BE THE SAME.

LET’S SEE: SQUARE: 16 X 16 = 256 RECTANGLE: 26 X 10 = 260

THEY SHOULD HAVE THE SAME AREA.

BEFORE I EXPLAIN, LET’S DRAW NOT AGAIN…! SOMETHING:

DOESN’T IT LOOK LIKE SEGMENT “A” IS SHORTER THAN SEGMENT “B”? I SWEAR THEY ARE THE SAME AND YOU CAN CHECK IT YOURSELVES. LIKEWISE, IF WE MEASURE THE FIGURES FROM THE PREVIOUS DRAWING WE WILL FIND OUT THAT THE FIGURES DO NOT COINCIDE.

57 Chapter 7 - WILLIAM SEALEY GOSSET

Experiments

IN RED ARE THE 4 BOXES THAT ARE MISSING.

LET’S LET OUR SUPERHERO DRAW AND CUT OUT THE NOW WHAT? COULDN’T WE PLAY A SHAPES. THEN WE’LL SEE THAT GAME OF VOLLEYBALL? THEY DON’T COINCIDE. THAT WAY, WE WILL BE SURE TO ALWAYS STUDY THE GRAPHS ALONGSIDE THE NUMERICAL DATA.

OK, BUT ONLY IF I AM LET’S GO! NOT THE REFEREE.

58 CHAPTER 8

ETIENNE L. LASPEYRES AND HERMANN PAASCHE Creators of the “cost of living indexes” that are still used today. Chapter 8 - LASPEYRES AND PAASCHE

HEY! HERE’S A BOOK THAT TALKS ABOUT US! DON’T TELL ME WE ARE ALREADY FAMOUS.

LET’S SEE… WELL, IT’S TRUE. IT IS A STATISTIC ABOUT EDUCATION AND I AM SURE WE AND OUR CLASSMATES ARE INCLUDED IN THESE NUMBERS.

Statistical Education Balearic Islands:

School Year Preschool Index Elementary Index High School Index 88-89 19.957 1 96.772 1,1260 89-90 19.958 1,0001 95.596 1,1123 90-91 19.220 0,9631 92.481 1,0761 91-92 19.313 0,9677 89.024 1,0358 92-93 19.706 0,9874 85.944 1 4.432 1 93-94 20.123 1,0083 83.197 0,9680 8.683 1,9592 94-95 20.719 1,0382 80.008 0,9309 10.742 2,4237 95-96 22.063 1,1055 77.419 0,9008 13.609 3,0706 96-97 23.169 1,1609 64.165 0,7466 28.975 6,5377 97-98 23.982 1,2017 55.294 0,6434 38.872 8,7708 98-99 24.449 1,2251 55.600 0,6469 39.821 8,9849

WHEW! I WAS JUST IMAGINING ALL THE COMPLICATIONS OF BEING FAMOUS AND EVERYTHING. WE SURE SAVED OURSELVES A LOT OF GRIEF!

60 Chapter 8 - LASPEYRES AND PAASCHE

HMMM! I SEE THAT THESE ARE IT MUST BE THE THE NUMBERS OF STUDENTS PAGE NUMBER WE THAT STUDY IN PRE-SCHOOL, ARE ON. ELEMENTARY, AND HIGH SCHOOL EACH YEAR LIKE US. WHAT I DON’T UNDERSTAND IS WHAT INDEX MEANS.

NO, IT MUST BE A THAT’S RIGHT. THE INDEX IN YOU MEAN DIVIDING THE FIGURE STATISTICAL CONCEPT. STATISTICS IS A PERCENTAGE OF STUDENTS OF ONE YEAR FROM THAT IS DERIVED FROM A FIGURE ANOTHER THAT WE USE AS A THAT WE USE AS A BASE FIGURE. STARTING POINT.

WE SEE THAT WHAT WE HAVE I AM DOING SOME CALCULATIONS ON DISCOVERED UP UNTO THIS THE PAGE THAT WE FOUND AND POINT IS GOOD. IN PRESCHOOL I SEE THAT IN THE PRESCHOOL THE INDEX MEASURES THE COLUMN THE RESULTS ARE OBTAINED DIFFERENCES BETWEEN GROUPS, BY TAKING THE AMOUNT OF STUDENTS IN THE CLASS THAT YEAR WELL… BETWEEN SCHOOL AND DIVIDING IT BY THE NUMBER OF YEARS, BUT I ALREADY TOLD STUDENTS THERE WERE IN 89-90. YOU GUYS THAT YOU HAVE TO 19.958 DIVIDE BY A FIGURE THAT IS = 1,0001 19.957 USED AS A BASE FIGURE. 10.220 = 0,9631 19.957 23.169 = 1,1609 19.957 BUT I CAN’T FIGURE OUT THE ONES FROM ELEMENTARY SCHOOL.

ACTUALLY, IT WAS A VERY AHA! I GOT IT! THE ELEMENTARY SIGNIFICANT YEAR. LUCK COLUMN USED THE 92-93 SCHOOL REALLY VARIED. YEAR AS ITS BASE FIGURE…

YOU’RE RIGHT, THAT YEAR THINGS CHANGED AND SOME PLACES STARTED WITH THE NEW SYSTEM.

61 Chapter 8 - LASPEYRES AND PAASCHE

SO, IF WE TAKE A CAREFUL LOOK AT IT, THERE WAS A DECREASE IN PRESCHOOL STUDENTS DURING THE SCHOOL YEARS OF: AND THE HIGH 90-91, 91-92, 92-93. SCHOOL COLUMN ALSO USED THAT ON THE OTHER HAND, THERE WAS MORE YEAR AS A THAN A 22% INCREASE OF STUDENTS IN REFERENCE. 98-99 COMPARED TO 88-89.

SO, IF WE LOOK AT THE INDEXES FOR HIGH SCHOOL, THEY HAVEN’T INCREASED BY 22% BUT ARE 9 TIMES GREATER THAN WHAT THEY WERE IN 92.

PERFECT. I LIKE IT. FIRST THE BINOMIAL’S DISSERTATION WHO OF COURSE, AND DUE TO THIS KNEW HOW TO DISTINGUISH INCREASE AFTER STARTING THE FIRST WELL BETWEEN THE 22% SEMESTER OF COURSES IN THIS NEW SYSTEM, THERE WAS A DECREASE IN INCREASE AND THE 9 TIMES ELEMENTARY STUDENTS. IT’S BECAUSE INCREASE OF STUDENTS. 7th AND 8th GRADE.

SO WITH THE INDEXES WE CAN DRAW NOW THAT WE’VE FOUND OUT ABOUT CONCLUSIONS QUICKLY AND A LOT MORE THIS INDEX, WE CAN FIND OUT EASILY SINCE THE NUMBERS ARE THE INDEXES OF THE ITEMS IN OUR COMPARED TO ONE UNIT. SHOPPING CART, THE GROSS NATIONAL PRODUCT, AND PRACTICALLY ANYTHING.

62 Chapter 8 - LASPEYRES AND PAASCHE

I THINK THOSE ARE A BIT MORE COMPLICATED. BUT YOU GUYS CAN CONTINUE THINKING ABOUT THE MATERIAL WE WILL NEED FOR OUR EXPERIMENTS WHILE I GO AND ASK.

LOOK WHAT I’VE GOT!

IT LOOKS LIKE THEY’VE ALREADY TOLD YOU.

WELL, YOU SEE, THERE ARE MANY DIFFERENT TYPES OF INDEXES: SOME SIMPLE, LIKE THE ONE WE FOUND OUT ABOUT, AND OTHERS THAT ARE MORE COMPLICATED, LIKE THE CPI (CONSUMER PRICE INDEX). THEY DREW A CHART FOR ME AND WHEN WE LEARN MORE ABOUT IT, WE CAN COME AND ASK ABOUT THE WAYS OF DOING IT AND CORRESPONDING STATISTICS.

GIVE ME A MOMENT, I’M GOING WHILE THEY ARE MAKING THEIR TO TRY TO CLARIFY THESE CHARTS, I READ THAT THE FRENCH NOTES OF YOURS… KING LUDOVICO XV HAD AN ANNUAL INCOME OF 100 MILLION.

63 Chapter 8 - LASPEYRES AND PAASCHE

BUT TWO HUNDRED YEARS EARLIER,

¡¡¡ !!! LUDOVICO XII BROUGHT IN 8 MILLION.

¡¡¡ !!! WHO DO YOU THINK EARNED MORE?

¡¡¡ !!! ¡¡¡ ¡¡¡ !!! ¡¡¡

NOPE! BECAUSE A SCIENTIST OF THE THIS IS LIKE THE BLACK XVII CENTURY USED TO DO EXPERI- STALLION. MENTS LIKE US AND CALCULATE THINGS WITH A SPECIAL SHOPPING BASKET (WHAT WE WOULD CALL TODAY “CONSUMER PRODUCT INDEX) AND SHOWED THAT FROM THE TIME OF LUDOVICO XII TO THE TIME OF LUDOVICO XV THE VALUE OF MONEY WENT DOWN IN VALUE AT A 1 RATE OF 22 ; DO THE MATH AND YOU WILL SEE THAT LUDOVICO XII EARNED MORE.

MONEY, JEWELS, ETC…

MONEY, JEWELS, ETC…

LUDOVIC XII LUDOVIC XV

0,5 Euros 1,25 Euros 0,020 pence

0,70 pence 1 Euro

0,40 pence

64 OLD FOOD (PRICES IN POUNDS)MODERN FOOD (PRICES IN EUROS) Chapter 8 - LASPEYRES AND PAASCHE

IF YOU REALLY TAKE A CLOSE LOOK AT IT, IT’S IN EUROS. IT THAT IS WHY YOU HAVE TO LOOKS LIKE THE PRICES IN LIFE KEEP IN MIND THE VALUE HAVEN’T GONE UP OVER THE OF MONEY AS IN THE CASE LAST 60 YEARS. WITH LUDOVICO…

I HEARD THE STORY AND IT SEEMS PERFECT FOR WHAT WE ARE DEALING WITH. LOOK AT THE DRAWINGS THAT GRAPH DID FOR US.

LASPEYRES INDEX

NEW OLD PRICES AMOUNTS

OLD OLD AMOUNTS PRICES

OLD AMOUNTS X NEW PRICES IL= OLD AMOUNTS X OLD PRICES

65 Chapter 8 - LASPEYRES AND PAASCHE

PAASCHE INDEX

NEW NEW AMOUNTS PRICES

NEW AMOUNTS OLD PRICES

NEW AMOUNTS X NEW PRICES IP= NEW AMOUNTS X OLD PRICES

SO AS YOU CAN SEE SOME OTHER INDEXES BESIDES THE ONE WE FOUND OUT ABOUT EXIST OUT THERE. SOME YOU CAN GET BY USING THE AVERAGE OF OTHERS…

=

THAT WORD AVERAGE SOUNDS FAMILIAR BUT I DON’T REMEMBER WHAT IT MEANS…

66 Chapter 8 - LASPEYRES AND PAASCHE

I’LL WRITE IT DOWN FOR THE NEXT EXPERIMENT.

MOVING ON, THESE DRAWINGS ARE USEFUL TO US AS A GUIDE. WE’LL GO BACK TO THE INSTITUTE ANOTHER DAY AND THEY WILL GIVE US ALL THE DATA.

REMEMBER THAT WE HAVE A BIRTHDAY PARTY ON WEDNESDAY. TOMORROW YOU HAVE TO BRING THE MONEY FOR THE GIFT AND THE PARTY, EVERY- ONE EXCEPT FOR 55. EACH ONE SHOULD BRING OK. BRING IT IN TWO WHAT THEY CAN, IF YOU CAN’T, YOU’LL HAVE TO SEALED ENVELOPES CONTRIBUTE MORE THE NEXT TIME. AND WE CAN USE IT TO TALK ABOUT AVERAGES, MEDIANS, AND MODES.

WHAT ARE YOU GUYS UP TO?

SOME OTHERS ARE COMING TOO. TELL THEM TO DO THE SAME. SHHH, HERE COMES 55.

OH NOTHING! YOU’RE ALWAYS SO SHARP…

67 CHAPTER 9

ABRAHAM DE MOIVRE AND CARL FRIEDRICH GAUSS

Abraham De-Moivre and Carl Friedrich Gauss De Moivre (1667-1754). Important scientist in various fields of mathematics. He participated in significantly changing statistics by going from binomial to normal distribution. Among his works is The Doctrine of Luck in which he calculates probabilities. Gauss, a German mathematician and statistician (1777-1855), did some important work relating to normal distribution, the theory of error, scattering, squared minimums… Chapter 9 - DE MOIVRE AND GAUSS

HURRY UP, 55’LL BE HERE ANY MINUTE.

YES. THERE ARE 15 ENVELOPES FOR THE WHILE YOU GUYS ARE DECORATING GIFT. WE’LL CALL THEM “G” ENVELOPES THE BACKYARD, I’LL GO BUY THE AND THERE ARE 16 ENVELOPES TO BUY ALL FOOD AND DRINKS FOR THE PARTY. THE PARTY SUPPLIES. DO YOU GUYS HAVE THE ENVELOPES WITH THE MONEY?

WHAT? 16 ENVELOPES?

THEN LET’S COUNT THE CONTENTS OF THE ENVELOPES AND DIVIDE THEM UP INTO TWO PILES, ONE THE “G” PILE AND WELL, I BUMPED INTO THE OTHER THE “P” PILE. HURRY, BEFORE THEY GET HERE. 55 AND CHANCE. SINCE CHANCE GAVE ME HER ENVELOPES, 55 WANTED TO GIVE ME ONE TOO. BUT DON’T WORRY, SHE DIDN’T KNOW THAT THE PARTY AND GIFT WERE FOR HER.

69 Chapter 9 - DE MOIVRE AND GAUSS

Experiments

G... P.... 100 100 300 200 350 250 400 300 425 400 475 500

G... P.... 1 ENVELOPE WITH 100 1 ENVELOPE WITH 100 5 ENVELOPES WITH 300 5 ENVELOPES WITH 200 3 ENVELOPES WITH 350 4 ENVELOPES WITH 250 2 ENVELOPES WITH 400 5 ENVELOPES WITH 300 1 ENVELOPE WITH 425 1 ENVELOPE WITH 400 2 ENVELOPES WITH 475 1 ENVELOPE WITH 500

WELL! I SEE YOU GUYS STARTED EVERYONE INSIDE, THE EXPERIMENT WITHOUT US. THEY’RE COMING.

70 Chapter 9 - DE MOIVRE AND GAUSS

NO, UH, WE JUST MADE UP A COUPLE OF TABLES WITH SOME DATA THAT WE HAD AND GAUSS WAS WAITING FOR YOU UH! AS YOU ALL KNOW, TODAY WE ARE GOING TO FIND GUYS TO ARRIVE SO WE COULD START. OUT THE ARITHMETICAL AVERAGE, THE MEDIAN, AND THE MODE. SO, WE MADE UP THESE TWO TABLES: THE FIRST IS A SECRET TABLE AND THE SECOND CONTAINS THE DATA OF THE CONTRIBUTIONS MADE FOR A PARTY. EVERYONE HAS CONTRIBUTED WHAT THEY COULD AND HERE WE WILL FIND THE FOLLOWING CENTRAL STATISTICAL PARAMETERS.

IS IT AN END OF WHO KNOWS? UH, ANY- THE YEAR PARTY? WAY, LET’S CONTINUE. I DREW THE FOLLOWING DRAWING. WE COULD WORK WITH THAT.

CHART

LET’S BEGIN WITH THE AVERAGE. LET’S FIND THE ARITHMETICAL AVERAGE AND IT WILL HELP US FOR THE FOLLOWING EXAMPLE:

WELL, WE START OFF BY SAYING THAT WE ARE A GROUP ON GOOD TERMS. EACH PERSON BROUGHT WHAT HE COULD AND LET’S ASSUME THAT EACH OF US HAS CONTRIBUTED EQUAL PARTS. SO, WE MUST SPLIT UP THE TOTAL AMOUNT EVENLY SO THAT EVERYONE HAS THE SAME AMOUNT. TO DO SO, WE WILL ADD UP ALL OF THE AMOUNTS AND DIVIDE THEM BY THE TOTAL NUMBER OF PARTICIPANTS.

THAT’S SOUNDS EASY.

71 Chapter 9 - DE MOIVRE AND GAUSS

LET ME WORK THE MATH OUT WAIT A SECOND. WHAT 55 DID WAS FOR “G”: VERY GOOD BUT WE COULD DO IT 100 + 300 + 300 + 300 + 300 + ANOTHER WAY. WHAT DO YOU 300 + 350 + 350 + 350 + 400 + THINK ABOUT THIS: 400 + 425 + 475 + 475 + 500. AGH! SOMEBODY GIVE ME A CALCULATOR! 100 X 1 + 300 X 5 + 350 X 3 + 400 X 2 + 425 X 1 + 475 X 2 + 500 X 1, AND THEN WE DIVIDE BY A TOTAL OF 15.

I’VE GOT MY COMPUTER AND I THINK THAT WITH A SPREADSHEET IT WOULD BE MUCH EASIER. Mode Median Average

TOTALS: Average

THAT’S RIGHT. I’LL MAKE THE TABLE FOR “P” ON THE COMPUTER. HOLD ON A SECOND. I’LL SEE IF THE COMPUTER CAN MAKE ONE FOR US.

72 Chapter 9 - DE MOIVRE AND GAUSS

GRAPH

Totals:

Average

IT WORKS GREAT. SO THEN, THE ONE FOR “P” WOULD BE LIKE THIS.

GRAPH

Totals:

Average

73 Chapter 9 - DE MOIVRE AND GAUSS

IT’D BE LIKE THIS.

GRAPH

THE ARITHMETICAL AVERAGE IS THE MOST COMMON CENTRAL

AVERAGE AND SERVES BEST FOR CALCULATING THINGS, ALTHOUGH

AT TIMES IT IS A GOOD IDEA TO DO IT ANOTHER WAY SINCE YOU

CAN GET A BETTER LOOK AT IT OR SIMPLY BECAUSE IT IS NOT POS-

SIBLE TO CALCULATE THE AVERAGE. LIKE IN THIS EXAMPLE:

30 WITH A 100 50 WITH A 200 20 WITH A 500 10 WITH MORE THAN 1000

THE MATH GETS HARDER BECAUSE WE DON’T KNOW HOW MUCH WE SHOULD MULTIPLY BY, 10, 1000, 2000, OR…

74 Chapter 9 - DE MOIVRE AND GAUSS

AND I WAS DOING FINE IT’S ALRIGHT. WE WILL USE BEFORE. THE MEDIAN WHICH IS A VALUE IN WHICH 50% OF THE VALUES ARE BELOW AND 50% OF THE VALUES ARE ABOVE. IN THE CASE OF THE PROBLEM ON THE PREVIOUS PAGE, IT WOULD BE 200. BUT WE COULD SEE IT A LOT BETTER IF WE DREW IT ON GRAPHS “G” AND “P”.

Median

Average

Median Average

IN THE SECOND CASE, THE AVERAGE AND THE MEDIAN COINCIDE. THAT’S BECAUSE YOU MEAN IF WE CALCULATE THEY WERE SYMMETRICALLY DISTRIBUTED. EXACTLY RIGHT RIDDLE. SO WE BOTH OF THEM, WE’LL FIND OUT ARE GOING TO ENTER ANOTHER SOMETHING ABOUT THE CENTRAL MEASUREMENT DISTRIBUTION OF THE VALUES. CALLED THE MODE. IT’S REALLY VERY SIMPLE AND IT REPRE- SENTS THE MOST FREQUENT VALUE OR VALUES.

75 Chapter 9 - DE MOIVRE AND GAUSS

THEN THERE COULD BE EXACTLY. THERE IS ONLY ONE AVERAGE VARIOUS MODES… AND ONE MEDIAN, BUT MODES DEPEND ON THE CIRCUMSTANCE. LET’S TAKE ANOTHER LOOK AT THE CHARTS:

Median

MODE Average

Median Average

MODE MODE

IN THIS CASE, THE I KNEW IT. IN THE DIS- YES, THE WINTER MODE AVERAGE AND MEDIAN ARE TRIBUTION OF “P” THERE AND SUMMER MODE. AROUND SPRINGTIME. ARE TWO MODES.

76 Chapter 9 - DE MOIVRE AND GAUSS

SO WHAT YOU’ RE SAYING IS THAT WE KNOW YEAH. WE KNOW A LOT ALREADY THAT FOR THE SUPPOSED PARTY WE CONTRIBUTED BUT WE STILL HAVE MORE TO AN AVERAGE OF 1 POUND AND THAT THE MOST LEARN. FOR EXAMPLE, WHY IS THE FREQUENTLY AMOUNT CONTRIBUTED WAS 75 OR 1 PARTY BEING THROWN? ALTHOUGH POUND. EASY STUFF WHEN IT COMES TO SMALL I DO BELIEVE, IF STATISTICS ARE GROUPS LIKE OURS. IT WOULD HAVE BEEN RIGHT, THAT ALMOST ALL OF THE SOMETHING ELSE IF IT WERE A PARTY FOR 3000 REASONS FOR THROWING A PARTY BOYS AND 5000 GIRLS. HAVE THE PROBABILITY OF 1 OF BEING GOOD.

WELL, THAT DOESN’T MATTER. OF COURSE. IT IT’S JUST A SUPPOSED PARTY COULD BE A END OF WHEN IS IT WHICH IS A GOOD THING FOR THE YEAR PARTY, OR GOING TO TAKE OUR EXPERIMENTS. A TRIBUTE, OR EVEN PLACE? A BIRTHDAY PARTY. BUT IMAGINE THAT TODAY WERE WEDNESDAY.

OF COURSE. NOW I WOULD LIKE TO TACKLE TWO OTHER PROBLEMS AND THEN WE CAN FORGET ABOUT THE PARTY. EVEN IF THE AVERAGE IS USUALLY THE MOST USEFUL MEASUREMENT, WE SAW THAT IN SOME CASES, WE COULDN’T CALCU- LATE IT, AND ALSO THAT IT A DRAWBACK: IT IS REALLY AFFECTED BY THE TWO EXTREMES.

WHAT?

77 Chapter 9 - DE MOIVRE AND GAUSS

THINK ABOUT THIS EXAMPLE:

Median= 11 Average 33 = 3 =11 Mode: none Totals: Concentrated

Median= 11 Average 33 = 3 =11 Mode: none Totals: Scattered

Median Average The modes would be all the numbers since each one appears once. In this case, we would say that there is no mode.

Median Average Concentrated

SCATTERED

IN THIS CASE YOU CAN SEE THAT WHAT WOULD DISTINGUISH ONE DISTRIBUTION FROM ANOTHER WOULD BE THE PARAMETER OF DEVIATION WHICH WOULD MEASURE THE CONCENTRATION OR SCATTERING.

78 Chapter 9 - DE MOIVRE AND GAUSS

HERE’S ANOTHER CASE:

Mode Median

Average Average

Mode Median Average

Average

Mode Median Average

Average

IN THIS CASE YOU CAN SEE THAT IF ONE VALUE OF THE LINES FROM BEGINNING TO END (VERY SCATTERED LIKE VALUE 48) IS CHANGED WITH ONE OF THE MORE CONCENTRATED ONES, 9, THE AVERAGE WOULD VARY A LOT, BUT THE MEDIAN AND MODE WOULDN’T.

ALTHOUGH IF WE WERE TO MAKE THE CHANGE IN CENTRAL VALUES, AS WE DID IN THE LAST TWO, THE MEDIAN AND MODE WOULD STAY THE SAME (ALTHOUGH THEY COULD VARY) WHILE THE AVERAGE CHANGES A BIT.

THIS IS WHY WE TRY TO FIND SOME MEASUREMENTS THAT INDICATE IF THE DISTRIBUTION IS MORE OR LESS SCATTERED. STANDARD DEVIATION IS THE MOST COMMON MEASUREMENTS USED FOR THIS. WE’LL TAKE A LOOK AT IT IN THE EXERCISES THAT WE ARE PREPARING. ALL IN ALL, WE’LL DO A LITTLE EXPERIMENT WITH STANDARD DEVIATION ON THE OLD TWO HAMS PROBLEM.

79 Chapter 9 - DE MOIVRE AND GAUSS

THERE ARE TWO PEOPLE WHO WILL BE THE DINERS AND WE HAVE TWO HAMS.

YEAH. ONE WON’T EAT A HAM AND THE OTHER WILL EAT THEM THE BUFFET BOYS. BOTH. THE AVERAGE IS ONE AND STATISTICALLY SPEAKING EACH ONE SHOULD EAT A HAM.

NO. IF WE STUDY STANDARD DEVIATION AND NOT ONLY THE AVERAGE, THINGS CHANGE:

Σ Sum Average

Standard Deviation =

STANDARD DEVIATION IS FOUND BY FOLLOWING THIS PROCEDURE:

1r. FIND THE DIFFERENCE BETWEEN EACH ELEMENT AND THE AVERAGE. 2n. SQUARE IT (SO THAT THE NEGATIVES DISAPPEAR). 3r. MULTIPLY BY THE FREQUENCY OF EACH ONE (HERE IT IS EASY, BECAUSE THE DINERS ACT AS THE FREQUENCY AND THEY ARE: ONE AND ONE). 4t. DIVIDE IT BY THE TOTAL FREQUENCY NUMBER (IN THIS CASE, TWO DINERS). 5è. TAKE THE SQUARE ROOT.

80 Chapter 9 - DE MOIVRE AND GAUSS

THIS IS PRETTY CAUGHT ME TRICKY. RED-HANDED GAUSS.

LOOK, DON’T BE SAYING THAT BECAUSE I KNOW YOU HAVE DONE ALL THE WORK ON THE COMPUTER.

Multiply each row

Subtract the average Average

Square it

Multiply by f Divide

Take the square root

Standard deviation = 1

I AM GOING TO CALCULATE IT ASSUMING THAT ONE EATS 1 2 AND THE OTHER EATS 1 1 . 2

81 Chapter 9 - DE MOIVRE AND GAUSS

HAMS DINERS

Σ Sum Average

Standard Deviation =

WELL, I AM GOING TO CALCULATE IT WHERE EACH ONE EATS 1 HAM.

HAMS DINERS

Σ Sum

Average

Standard Deviation =

I THINK OUR SUPERHERO WILL NOTICE THE DIFFERENCE BET- WEEN THE THREE CASES OF THE STAN- DARD DEVIATION. EVEN THOUGH THEY SUP- POSEDLY HAD AN AVERAGE OF 1, THE DISTRIBU- TION OF HAM IN EACH CASE COMES OUT TO A STANDARD DEVIATION OF “0”. WHILE THE GLUTTON GETS HIS FILL AND THE OTHER GETS RIPPED OFF, COME OUT TO A STAN- DARD DEVIATION OF “1”, VERY HIGH IN THIS CASE.

82 Chapter 9 - DE MOIVRE AND GAUSS

WHEN IT COMES TO FOOD, THE AVERAGE SHOULD BE MODERATE AND THE STANDARD DEVIATION SHOULD BE CLOSE TO ZERO. DOWN WITH ANOREXIA AND GLUTTONY.

NOW THAT YOU MENTION AVERAGE AND SCATTERING, CHANCE AND I HAVE BEEN OK, BUT LEAVE RESEARCHING ON OUR OWN AND FOUND OUT THE SOMETHING FOR FOLLOWING; UNAMBIGUOUS DISTRIBUTION IS LATER. DETERMINED BY THE AVERAGE AND THE SCATTERING VERY OFTEN TIMES WHEN EXPERIMENTS ARE DONE ON LARGE QUANTITIES OF DATA REGARDING AGE, WEIGHT, PEOPLE’S HEIGHT, ETC.

YEAH, HERE’S THE GOOD PART. “NORMAL” DISTRIBUTION IS BELL-SHAPED. LIKE THIS:

Standard Standard Standard deviation deviation deviation 2 0,5 1

83 Chapter 9 - DE MOIVRE AND GAUSS

AND IT’S CALLED…

BOOM BADA BOOM BADA BOOM!!!

GAUSS’ BELL CURVE!!!

GAUSS’ BELL CURVE IS VERY “NORMAL”, BUT OUR GAUSS NORMAL, IN ITS PUREST SENSE, NOT THAT HE’S REALLY JUST WAIT UNTIL I TEACH NORMAL, BUT HE IS… GOOD AND A CLASSES AT THE UNIVERSITY HARD WORKER. AND DISCOVER NEW THEORIES.

84 Chapter 9 - DE MOIVRE AND GAUSS

2 1 2 X 1 X

AND RIDDLE HELPING OUT WITH THE AND WHEN CHANCE HAS STATISTIC HANDLING A PROBABILITY OF RESEARCH ON A NEW RESEARCH CENTER AND VACCINE AND 55 DOING A SOCCER POOL CLUB. HER DETAILED SOCIAL ANALYSES!!!

AND WHEN GRAPH PRESENTS HER MARKET RESEARCH BEFORE THE BOARD OF DIRECTORS OF HER COMPANY AND BINOMIAL, THE MATHEMATICAL FORMULAS WHICH AID THE STATISTICS ADVANCE.

OK, THAT’S ENOUGH. WELL, I WOULD BECAUSE IN CLASS WE ARE LIKE TO SAY THAT STUDYING SOCRATES AND I STATISTICALLY NOW KNOW THAT I KNOW SPEAKING, I AM SURE NOTHING. EVEN THOUGH I THAT I HAVE A 0.3 UNDERSTAND MORE THINGS CHANCE OF BEING AND UNDERSTAND THEM WRONG. BETTER.

85 Chapter 9 - DE MOIVRE AND GAUSS

OKAY, OKAY. IT’S ALL CLEAR NOW. THERE’S ONLY ONE THING LEFT. SOMETHING MORE LET’S GO INTO THE IMPORTANT THAT WE HAVEN’T FORGOTTEN. BACKYARD FOR A MINUTE.

WHAT IS THIS? A PARTY!

YOU DIDN’T FORGET DID YOU? WE DIDN’T! HOW COULD WE FORGET YOUR BIRTHDAY!

HAPPY BIRTHDAY!!

THE END

86 ANNEXES ANNEX 1

COIN TOSS 8 TIMES

HEADS: _____ TAILS: _____ Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue) Heads Tails (in red) (in blue)

88 ANNEX 2

COIN TOSS 50 TIMES Successes (in red) Failures (in black)

Successes (in red) Failures (in black)

Successes (in red) Failures (in black)

Successes (in red) Failures (in black)

Successes (in red) Failures (in black)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

DO YOU DRAW ANY CONCLUSIONS FROM THE COLORS? 89 ANNEX 3

PROBABILITIES CALCULATOR

PROBABILITY CONSTRUCTION

SUCCESS 2 1 try Square (a+b) 3 2 1 try 2 tries (a+b) = (a+b) x (a+b) 3 2 FAILURE 1 3 tries (a+b) = (a+b) x (a+b) 3 4 tries (a+b)4 = (a+b)2 x (a+b)2

5 tries (a+b)5 = (a+b)4 x (a+b) = (a+b)3 x (a+b)2 ab Divide the number of squares of the pertinent 1 try monomial between the total of squares.

a b ab a2 ab ab b a a2 ab a2b a2b2 a2 a3 a2 a4 a3ba3b b ab b2

2 tries ab2 ab3 ab a2b aba3b a2b2 a2b2 ab2 ab3 ab a2b aba3b a2b2 a2b2 b3 b4 b2 ab2 baba2b2 3 ab3

3 tries 4 tries

a3 a2ba2ba2bab2 ab2 ab2 b3

a2b3 a2 a5 a4b a4ba4ba3b2 a3b2 a3b2

ab4 ab a4b a3b2 a3b2 a3b2 a2b3 a2b3 a2b3 ab4 ab a4b a3b2 a3b2 a3b2 a2b3 a2b3 a2b3 b5 b2 a3b2 a2b3 a2b3 a2b3 ab4 ab4 ab4 5 tries

SQUARES SQUARES OF a4 TOTAL SQUARES PROBABILITY

EXAMPLE: Probability of 4 successes in 4 tries a x a x a x a = a4 1 x 16 = 16 81= 0,1975 Probability of 3 successes and 1 failure in 4 tries a x a x a x b = a3b 4 x 8 = 32 81= 0,3950 Probability of 2 successes and 2 failures in 4 tries a x a x b x b = a2b2 6 x 4 = 24 81= 0,2962 Probability of 1 success and 3 failures in 4 tries a x b x b x b = ab3 4 x 2 = 8 81= 0,0098 Probability of 4 failures in 4 tries b x b x b x b = b4 1 x 1 = 1 81= 0,0123 TOTAL=1

90 ANNEX 4

PROBABILITY 123456 1 2 SUCCESS 2 3 a a2 ab 1 try 3 FAILURE 1 4 3 5 b ab b2 6

2 tries

a2 ab ab b2

a2 a4 a3b a3b a2b2

ab a3b a2b2 a2b2 ab3

ab a3b a2b2 a2b2 ab3

b2 a2b2 ab3 ab3 b4

4 tries

91 ANNEX 5

PROBABILITY

SUCCESS 5 6 a b 1 try 123456 FAILURE 1 1 6 2 a 3 a2 ab 4 5 b 6 ab b2

2 tries

a2 ab ab b2

a2b2 a2 a4 a3b a3b

ab3 ab a3b a2b2 a2b2

ab3 ab a3b a2b2 a2b2

b4 b2 a2b2 ab3 ab3

4 tries

92 ANNEX 6

PROBABILITY

SUCCESS 1 3 1 try FAILURE 2 3

e f e fee ef ef ff ee ef ef ff eee ee ff 1 try ef ef 2 tries 3 tries

ff

4 tries

ee ef ef ff eee eef eef eef eff eff eff fff eee eef eef eef

eff

eff

eff

fff

5 tries 6 tries

93 ANNEX 7/1

PROBABILITY

SUCCESS 1 2 1 try FAILURE 1 2

SAMPLE SPACES

HEADS TAILS (Success) (Failure) FOR

1, 2, 3, 4, 5, 6, 7, 8, AND 9 TRIES

94 ANNEX 7/2

95 ANNEX 8

Balearic Islands Students School Preschool Index Elementary Index High Index year school 88-89 19957 1.012737237 96772 1.125989016 89-90 19958 1.012787983 95596 1.112305687 90-91 19220 0.975337461 92481 1.076061156 91-92 19313 0.980056835 89024 1.035837289 92-93 19706 1 85944 1 4432 1 93-94 20123 1.021161068 83197 0.968037327 8683 1.95916065 94-95 20719 1.051405663 80008 0.93093177 10742 2.423736462 95-96 22063 1.119608241 77419 0.900807503 13609 3.070622744 96-97 23169 1.175733279 64165 0.746590803 28975 6.537680505 97-98 23982 1.216990 55294 0.643372429 38872 8.770758123 98-99 24449 1.240688115 55600 0.646932887 39821 8.984882671

Balearic Islands Students School year BUP-COU Bach-LOGSE Bach-Exper. Total Index

88-89 21209 813 22022 89-90 21982 1240 23222 90-91 22185 2664 24849 91-92 22590 4922 27512 92-93 21038 921 2567 24526 1 93-94 19926 2646 237 22809 0.929992661 94-95 18513 3911 22424 0.914295034 95-96 15571 5019 20590 0.839517247 96-97 11772 5551 17323 0.706311669 97-98 8901 7100 16001 0,652410 98-99 4946 9083 14029 0.572005219

Balearic Islands Students School year FP1 Index FP2 Index CFGM Index CFGS Index 88-89 7108 1.1176 4056 0.8793 89-90 7049 1.1083 4172 0.9044 90-91 6360 1 4613 1 75 1 76 1 91-92 5114 0.8041 4996 1.0830 180 2.4000 243 3.1974 92-93 3919 0.6162 5062 1.0973 553 7.3733 226 2.9737 93-94 2662 0.4186 4697 1.0182 813 10.8400 300 3.9474 94-95 2318 0.3645 3826 0.8294 1050 14.0000 394 5.1842 95-96 1789 0.2813 2588 0.5610 1225 16.3333 698 9.1842 96-97 1360 0.2138 1738 0.3768 1763 23.5067 993 13.0658 97-98 686 0.1079 1016 0.2202 2466 32.8800 1481 19.4868 98-99 184 0.0289 480 0.1041 2903 38.7067 1774 23.3421

96 ANNEX 9

Population by age group and sex

Total Men Women

Total

Population by age group and sex Women Population Women Men Men

Men Women

97 ANNEX 10

CITY TOTAL MEN WOMEN

Balearic Islands 796483 392835 403648

Alaró 3834 1834 2000 Alcúdia 10581 5345 5236 Algaida 3542 1766 1776 Andratx 8333 4164 4169 Artà 5936 2963 2973 Banyalbufar 503 264 239 Binissalem 5019 2424 2595 Búger 951 470 481 Bunyola 4338 2144 2194 Calvià 32587 16293 16294 Campanet 2277 1115 1162 Campos 6944 3478 3466 Capdepera 6752 3374 3378 Consell 2210 1090 1120 Costitx 849 415 434 Deià 625 311 314 Escorca 275 148 127 Esporles 3811 1900 1911 Estellencs 338 176 162 Felanitx 14600 7268 7332 Fornalutx 580 290 290 Inca 21103 10425 10678 Lloret de Vistalegre 837 415 422 Lloseta 4529 2231 2298 Llubí 1893 926 967 Llucmajor 21771 10804 10967 Manacor 30177 14988 15189 Mancor de la Vall 936 453 483 Maria de la Salut 1733 861 872 Marratxí 18084 9101 8983 Montuïri 2235 1105 1130 Muro 6028 2979 3049 Palma 319181 154748 164433 Petra 2571 1244 1327 Pollença 13450 6713 6737 Porreres 4226 2102 2124 ANNEX 11

sa Pobla 10064 5169 4895 Puigpunyent 1163 576 587 1969 1009 960 Sant Joan 1662 826 836 Sant Llorenç 5594 2793 2801 Santa Eugènia 1114 548 566 Santa Margalida 7107 3532 3575 Santa Maria del Camí 4558 2243 2315 Santanyí 7974 4026 3948 Selva 2918 1425 1493 ses Salines 3240 1642 1598 Sineu 2616 1278 1338 Sóller 11207 5565 5642 Son Servera 8065 4061 4004 Valldemossa 1599 779 820 Vilafranca de Bonany 2249 1101 1148 Ariany 772 379 393 MALLORCA 637510 313279 324231

Alaior 7046 3490 3556 Ciutadella 21785 10853 10932 Ferreries 3921 2050 1871 Maó 22358 10878 11480 es Mercadal 2723 1353 1370 Sant Lluís 4106 2058 2048 es Castell 6005 3017 2988 es Migjorn Gran 1126 576 550 MENORCA 69070 34275 34795

Formentera 5859 2966 2893

Eivissa 31582 15728 15854 Sant Antoni 14849 7507 7342 Sant Josep 13364 6815 6549 Sant Joan 3943 1991 1952 Santa Eulàlia 20306 10274 10032 EIVISSA 84044 42315 41729 ANNEX 12

Height table in cm of 40 elementary students:

145 148 152 167 170 132 120 139 160 162 167 171 170 148 168 175 18 149 151 155 172 167 163 151 142 144 147 141 150 140 152 161 170 17 169 149 151 152 163 164 170 142 16 Table order: ascending

15 120 132 139 140 141 142 142 144 12345678 145 147 148 148 149 149 150 151 14 9 10111213141516 13 151 151 152 152 152 155 160 161 17 18 19 20 21 22 23 24 12 162 163 163 164 167 167 167 168 25 26 27 28 29 30 31 32 169 170 170 170 170 171 172 175 33 34 35 36 37 38 39 40

Median : 152 (2020) “Stems and Leaves” Diagram

Tens of 120 cm 12: 0 Tens of 130 cm 13: 29 Tens of 140 cm 14: 0 1224578899 Tens of 150 cm 15: 01112225 Tens of 160 cm 16:0 1233477789 Tens of 170 cm 17: 0 0 0 0 1 2 5

100 ANNEX 13

x f xf x-x (x-x)22 (x-x) f 120 1 120 -35.225 1240.801 1240.801 132 1 132 -23.225 539.401 539.401 139 1 139 -16.225 231.801 231.801 140 1 140 -15.225 202.351 202.351 141 1 141 -14.225 202.351 202.351 142 2 284 -13.225 174.901 349.801 50% 144 1 144 -11.225 126.001 126.001 145 1 145 -10.225 104.551 104.551 147 1 147 -8.225 67.651 67.651 148 2 296 -7.225 52.201 104.401 149 2 298 -6.225 38.751 77.501 150 1 150 -5.225 27.301 27.301 151 3 453 -4.225 17.851 53.552 Median 152 3 456 -3.225 10.401 31.202 155 1 155 -0.225 0.051 0.051 160 1 160 4.775 22.801 22.801 161 1 161 5.775 33.351 33.351 162 1 162 6.775 45.901 45.901 163 2 326 7.775 60.451 120.901 164 1 164 8.775 77.001 77.001 167 3 501 11.775 138.651 415.952 50% 168 1 168 12.775 163.201 163.201 169 1 169 13.775 189.751 189.751 Mode=170 170 4 680 14.775 218.301 873.203 171 1 171 15.775 248.851 248.851 172 2 172 16.775 281.401 281.401 175 1 175 19.775 391.051 391.051

Sum 40 6209 6282.975

Average = 6209 = 155.225 Variance = 6282.975 = 157.074 40 40

Standard Deviation = Variance = 12.533

101 Printed on recycled paper Dice and Data