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Mendenhall : Introduction To Probability And Statistics
Solutions to exercices
[ 1.3 #1.01] a) The student.b) The exam.
c) The patient.
d) The azalea plant.
e) The car.
[ 1.3 #1.02] a) The time to assemble is quantitative.b) The
number of students is quantitative.
c) The rating of a politician is qualitative.
d) The province or territory of residence is qualitative.
[ 1.3 #1.03] a) The population is a discrete variable.b) The
weight is a continuous variable.
c) The time is a continuous variable.
d) The number of consumers is a discrete variable.
[ 1.3 #1.04] a) The number of boating acccidents is a discrete
variable.b) The time is a continuous variable.
c) The choice of colour is a qualitative variable.
d) The number of brothers and sisters is a discrete
variable.
e) The yield in kilograms is a continuous variable.
[ 1.3 #1.05] a) The vehicule is the experimental unit.b) The
variables are : the type (qualitative), the make (qualitative),
carpool or not
(qualitative), the one-way commute distance (quantitative -
continuous) and theage of vehicule (quantitative - discrete).
c) This is a multivariate data.
[ 1.3 #1.06] a) This is a population.b) The variable is the age
at death.
c) This is a quantitative variable.
[ 1.3 #1.07] The population is the voters opinions (for or
against the candidate) at the time ofthe election for all persons
voting during the election. It depends on time by thefact that
voters opinions may change frequently during election time or prior
to theelection time.
[ 1.3 #1.08] a) The variable is the survival time.b) This is a
quantitative continuous variable.
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c) The survival time for all patients having a particular type
of cancer and havingundergone a particular type of
radiotherapy.
d) He could select a significative number of people from
different ages and differenthealth conditions from this population
to get a representative sample.
e) Since this is only a sample, the result could be wrong. If we
sample fromall patients having cancer and radiotherapy, some may
still be living and theirsurvival time will not be measurable.
Hence, we cannot sample directly from thepopulation of interest,
but must arrive at some reasonable alternate populationfrom which
to sample.
[ 1.3 #1.09] a) The variable is the reading score. This is a
quantitative discrete variable (sinceit is an interger score).
b) The student is the experimental unit.
c) The population is the reading score for all students who
could possibly be taughtby this method.
[ 1.3 #1.10] a) The experimental unit is the people.b) The
variable is the category and is qualitative (A, B, C, and D).
c)
d)
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e) The shape of the bar chart changes, but it is not
important.
f) We have : 1450 = 0, 28 for B,2050 = 0, 4 for C, and
550 = 0, 1 for D.
g) There is : (5014)50 100 = 3650 100 = 72%.[ 1.3 #1.11] a) The
experimental unit is the pair of jeans.
b) The variable is province and it is qualitative (QC, ON,
MB).
c)
d)
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e) 825 = 0, 32.f) Since we have 0, 32 in QC, 925 = 0, 36 in ON,
and
825 = 0, 32 in MB, then ON
produce the most jeans.
g) By looking at the bars, we can easily know that ON produces
the most jeans.Also, we can conclude that since the bars and the
sectors are almost equal insize, the three provinces produced
roughly the same number of pairs of jeans.
[ 1.3 #1.12] a) The opinion about the Islam-West clash for all
Canadian adults.b) All the Canadians adults of 2007.
c) No, we could add the other reason category.
d) No, since peoples opinion can change greatly in one year, the
sample taken in2007 could be not representative in 2008.
[ 1.3 #1.13] a) No, we could add Morocco, Iran, etc.b) A bar
chart is appropriate.
c)
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d) Answers may vary.
[ 1.3 #1.14] a) The variable is the educational attainment.b)
This variable is qualitative.
c) The numbers represent the percentage of the population type
for each category.
d)
e)
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f) 6% of the Aboriginal population has a university degree.
Since 17, 70% of thenon-Aboriginal population has a university
degree, there is an important gapbetween these two population : we
have almost 3 times more educated non-Aboriginal people than
educated Aboriginal.
[ 1.3 #1.15] a) Yes. The total percentage of education level in
each bar graph is 100%.b) Yes. There is a significant increase
(from 39% to 46%) in the post secondary
education attainment over the years.
c)
[ 1.5 #1.16] Range: 1 0, 5 = 0, 5; Min. Class Width: 0,58 = 0,
0625; Conv. Class Width:0, 06Range: 100 0 = 100; Min. Class Width:
1006 = 16, 66667; Conv. Class Width:17
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Range: 1500 1200 = 300; Min. Class Width: 3009 = 33, 33333;
Conv. ClassWidth: 33
[ 1.5 #1.17] Convenient Starting Point: 0, 5; First Two Classes:
0, 5 to < 0, 56, 0, 56 to< 0, 62Convenient Starting Point: 0;
First Two Classes: 0 to < 16, 16 to < 32Convenient Starting
Point: 1200; First Two Classes: 1200 to < 1233, 1233 to1266
[ 1.5
#1.18]1|682|12555788993|11455666777789994|0001223456789995|116676|12
a) The stem and leaf display has a mound shaped
distribution.
b) From the stem and lead display, the smallest observation is
1.6.
c) The eight and ninth largest observations are both 4.9.
[ 1.5 #1.19] a) For 50 values, we should use 7 or 8 class
intervals.b)
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c) There is 2+5+5+5+14+6+650 =4350 of the measurements that are
less than 5.1.
d) There is 14+6+6+2+5+250 =3550 =
710 of the measurements that are larger than 3.6.
e) The stem and leaf display has more peaked mound-shaped
distribution than therelative frequency histogram because of the
smaller number of groups.
[ 1.5 #1.20] a)3|34555667799994|00223334458
b)
3|343|555667799994|0022333444|58
Yes. With this method, its more easy to know where is the center
of distributionand where are the peaks.
[ 1.5 #1.21] a)
b) There is 620 =310 = 0.3 measurements greater than 1.
c) There is 9+520 =1420 =
710 = 0.7 measurements less than 2.
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d) The probability is 620 =310 = 0.3.
e) It is almost symmetric and there is no outliers.
[ 1.5 #1.22] a)
b) With a leaf unit of 0.1.c)
0|000001|000000002|000000
d) Yes they do. They have the same three bars and the same
peaks.
[ 1.5 #1.23] As seen on the line chart, the time to resolve the
maze decrease a lot since day 3. Wecan assume that the rat has
learned.
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[ 1.5 #1.24] a)
b) The measurement increases during the first year and decreases
during the nextthree years. It increases between year 5 and 6,
decreases during one year, is
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stable for one year and finally decreases the next two
years.
[ 1.5 #1.25] a) We could use a frequency histogram :
We could also use a stem and leaf plot with a leaf unit of 1.0
:
5|576|1236|5787|27|568|248|66799|134
b) The distribution is not mound-shaped, but is rather bimodal
with two peakscentred around the scores 65 and 85.
c) This might indicate that the students are divided into two
groups : those whounderstand the material and do well on exams,
annd those who do not have athorough command of the material.
[ 1.5 #1.26] a)
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b) The shape is skewed left.
c) 3650 =1825
[ 1.5 #1.27] a) A bar chart or a pie chart would be appropriate,
since we have qualitativevariables.
b) The bar chart :
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c) The average annual income is growing with the education
level.
[ 1.5 #1.28] a)
3|0001122233443|5556666677889994|0000111122334|556667885|005|5
b)
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c) The graphs look almost the same, but the stem and leaf plot
has the advantageto show all the measurements.
d) 2750e) The probability is 4750 .
[ 1.5 #1.29] a)
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b) The pareto chart is more effective, since it is more easy to
compare the data.
c) The number of passengers; also, we could consider the size of
the airline. Ina big airline, it could be difficult to serve the
passengers; if we have a lot of
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passengers, it could be more difficult to give a good
service.
[ 1.5 #1.30] a)
0|223334440|5566667778888991|001111111222333441|66778888992|1232|583|113|64|4|55|2
b) 2560 =512
c) 0.2
[ 1.5 #1.31] a)
This is a skewed left distribution. We can consider 4, 4.5 and
5.2 as outliers.b) Maybe the supermarket was understaffed that day,
or there may have been an
unusuallu large number of customers in the store.
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c) The two graphs convey the same information. The stem and leaf
plot allows usto actually recreate the actual data set, while the
histogram does not.
[ 1.5 #1.32] a)
b) The leaf unit is 0.001 :
26|8927|1127|56928|112
c) No, but some measurements are identical.
[ 1.5 #1.33] a) It would be roughly mound-shaped.
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b) The leaf unit is 1 :
3|94|4|56667785|1245|5796|016|56697|04
c) J. Clark, B. Mulroney, A. Meighen, K. Campbell and S. Harper.
They are allconservative; maybe the trend of transferring the
leadership to younger genera-tions in that party is faster.
[ 1.5 #1.34] a)
b) This is skewed to the right.
c) Its not far enough to the main distribution to be considered
as an outlier.
[ 1.5 #1.35] a)
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b) 420 =15
[ 1.5 #1.36] a) This is clearly skewed to the left. There is two
outliers : 16.4 and 17.1.
0|9991|2242|01243|44|055|06|487|13
b) The stem and leaf plot is clearly better, because we can see
every measurement.
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[ 1.5 #1.37] a) The variable is the number of contaminated Waste
Sites in each province. Thisis a discrete variable.
b)
1|72|13|44|445|5676|7|88
The distribution is skewed to the right with three outliers. The
unusually largemeasurements are : 205, 235 and 300.
b) Its very difficult to know. This is the largest provinces,
but its also the caseof Nunavut, Yukon and Northwest Territories
which have not much waste Sites.Some other variables like
population size, numbers of industries, etc. can helpto explain the
data.
[ 1. Supp. #1.38] a) The ethnic origin is a qualitative
variable.b) The score is a quantitative variable.
c) The type of establishment is a qualitative variable.
d) The mercury concentration is a quantitative variable.
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[ 1. Supp. #1.39] a) The distribution of non-secured loan sizes
might be skewed (a few extremelylarge loans are possible).
b) The distribution of secured loan sizes is not likely to
contain unusually large orsmall values.
c) Not likely to be skewed.
d) Not likely to be skewed.
e) If a package is dropped, it is likely that all the shells
will be broken. Hence, afew large number of broken shells is
possible. The distribution will be skewed.
f) If an animal has one tick, he is likely to have more than
one. There will besome 0s with uninfected rabbits, and then a
larger number of large values.The distribution will not be
symmetric.
[ 1. Supp. #1.40] a) Discrete.b) Continuous.
c) Discrete.
d) Discrete.
e) Continuous.
[ 1. Supp. #1.41] a) Continuous.b) Continuous.
c) Discrete.
d) Discrete.
e) Discrete.
[ 1. Supp. #1.42] a) Discrete.b) Continuous.
c) Continuous.
d) Discrete.
[ 1. Supp. #1.43]
7|898|0179|01244566688
10|17911|2
The distribution is symmetric, mound-shaped.
[ 1. Supp. #1.44] a) The graphical technique used is the bar
chart.b) This could be significative, but we have to be careful.
Some student may have
good score for other reasons that have no link with this method
: if they knowthat they are on study, they could work harder and
get better grades.
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[ 1.Supp. #1.48] a)
b) Since Prince Rupert is located west of the Coastal mountain,
it is not unusualthat the average rainfall would be very high.
c) The value x = 1154.66 (Vancouver) does not lie far from the
centre of thedistribution. It would not be considered unusually
rainy.
[ 1. Supp. #1.58] a) There is no unusual features. The
distribution is roughly mound shaped.
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b)
4|395|6|3567|02348|3446789|125
10|01211|067912|13|1
With 1.0 as leaf unit.c) Since the data appears in Mercers site,
this global investment consulting agency
may have chosen the cities of its business priorities.
[ 1. Supp. #1.67] a) Cest approximativement en forme de
cloche.b) Il y a quelques donnees aberranets a` 38.2. Probablement
que la personne dont
la temperature etait de 38.22 avait une maladie.c) Oui, mais
quelque peu a` droite.
[ 2.2 #2.01] a)
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b) La moyenne : x = 0+5+1+1+35 = 2. La mediane : si on met les
donnees en ordrecroissant, on obtient : 0, 1, 1, 3, 5; la position
de la mediane est (n+1)2 =
(5+1)2 = 3
cest-a`-dire la position 3, donc m = 1. Le mode est 1.c) x 6= m,
alors la distribution est asymetrique. Ici, m < x, alors cest
asymetrique
a` droite.
[ 2.2 #2.02] a) x = 3+2+5+6+4+4+3+58 = 4b) Si on met les donnees
en ordre croissant, on obtient : 2, 3, 3, 4, 4, 5, 5, 6; la
posi-
tion de la mediane est (n+1)2 =(8+1)
2 = 4.5, cest-a`-dire entre les positions 4 et5, donc m = 4+42 =
4.
c) x = m, alors la distribution est symetrique. Le graphique est
clairementsymetrique :
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[ 2.2 #2.03] a) x = 3+5+4+6+10+5+6+9+2+810 = 5.8b) Si on met les
donnees en ordre croissant, on obtient : 2, 3, 4, 5, 5, 6, 6, 8, 9,
10; la
position de la mediane est (n+1)2 =(10+1)
2 = 5.5, cest-a`-dire entre les positions5 et 6, donc m = 5+62 =
5.5.
c) Les modes sont 5 et 6. Lensemble est bimodal.
[ 2.2 #2.04] a) x =
825+883+947+988+1036+1044+1152+1197+1347+140410 = 825b) x =
847+833+1014+983+1126+1121+1157+1174+1396+137410 = 847c) Si le fait
davoir une voiture est important, le prix moyen de lassurance est
une
mesure interessante a` considerer pour negocier lachat dune
assurance. Cettemesure est egalement interessante pour mesurer une
partie du cout de la vieet comparer les provinces entre elles. A`
cet egard, dautres mesures moyennesauraient ete interessantes : le
cout moyen de leducation post-secondaire, le coutdes habitations,
le cout moyen du panier depicerie, etc.
[ 2.2 #2.05] a)0|0000001|00000000000002|00003|00
La feuille a pour unite 1.0 et la distribution est asymetrique
a` droite.b) Le mode est 1.
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c) x = 1+1+0+1+3+0+0+1+1+1+2+2+2+1+0+1+1+3+0+1+1+0+2+1+125 =
1.08. On meten ordre les donnees : 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, , 2, 2, 2, 2, 3, 3;(n+1)
2 =(25+1)
2 = 13, la mediane est a` la position 13, alors m = 1. Le mode
est1.
d) La moyenne est lege`rement plus a` droite que la mediane, ce
qui correspond aufait que la distribution est asymetrique a`
droite.
[ 2.2 #2.06] a)
1|788882|000012342|663|233|4|4|95|25|6
La distribution est asymetrique a` droite.
b) x =
56+52+49+33+32+26+26+24+23+22+21.5+20.7+20.3+20.1+20+18.7+18.4+18.2+18+17.520
=26.82. Pour la mediane, la position est (n+1)2 =
(20+1)2 = 10.5, cest-a`-dire entre
la position 10 et 11, donc m = 21.5+222 = 21.75.
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c) Comme la moyenne est fortement influencee par les donnees
aberrantes, la me-diane serait une mesure plus adequate du centre
de la distribution.
[ 2.2 #2.07] Comme le nombre denfants doit etre une variable
discre`te, il est evident que 2.5represente le nombre moyen
denfants par famille pour toutes les familles aux Etats-Unis durant
les annees 30.
[ 2.2 #2.08] a) La moyenne : x =
0.99+1.12+1.92+0.63+1.23+0.67+0.85+0.69+0.65+0.6+0.53+0.6+1.41+0.6614
=0.8964.
b) Pour la mediane, on met en ordre les donnees : 0.53, 0.6,
0.6, ..., 1.12, 1.23, 1.41, 1.92;la position est (n+1)2 =
(14+1)2 = 7.5, cest-a`-dire entre la position 7 et 8, donc
m = 0.67+0.692 = 0.68.c) Yes, the distribution is skewed to the
right, because x > m.
[ 2.2 #2.09] The distribution of sports salaries will be skewed
to the right, because of the veryhigh salaries of some sports
figures. Hence, the median salary would be a bettermeasure of
centre than the mean.
[ 2.2 #2.10] a) La moyenne : x =
175+200+190+185+250+190+230+225+240+26510 = 215.b) Pour la mediane,
on met en ordre les donnees : 175, 185, 190, 190, 200, 225, 230,
240, 250, 265;
la position est (n+1)2 =(10+1)
2 = 5.5, cest-a`-dire entre la position 5 et 6, doncm = 200+2252
= 212.5.
c) Comme il ny a pas de donnees aberrantes, la moyenne demeure
la mesure laplus fiable.
[ 2.2 #2.11] a) La moyenne : x =
22+16+12+19+40+40+21+10+20+19+4+65+39+20+6+34+23+718 =
23.16667.Pour la medianne, on met en ordre croissant les donnees :
4, 6, 7, 10, 12, 16, ..., 40, 40, 65;la position est (n+1)2 =
(18+1)2 = 9.5, cest-a`-dire entre la position 9 et 10, donc
m = 20+202 = 20. Les modes sont 19, 20 et 40.b) x > m. La
distribution est donc asymetrique a` droite.
c) Oui, la distribution est asymetrique a` droite.
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[ 2.2 #2.12] a) x = 1200+800+1050+750+700+500+600+670+1200+65010
= 812.b) On met en ordre croissant les donnees : 500, 600, 650,
670, 700, 750, 800, 1050, 1200, 1200;
la position est (n+1)2 =(10+1)
2 = 5.5, cest-a`-dire entre la position 5 et 6, doncm = 700+7502
= 725.
c) Ces donnees permettent de faire un achat plus eclaire. La
qualite techniquede lappareil, la resolution de lecran, sa
durabilite, etc., sont des donnees quiauraient pu egalement etre
prises en consideration.
[ 2.2 #2.13] a) x = 8+6+3+23+2+6+5+4+1+0+2+7+11+5+8+8+1017 =
6.411765.b) On met en ordre croissant les donnees : 0, 1, 2, 2, 3,
4, 5, 5, 6, 6, 7, 8, 8, 8, 10, 11, 23;
la position est (n+1)2 =(17+1)
2 = 9, donc m = 6. Le mode est 8.c) x > m (mais faiblement).
La distribution est donc lege`rement asymetrique a`
droite.
d) Il y a une mesure aberrante, 23. Pour cette raison, la
mediane est une valeurplus fiable, car elle est moins sensible aux
valeurs aberrantes.
[ 2.3 #2.14] a) La moyenne de lechantillon : x = 2+1+1+3+55 =
2.4.b) La variance de lechantillon :
s2 =(xi x)2n 1
= (2 2.4)2 + (1 2.4)2 + (1 2.4)2 + (3 2.4)2 + (5 2.4)2
5 1= 2.8
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c) Lecart-type de lechantillon : s =s2 =
2.8 = 1.67332.
d) On prend la formule computationnelle :
s2 =x2i (
xi)2n
n 1
=(22 + 12 + 12 + 32 + 52) (2+1+1+3+5)25
5 1=
40 14454
= 2.8
Et s =s2 =
2.8 = 1.67332. Les resultats sont identiques.
[ 2.3 #2.15] a) Letendue : R = 4 1 = 3.b) La moyenne de
lechantillon : x = 4+1+3+1+3+1+2+28 = 2.125.c) On prend la formule
computationnelle pour trouver la variance :
s2 =x2i (
xi)2n
n 1
=(42 + 12 + 32 + 12 + 32 + 12 + 22 + 22) (4+1+3+1+3+1+2+2)28
8 1=
45 28987
= 1.267857
Et, pour lecart-type : s =s2 =
1.267857 = 1.125992.
d) On calcule la variance de lechantillon :
s2 =(xi x)2n 1
= (4 2.125)2 + (1 2.125)2 + (3 2.125)2 + ...+ (2 2.125)2 + (2
2.125)2
8 1= 1.267857
Lecart-type de lechantillon : s =s2 =
1.267857 = 1.125992. Les valeurs
sont identiques.
[ 2.3 #2.16] a) Letendue : R = 6 1 = 5.b) La moyenne de
lechantillon : x = 3+1+5+6+4+4+3+58 = 3.875.
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c) La variance de lechantillon :
s2 =(xi x)2n 1
= 2(3 3.875)2 + (1 3.875)2 + 2(5 3.875)2 + (6 3.875)2 + 2(4
3.875)2
8 1= 2.410714
Lecart-type de lechantillon : s =s2 =
2.410714 = 1.552647.
On peut aussi calculer la variance de lechantillon avec la
formule computation-nelle :
s2 =x2i (
xi)2n
n 1
=(32 + 12 + 52 + 62 + 42 + 42 + 32 + 52) (3+1+5+6+4+4+3+5)28
8 1=
137 96187
= 2.410714.
c) 51.552647 = 3.220307. Alors letendue est approximativement 3
fois lecart-type.
[ 2.3 #2.17] a) Letendue : R = 2.39 1.28 = 1.11.b) On calcule la
variance :
s2 =x2i (
xi)2n
n 1
=(1.282 + 2.392 + 1.52 + 1.882 + 1.512)
(1.28+2.39+1.5+1.88+1.51)25
4
=15.415 73.27365
4= 0.19007
Alors lecart-type est : s =s2 =
0.19007 = 0.4359702
c) 1.110.4359702 = 2.546046. Alors letendue est
approximativement 2.5 fois lecart-type.
[ 2.3 #2.18] a) Letendue : R = 312.4 165.12 = 147.28.b) La
moyenne : x =
204.94+180+178.23+176.43+...+312.4+238.66+225.47+222.2312 =
227.2167.
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c) On calcule dabord la variance :
s2 =x2i (
xi)2n
n 1=
647847.1 74343481211
= 2574.373
Alors lecart-type est : s =s2 =
2574.373 = 50.73828
[ 2.5 #2.21] a) x = 50, s = 10. Linterval est defini en tant que
x = ks; on a 50 k10 = 40 et50 + k10 = 60, alors k = 1, donc la
proportion des donnees entre 40 et 60, seloncette re`gle, est de
68%.
b) x = 50, s = 10. Linterval est defini en tant que x = ks; on a
50 k10 = 30 et50 + k10 = 70, alors k = 2, donc la proportion des
donnees entre 30 et 70, seloncette re`gle, est de 95%.
c) Comme 68% des donnees sont entre 40 and 60, alors 34% des
donnees sont entre50 et 60. Comme 95% des donnees sont entre 30 et
70, alors 47.5% des donneessont entre 30 and 50. On conclut que 34%
+ 47.5% = 81.5% des donnees sontentre 30 et 60.
d) Comme 5% des donnees sont a` lexterieur de la distribution
entre 30 et 70, il ya 2.5% de la distribution qui se trouve entre 0
et 30. Il y a alors 81.5%+2.5% =84% de la distribution qui se
trouve entre 0 et 60. On conclut que la probabiliteque la
distribution se situe au-dela` de 60 est de 16%.
[ 2.5 #2.35] a) Avec une feuille dunite 1.0 :
0|91|162|3353|184|00165|12456|27|138|79|2
On voit que la distribution est relativement en forme de
cloche.
b) La moyenne : x =
46+73+41+25+51+52+31+23+55+62+16+9+92+40+38+71+54+11+87+40+2321
=
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44.7619. Pour obtenir lecart-type, on calcule dabord la variance
:
s2 =(xi x)2n 1
= (46 44.7691)2 + (73 44.7619)2 + (41 44.7619)2 + ...+ (23
44.7619)
21 1= 547.9905
Lecart-type : s =s2 =
547.9905 = 23.4092.
c) Comme la distribution est relativement en forme de cloche, on
peut utiliser lare`gle empirique : pour un interval defini en tant
que x2s, cest-a`-dire allant de44.76192(23.4092) = 2.0565 a`
44.7619 + 2(23.4092) = 91.5803, on sait quaumoins 95% de la
distribution sera situee dans cet interval. On peut verifier quily
a bien 2021 = 0.952381 = 95% des donnees dans cet intervale (toutes
les donneessauf 92).
[ 2.5 #2.39] a) La moyenne : x =
0(10)+1(5)+2(3)+3(2)+4(1)+6(1)+9(1)+10(1)24 = 1.916667;
autrementdit, on a x =
0+0+0+0+0+0+0+0+0+0+1+1+1+1+1+2+2+2+3+3+...+9+1024 = 1.916667.Pour
obtenir lecart-type, on calcule dabord la variance :
s2 =(xi x)2n 1
= 10(0 1.916667]2 + 5(1 1.916667)2 + ...+ (9 1.9166667)2 + (10
1.9166667)2
24 1= 7.818841
Lecart-type : s =s2 =
7.818841 = 2.796219.
b) x s = [1.916667 2.796219, 1.916667 + 2.796219] = [0.879552,
4.712886], x2s = [1.9166672(2.796219), 1.916667+2(2.796219)] =
[3.675771, 7.509105] etx3s = [1.9166673(2.796219),
1.916667+3(2.796219)] = [6.47199, 10.30532]
c) 21/24 = 0.875 = 87.5% des donnees sont dans le premier
interval (ce qui estconforme au theore`me de Tchebysheff qui
stipule quil doit y en avoir au moins0%, et la re`gle empirique,
qui en prevoit 68%) , 22/24 = 0.9166667 = 91.66667%dans le second
(ce qui est conforme au theore`me de Tchebysheff qui stipulequil
doit y en avoir au moins 75%, et la re`gle empirique, qui en
prevoit 95%),et 24/24 = 1 = 100% des donnees dans le troisie`me (ce
qui est conforme autheore`me de Tchebysheff qui stipule quil doit y
en avoir au moins 89%, et lare`gle empirique, qui en prevoit 99,
7%).
[ 2.7 #2.42] a) On met en ordre les donnees : 0, 1, 3, 4, 4, 5,
6, 6, 6, 7, 7, 8; la position de la me-diane est (n+1)2 =
(12+1)2 = 6.5, cest-a`-dire entre la position 6 et 7, donc m
=
5+62 = 5.5. On a la position de Q1 =
(n+1)4 = 3.25, alors Q1 = 3 +
(43)4 = 3.25.
On a la position de Q3 = 3(n+1)4 = 9.75, alors Q3 = 6+3(76)
4 = 6.75. On a doncIQR = Q3 Q1 = 6.75 3.25 = 3.5 et les cinq
nombres : Min = 0, Q1 = 3.25,m = 5.5, Q3 = 6.75 et Max = 8.
32
-
b) La moyenne : x = 8+7+1+4+6+6+4+5+7+3+012 = 4.75. Pour obtenir
lecart-type,on calcule dabord la variance :
s2 =(xi x)2n 1
= (0 4.75)2 + (1 4.75)2 + (3 4.75)2 + ...+ (8 4.75)
12 1= 6.022727
Lecart-type : s =s2 =
x2i
(
xi)2
nn1 =
x2inx2n1 =
6.022727 =
2.454124.c) Avec une courbe normale en cloche, on se sert
souvent de la cote z pour verifier
si une donnee est aberrante (situee dans la zone qui, selon la
re`gle empirique,est situee dans x 3s). z score = xxs . Pour la
plus petite valeur, on azscore = 04.752.454124 = 1.935518. Pour la
plus grande valeur, on a : zscore =84.752.454124 = 1.324301. Comme
aucune valeur absolue de ces deux cotes z nexce`de2, ces valeurs ne
sont pas exceptionnellement elevees ou faibles.
[ 2.7 #2.43] On met en ordre les donnees : 0, 1, 5, 6, 7, 8, 9,
10, 12, 12, 13, 14, 16, 19, 19; la po-sition de la mediane est
(15+1)2 = 8, cest-a`-dire a` la position 8, donc m = 10.On a la
position de Q1 = (n+1)4 = 4, alors Q1 = 6. On a la position deQ3 =
3(n+1)4 = 12, alors Q3 = 14. On a IQR = Q3Q1 = 8 et les cinq
nombres: Min = 0, Q1 = 6, m = 10, Q3 = 14 et Max = 19.
[ 2.7 #2.45] On met en ordre les donnees : 2, 3, 4, 5, 6, 6, 6,
7, 8, 9, 9, 10, 22; la position de lamediane est (13+1)2 = 7,
cest-a`-dire a` la position 7, donc m = 6. On a laposition de Q1 =
(n+1)4 = 3.5, alors Q1 = 4 +
(54)4 = 4.25. On a la position de
Q3 = 3(n+1)4 = 10.5, alors Q3 = 9 +3(99)
4 = 9. On a IQR = Q3 Q1 = 4.75et les cinq nombres : Min = 2, Q1
= 4.25, m = 6, Q3 = 9 et Max = 22. Laplus petite valeur adjacente
est Q1 1.5(IQR) = 4.25 1.5(4.75) = 2.875 etla plus grande valeur
adjacente est Q3 + 1.5(IQR) = 9 + 1.5(4.75) = 16.125. Onvoit
clairement que 22 est une donnee aberrante.
33
-
[ 2.7 #2.46] Il y a 69% des candidats qui ont fait moins bien
que moi, tandis que 31% des candidatsont fait mieux.
[ 2.7 #2.49] a) On met en ordre les donnees pour Mario Lemieux :
1, 6, 7, 17, 19, 28, 35, 44, 45, 50, 54, 69, 69, 70, 85;la position
de la mediane est (15+1)2 = 8, donc m = 44. On a la position deQ1 =
(n+1)4 = 4, alors Q1 = 17. On a la position de Q3 =
3(n+1)4 = 12, alors
Q3 = 69. On a donc les cinq nombres : Min = 1, Q1 = 17, m = 44,
Q3 = 69,Max = 85.
On met en ordre les donnees pour Brett Hull : 0, 1, 25, 29, 30,
32, 37, 39, 41, 42, 54, 57, 70, 72, 86;on a m = 39, Q1 = 29 et Q3 =
57. On a donc les cinq nombres : Min = 0,Q1 = 29, m = 39, Q3 = 57
et Max = 86.
b) Il ny a pas de donnees aberrantes :
34
-
c) La distribution semble assez symetrique pour Lemieux, tandis
quelle est davan-tage asymetrique pour Hull. La distribution de
Lemieux varie beaucoup plus,car elle a un IQR plus eleve, mais elle
a aussi une mediane plus elevee pour lenombre de buts.
[ 2.7 #2.50] On a m = 100. Il ny a pas de valeurs aberrantes a`
gauche, mais deux a` droite (169et 208). La distribution est
symetrique (car la mediane est au centre de IQR); si lamediane
avait ete a` droite, on aurait eu une distribution asymetrique a`
gauche.
[ 2.7 #2.54] a) La position de Q1 = n+14 = 3.5, alors Q1 =
97+982 = 97.5. La position deQ3 = 3n+14 = 10.5, alors Q1 =
106+1072 = 106.5
b) IQR = Q3 Q1 = 106.5 97.5 = 9c) La cloture de gauche : Q1
1.5(IQR) = 97.5 1.5(9) = 84.d) La cloture de droite : Q3 + 1.5(IQR)
= 106.5 + 1.5(9) = 120.e) On prend la mediane dont la position est
n+12 = 7: m = 104.
35
-
f) Oui, 126 est une donnee aberrante.g) On a z score = xxs . On
trouve x = 105.53 et s = 8.42. On a donc zmin =93103.53
8.42=1.25 et zmax =126103.53
8.42 = 2.67. On trouve ainsi que la valeur 126 est unedonnee
aberrante, car la valeur absolue de sa cote z est superieure a`
2.
h) On a zhamilton = 98103.538.42 = 0.66. Il sagit dune valeur
normale, car sa valeurabsolue est inferieure a` 2.
i) La ville de Victoria serait le bon choix, car comme le prix
de lessence y esteleve, on doit moins en consommer, ce qui nuit
moins a` lenvironnement.
[ 2.7 #2.55] a) x = 26.21429, s = 1.251373.b) x = 26.14286, s =
2.413333.c) Ils ont en general le meme nombre de raisins par bote,
mais dans la marque
Sunmaid, il y a certaines botes qui ont beaucoup plus de raisins
que la moyenne.Cest ce quindique lecart-type.
[ 2.7 #2.63] Pour n = 400, la moyenne est x = 600 et la variance
est s2 = 4900. On a donc lecarttype s =
s2 = 70. Comme la distribution est en forme de cloche, on peut
presumer,
selon la re`gle empirique, que 68% des donnees se trouvent dans
lintervalle xs, cest-a`-dire (530, 670), donc 400(0.68) = 272
resultats. On sait aussi que 95% des donneesse trouvent dans
lintervalle x 2s, cest-a`-dire (460, 740), donc 400(0.95) =
380resultats.
[ 2.7 #2.64] a) x =
xin = 6.85. s =
s2 =
x2i
(
xi)2
nn1 = 1.008299
b) zmax = xxs =8.56.851.008299 = 1.636419. Cette valeur est
habituelle et nest pas
aberrante (car sa valeur absolue est inferieure a` 2.)
36
-
c) On cherche le mode : 7.d) On met en ordre les donnees : 5, 6,
6, 6.75, 7, 7, 7, 7.25, 8, 8.50. On trouve la
position de la mediane : (n+1)2 = 5.5, alors m = 7. On trouve la
position deQ1 = (n+1)4 = 2.75, alors Q1 = 6. On trouve la position
de Q3 =
3(n+1)4 = 8.75,
alors Q3 = 7.25 + 0.25(87.25) = 7.4375. On a IQR = Q3Q1 =
7.43756 =1.4375. On trouve la cloture de gauche : Q1 1.5(1.4375) =
3.84375. Ontrouve la cloture de droite : Q3 + 1.5(1.4375) =
9.59375. Il ny a aucune donneeaberrante, comme le laissait predire
la cote z. Dans ce cas, les clotures peuventetre les valeurs
minimales et maximales :
[ 2.7 #2.78] Brett Hull a une distribution assez symetrique; il
y a une donnee aberrante, illustrantune saison avec un nombre de
but exceptionnel. Les distributions de Mario Lemieuxet de Bobby
Hull sont asymetriques a` gauche, tandis que celle de Wayne Gretzky
estasymetrique a` droite. La variabilite est sensiblement la meme
entre les deux Hull,celle de Gretzky et Lemieux sont pratiquement
identique et superieur a` celle des Hull.Gretzky a la cloture de
droite la plus eloignee, ce qui laisse penser quil a un nombrede
buts tre`s eleve lors dune saison exceptionnelle. Lemieux est celui
avec le IQR leplus eleve et la mediane la plus elevee; cest Brett
Hull qui a la plus faible mediane(autour de 38) alors que les
autres ont une mediane qui se situe approximativemententre 41 et
44.
[ 3.2 #3.05] a) La population dinteret est lopinion des parents
et des enfants a` propos du tempslibre dont dispose les enfants.
Lechantillon, cest 198 opinions des parents et200 opinions des
enfants a` propos du temps libre dont dispose les enfants.
b) Il sagit de la variable qualitative opinion. Les donnees sont
bivariees dans lamesure ou`, pour chaque personne, on lui demande
sil est un adulte ou un enfant
37
-
et sa reponse a` la question (juste assez, pas assez, trop, ne
sait pas)
c) Chaque entree dans le tableau represente le nombre de
personnes qui entre danschaque categorie dopinions.
d)
38
-
e) On pourrait utiliser un diagramme comparatif a` barres ou un
diagramme enpiles. Comme les donnees representent la frequence
relative, il est plus represen-tatif dutiliser les diagrammes
circulaires.
[ 3.4 #3.11] a)
b) Il y a une relation approximativement lineaire.
c) r = sxysxsy . On trouve dabord la covariance : sxy =
xy
x
y
nn1 =
126 193765 =
1.7666. On trouve ensuite les ecarts-type : sx = 1.47196 et sy =
1.32916. On adonc r = 1.76661.471961.32916 = 0.9029526.
d) On trouve dabord x = 3.166667 et y = 6.166667. Ensuite, on
trouve b =r sysx = 0.9029526 1.329161.47196 = 0.815354. On trouve a
= y bx = 6.166667 0.815354(3.166667) = 3.584712. Lequation est donc
approximativement : y =3.58 + 0.815x. On voit quelle correspond aux
donnees.
39
-
[ 3.4 #3.15] a) On doit trouver a = y bx. y = 73.61167, x =
2.833333. On trouve b = r sysx .On a sy = 35.79161 et sx = 1.47196.
On trouve la covariance : sxy = 51.62233et r = sxysxsy =
51.6223335.791611.47196 = 0.9798517. Donc b = 0.9798517
35.791611.47196 =
23.82569 et donc a = 73.61167 (23.82569 2.833333) = 6.10556.
Lequation estapproximativement y = 6.106 + 23.826x.
b) Lequation est assez fide`le aux informations :
40
-
c) Pour x = 6, on a y = 6.106 + 23.826(6) = 149.06. Il est
risque destimer unevaleur de y a` partir dune valeur de x qui se
situe a` lexterieur de la zone decollecte des donnees.
[ 3.4 #3.17] a)
41
-
b) Il ny a pas vraiment de tendance prononcee, mais on remarque
quil y a unecorrelation assez positive entre le pre-test et le
post-test.
c) r = sxysxsy . On a sxy = 78.07143, sx = 9.28644 et sy =
11.05613. On a doncr = 78.071439,2864411.05613 = 0.7603958. Cest
une valeur positive, ce qui confirme notreinterpretation du
graphique.
[ 3.4 #3.19] a) La variable dependance est la taille de la
television x, tandis que le prix de latelevision y est la variable
dependante. En effet, on veut predire le prixa` partirde la
taille.
b) La relation est somme toute assez lineaire, quoiquelle est
assez curviligne.
42
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[ 3.4 #3.20] a) r = sxysxsy . On a sxy = 1313.778, sx = 6.219146
et sy = 250.1022. On a doncr = 1313.7786.219146250.1022 =
0.8446014. Cest une valeur fortement positive. Ceciconfirme que la
relation est lineaire.
b) On a b = r sysx = 0.8446014250.10226.219146 = 33.96554. Pour
trouver a = y bx, il faut
y = 812 et x = 43.3, donc a = 812 33.96554(43.3) = 658.7079. On
a donclequation : y = 658.71 + 33.97x.
c) Avec y = 658.71 + 33.97(46), on trouve y = 903.91.d) Faire
une extrapolation, cest-a`-dire utiliser des donnees qui vont
au-dela` de
letendue de la recherche, cela risque de fournir des resultats
qui ne sont pasfiables.
[ 4.3 #4.01] a) = {E1, E2, E3, E4, E5, E6}. On a : E1 =
{Observer un 1}, E2 = {Observerun 2}, E3 = {Observer un 3}, E4 =
{Observer un 4}, E5 = {Observer un 5} etE6 = {Observer un 6}.
b) A = {E2}, B = {E2, E4, E6}, C = {E3, E4, E5, E6}, D = {E2}, E
= {E2, E4, E6},F =
c) P (Ei) = 16d) On a P (A) = P (D) = 16 , P (B) = P (E) = P
(E2)+P (E4)+P (E6) =
12 , P (C) =
P (E3) + P (E4) + P (E5) + P (E6) = 23 , P (F ) = 0. On trouve
:P (Ei) = 1.
[ 4.3 #4.02] a) = {E1, E2, E3, E4, E5}. Ei P (Ei) = P (E1) + P
(E2) + P (E3) + P (E4) +P (E5) = 1. Alors 3P (E5) = 0.3 et P (E5) =
0.1. Et P (E4) = 0.2
b) P (A) = P (E1) + P (E2) + P (E3) = 0.15 + 0.4 + 0.2 = 0.75, P
(B) = P (E2) +P (E3) = 0.15 + 0.4 = 0.55
43
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c) (A B) = {E1, E2, E3, E4}d) (A B) = {E3}
[ 4.3 #4.03] = {E1, E2, ..., E10}. P (E1) = 0.45, P (E2) = 0.453
= 0.15. On aP (Ei) =
0.6 + P (E3) + P (E4) + ...+ P (E10) = 1. Donc, P (E3) + P (E4)
+ ...+ P (E10) = 0.4.Comme ces derniers sont equiprobables, on a P
(E3) = P (E4) = P (E5) = P (E6) =P (E7) = P (E8) = P (E9) = P (E10)
= 0.48 = 0.05.
[ 4.3 #4.04] = {E1, E2, E3, E4}. On a E1 = {HH} = 0.49, E2 =
{HM}, E3 = {MH} = 0.21et E4 = {MM} = 0.09.
a) Sachant queP (Ei) = 1, on a P (E2) = 1 0.49 0.21 0.09 =
0.21.
b) A = {E1, E2, E3}. P (A) = {la probabilite que le joueur
atteigne la cible aumoins une fois} = P (E1) + P (E2) + P (E3) =
0.49 + 0.21 + 0.21 = 0.91.
[ 4.3 #4.05] a) = {E1, E2, ..., E24}. On a : E1 = {DNQ}, E2 =
{DNL}, E3 = {DQN},E4 = {DQL}, E5 = {DLN}, E6 = {DLQ}, E7 = {NDQ},
E8 = {NDL},E9 = {NQL}, E10 = {NQD}, E11 = {NLQ}, E12 = {DLN}, E13 =
{QDN},E14 = {QDL}, E15 = {QNL}, E16 = {QND}, E17 = {QLN}, E18 =
{QLD},E19 = {LDN}, E20 = {LDQ}, E21 = {LNQ}, E22 = {LND}, E23 =
{LQN},E24 = {LQD}.
b) A = {E2, E4, E5, E6, E8, E9, E11, E12, E14, E15, E17, E18,
E19, E20, E21, E22, E23, E24}.Et P (A) = 1824 =
34 .
c) Si B = {La somme des piges egale ou depasse 1.10 $}, on voit
que cela nestpossible que sil y a un dollars dans la pige, donc P
(B) = P (A) = 34 .
[ 4.3 #4.07] = {E1, E2, ..., E20}. On a : E1 = {R1R2}, E2 =
{R1R3}, E3 = {R1J1}, E4 ={R1J2}, E5 = {R2R1}, E6 = {R2R3}, E7 =
{R2J1}, E8 = {R2J2}, E9 = {R3R1},E10 = {R3R2}, E11 = {R3J1}, E12 =
{R3J2}, E13 = {J1R1}, E14 = {J1R2}, E15 ={J1R3}, E16 = {J1J2}, E17
= {J2R1}, E18 = {J2R2}, E19 = {J2R3}, E20 = {J2J1}.
[ 4.3 #4.08] E21 = {R1R1}, E17 = {R2R2}, E18 = {R3R3}, E19 =
{J1J1}, E20 = {J2J2}.[ 4.3 #4.09] = {E1, E2, E3, E4}. E1 = {BeUt},
E2 = {BeNUt}, E3 = {SuperfUt}, E4 =
{SuperfNUt}.a) On a : P (E1) = 0.44, P (E2) = 0.14, P (E3) =
0.02 et P (E4) = 0.4. On
definit A = {Ladulte a besoin de lunettes} = {E1, E2}. On a donc
: P (A) =P (E1) + P (E2) = 0.44 + 0.14 = 0.58.
b) On cherche P (E2) = 0.14.c) C = {Ladulte utilise des
lunettes, quil en ait ou non besoin} = {E1, E3}. On
a donc P (C) = P (E1) + P (E3) = 0.44 + 0.02 = 0.46.
[ 4.3 #4.14] a) Il y a 4 3 2 1 = 24 evenements dans = {JBED,
JBDE, JEBD, JEDB,JDEB,
JDBE,BJED,BJDE,BDJE,BDEJ,BEJD,BEDJ,EJBD,EJDB,EBJD,EBDJ,EDBJ,EDJD,DJBE,DJEB,DBJE,DBEJ,DEBJ,DEJB}
b) P (Ei) = 124 .c) A = {Dave gagne la course} = {DJBE, DJEB,
DBJE, DBEJ, DEBJ, DEJB}.
P (A) = 624 =14 .
44
-
d) B = {Dave gagne et John est second} = {DJBE, DJEB}. P (B) =
224 = 112 .e) C = {Ed arrive dernier}. Or, P (C) = P (A) = 14 .
[ 4.3 #4.15] = {E1, E2, E3, E4}. E1 = {AnOn}, E2 = {AmOn}, E3 =
{AnOv}, E4 = {AmOv}.a) On a P (E1) = 140300 , P (E2) =
6300 , P (E3) =
3300 , P (E4) =
151300 . A = {E1}. On a
P (A) = 140300 = 0.467.b) B = {La mouche a des yeux vermillons}
= {E3, E4} On a P (B) = P (E3) +
P (E4) = 3+151300 =154300 = 0.513.
c) C = {La mouche a des yeux vermillons ou des ailes miniatures}
= {E2, E3, E4}.On a P (C) = P (E2) + P (E3) + P (E4) = 6+3+151300
=
160300 = 0.533
[ 4.6 #4.41] a) Pour P (A) = 0.3 et P (B) = 0.4 et A et B
disjoints, on a : P (AB) = P (A|B) =0 et P (A B) = P (A) + P (B) =
0.3 + 0.4 = 0.7.
b) Pour P (A) = 0.3 et P (B) = 0.4 et A et B independants, on a
: P (A B) =P (A)P (B) = 0.3 0.4 = 0.12, P (A B) = P (A) + P (B) P
(A B) = 0.3 +0.4 0.12 = 0.58
c) Si P (A) = 0.1, P (B) = 0.5 et P (A|B) = 0.1 et que P (A) = P
(A|B), cest queles evenements A et B sont independants. Dans ce
cas, on trouve : P (AB) =P (A)P (B) = 0.1 0.5 = 0.05 et P (A B) = P
(A) + P (B) P (A B) =0.1 + 0.5 0.05 = 0.55.
d) Si P (A B) = 0, alors que P (A) = 0.2 et P (B) = 0.5, cest
que les evenementsA et B sont disjoints. On trouve : P (A B) = P
(A) + P (B) = 0.2 + 0.5 = 0.7et P (A|B) = 0.
[ 4.6 #4.42] Chaque evenement a une probabilite equivalente de
1/5.a) On a : Ac = {E2, E4, E5}. P (Ac) = P (E1) + P (E2) + P (E3)
= 35 = 0.6b) On a : (A B) = {E1}. P (A B) = P (E1) = 15c) On a : (B
C) = {E4}. P (B C) = P (E4) = 15d) On a : (A B) = {E1, E2, E3, E4,
E5}. P (A B) = P (E1) + P (E2) + P (E3) +
P (E4) + P (E5) = 1e) On a : (B|C) = {E4}. On a P (B|C) = P (E4)
= 12 (car il y a une chance sur
deux que E4 apparaisse dans C).
f) On a : (A|B) = {E1}. P (A|B) = P (E1) = 14 (car il y a une
chance sur quatreque E1 apparaisse dans B).
g) On a : (A B C) = {E1, E2, E3, E4, E5}. P (A B C) = P (E1) + P
(E2) +P (E3) + P (E4) + P (E5) = 1
h) On a : (A B)c = {E2, E3, E4, E5}. P ((A B)c) = P (E2) + P
(E3) + P (E4) +P (E5) = 45
[ 4.6 #4.43] a) P (Ac) = 1 P (A) = 1 0.4 = 0.6b) P ((A B)c) = 1
P (A B) = 1 15 = 45
Les resultats sont les memes.
[ 4.6 #4.44] a) P (A|B) = P (AB)P (B) =1545
= 14
45
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b) P (B|C) = P (BC)P (C) =1525
= 12Les resultats sont les memes.
[ 4.6 #4.45] a) P (A B) = P (A) + P (B) P (A B) = 25 + 45 15 =
1b) P (A B) = P (A|B)P (B) = 14 45 = 15c) P (B C) = P (B|C)P (C) =
12 25 = 15
Les resultats sont les memes.
[ 4.6 #4.46] a) Ici, P (A|B) 6= P (A), car 14 6= 25 . Aussi P
(AB) 6= P (A)P (B), car 15 6= 25 45 . Onconclut que A et B ne sont
pas des evenements independants, mais dependants.
b) Ils ne sont pas disjoints, car P (A B) 6= 0 et P (A|B) 6= 0.[
4.6 #4.47] = {E1, E2, E3, E4, E5, E6} et E1 = {1}, E2 = {2}, E3 =
{3}, E4 = {4}, E5 = {5}
et E6 = {6}. On definit : A = {E1, E2, E3}, B = {E1, E2} et C =
{E4, E5, E6}.Comme les evenements elementaires sont equiprobables,
on trouve : P (A) = 12 ,P (B) = 13 et P (C) =
12 .
a) = Ei = 1.b) On a : (A|B) = {E1, E2} et P (A|B) = 1. Autre
solution : P (A|B) = P (AB)P (B) =
1313
= 1
c) P (B) = P (E1) + P (E2) = 13d) On a : (A B C) = , alors P (A
B C) = 0e) On a : (A B) = {E1, E2} = 13 . Autre solution : (A B) =
P (B)(A|B) = 13f) On a : (A C) = , alors P (A C) = 0g) On a : (B C)
= , alors P (B C) = 0h) On a : (A C) = {E1, E2, E3, E4, E5, E6},
alors P (A C) = 1.i) On a : (B C) = {E1, E2, E4, E5, E6}, alors P
(B C) = 56 .
[ 4.6 #4.48] a) P (A|B) 6= P (A), car 1 6= 12 . Aussi P (A B) 6=
P (A)P (B), car 13 6= 16 . Onconclut que A et B ne sont pas des
evenements independants, mais dependants.Ils ne sont pas non plus
disjoints, car (A B) 6= 0 et (A|B) 6= 0.
b) On a P (A C) = 0. Ils sont disjoints et donc dependants.[ 4.6
#4.49] a) P (A B) = P (A)P (B) = 0.4 0.2 = 0.08
b) P (A B) = P (A) + P (B) P (A)P (B) = 0.4 + 0.2 0.08 = 0.52[
4.6 #4.50] a) P (A B) = 0
b) P (A B) = P (A) + P (B) = 0.3 + 0.5 = 0.8
[ 4.6 #4.51] a) P (B|A) = P (BA)P (A) = 0.120.4 = 0.3b) Non, car
P (A B) 6= 0.c) P (B|A) = P (B), alors A et B sont des evenements
independants. On peut
egalement verifier que : P (A B) = P (A)P (B).
46
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[ 4.6 #4.52] a) P (A) = P (A B) + P (A Bc) = 0.34 + 0.15 =
0.49b) P (B) = P (A B) + P (B Ac) = 0.34 + 0.46 = 0.8c) P (A B) =
0.34d) P (A B) = P (A) + P (B) P (A B) = 0.49 + 0.8 0.34 = 0.95e) P
(A|B) = P (AB)P (B) = 0.340.8 = 0.425f) P (B|A) = P (BA)P (A) =
0.340.49 = 0.6938776
[ 4.6 #4.53] a) Non, car P (A B) 6= 0 et P (A|B) 6= 0.b) Non,
car P (A) 6= P (A|B). On a aussi que P (A B) 6= P (A)P (B).
[ 4.6 #4.54] On a A = {la personne utilise la drogue}, B = {le
test reve`le que oui}. On aP (B|A) = P (Bc|Ac) = 0.98 et P (Bc|A) =
(B|Ac) = 0.02.
a) On a C = {Le non-utilisateur de drogue echoue a` se faire
reperer aux deuxtests.}. On cherche : P (C) = P (B|Ac) P (B|Ac).
Comme les tests sontindependants, on a P (C) = P (B|Ac) P (B|Ac) =
P (B|Ac)P (B|Ac) = 0.02 0.02 = 0.0004.
b) On a D = {Lutilisateur de drogue est repere a` au moins un
des deux tests.}.On cherche : P (D) = (P (Bc|A) P (B|A)) (P (B|A) P
(Bc|A)) (P (B|A) P (B|A)). Comme les tests sont independants et que
ces trois evenementssont disjoints, on trouve : P (D) = (P (Bc|A)P
(B|A)) + (P (B|A)P (Bc|A)) +(P (B|A)P (B|A)) = (0.02)(0.98) +
(0.98)(0.02) + (0.98)(0.98) = 0.9996
c) On a E = {Lutilisateur de drogue passe les deux tests sans se
faire reperer.}.On cherche : P (E) = P (B|A) P (B|A). Comme les
tests sont independants,on a P (B|A) P (B|A) = P (Bc|A)P (Bc|A) =
0.02 0.02 = 0.0004
[ 4.6 #4.60] On a A = {Choisir Tim Horton}, B = {Choisir
Starbucks} et C = {Acheter un cafesans cafeine}. On a P (A) = 0.7,
P (B) = 0.3 et P (C) = 0.6.
a) On cherche P (A C) = P (A)P (C) = 0.7 0.6 = 0.42b) Les
evenements sont independants, car il est specifie que le choix de
cafe ne
depend pas de letablissement choisi.
c) On a P (BC) = P (B)P (C) = 0.30.6 = 0.18. On cherche P (B|C)
= P (BC)P (B) =0.180.6 = 0.3.
d) On cherche P (P (A)P (C) = P (A)+P (C)P (AC) = 0.7+0.60.42 =
0.88[ 4.6 #4.61] On a A = {Le premier inspecteur decouvre la
defaillance}, B = {Le deuxie`me in-
specteur decouvre la defaillance}. On donne P (A) = 0.1, P
(Bc|A) = 0.5. On trouveP (B|A) = 10.5. On cherche P (A) P (B|A).
Comme ces deux evenements sontindependants, on a : P (A) P (B|A) =
P (A)P (B|A) = 0.1 00.5 = 0.05
[ 4.6 #4.62] A = {Fumeur}, B = {Mort du cancer}. On donne P (A)
= 0.2, P (Ac) = 10.20.8,P (B) = 0.006. On a P (B|A) = 10P (B|Ac).
On a P (B) = P (BA)(BAc). Donc,comme il sagit de deux evenements
independants, on trouve P (B) = P (A)P (B|A)P (Ac)P (B|Ac) = P (A)P
(B|A) + P (Ac)P (B|Ac) = 0.2P (B|A) + 0.8P (B|A)10 = 0.006.Donc P
(B|A) = 0.0214.
47
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[ 4.6 #4.63] On a : A = {A detecte}, B = {B detecte}. On a P (A)
= 0.95, P (B) = 0.98 etP (A B) = 0.94.
a) On cherche P (A B) = P (A) + P (B) P (A B) = 0.95 + 0.98 0.94
= 0.99b) On cherche P (Ac Bc) = P ((A B)c) = 1 P ((A B)) = 0.01
[ 4.6 #4.65] a) P (A) = 541029b) P (F ) = 5171029c) P (A F ) =
371029d) P (F |A) = 3754e) P (F |B) = 6499f) P (F |C) = 138241g) P
(C|M) = 103512h) P (Bc) = 1 991029 = 9301029 = 310343
[ 4.6 #4.69] On a P (S1) = 0.7, P (S2) = 0.3, P (A|S1) = 0.2 et
P (A|S2) = 0.3.a) On cherche P (A) = P (S1)P (A|S1) + P (S2)P
(A|S2) = 0.7 0.2 + 0.3 0.3 = 0.23b) On cherche P (S1|A) = P (S1)P
(A|S1)P (A) = 0.70.20.23 = 0.609. On cherche P (S2|A) =
P (S2)P (A|S2)P (A) =
(0.3)(0.3)0.23 = 0.391.
[ 4.6 #4.70] On a P (S1) = 0.2, P (S2) = 0.5, P (S3) = 0.3, P
(A|S1) = 0.2, P (A|S2) = 0.1 etP (A|S3) = 0.3. On trouve P (A) = P
(S1)P (A|S1)+P (S2)P (A|S2)+P (S3)P (A|S3) =(0.2)(0.2) + (0.5)(0.1)
+ (0.3)(0.3) = 0.18. On cherche P (S1|A) = P (S1)P (A|S1)P (A)
=(0.2)(0.2)
0.18 = 0.222. On cherche P (S2|A) = P (S2)P (A|S2)P (A) =
(0.5)(0.1)0.18 = 0.278. Oncherche P (S3|A) = P (S3)P (A|S3)P (A) =
(0.3)(0.3)0.18 = 0.5.
[ 4.7 #4.71] On a P (A) = P (S1)P (A|S1) + P (S2)P (A|S2) =
(0.6)(0.3) + (0.4)(0.5) = 0.38[ 4.7 #4.72] On definit V = {Crime
violent}, N = {Crime non-violent} etE = {Crime enregistre}.
On a P (V ) = 0.2, P (N) = 1P (V ) = 10.2 = 0.8, P (E|V ) = 0.9
et P (E|N) = 0.7.a) On cherche P (E) = P (V )P (E|V ) + P (N)P
(E|N) = (0.2)(0.9) + (0.8)(0.7) =
0.74b) On utilise la re`gle de Bayes. On cherche P (V |E) = P (V
)P (E|V )P (E) = (0.2)(0.9)(0.74) =
0.243. On cherche P (N |E) = P (N)P (E|N)P (E) =
(0.8)(0.7)(0.74) = 0.757 On a aussiP (N |E) = 1 P (V |E) = 1 0.243
= 0.757
c) Comme la proportion de crimes non-violents est plus elevee au
depart (P (A)
2) = 0.2 + 0.1 = 0.3e) P (x 3) = 0.1 + 0.3 + 0.3 + 0.2 = 1 0.1 =
0.9
[ 4.8 #4.85] = xp(x) = (0.1)(0) + (1)(0.4) + (2)(0.4) + (3)(0.1)
= 1.5[ 4.8 #4.86] On definit D : (la personne aime David
Letterman), J : (la personne aime Jay Leno).
On sait que P (J) = 0.52
a) On a = (DDD,DJD,DJJ,DDJ,JDD,JDJ,JJD,JJJ). Si x =(le nombre de
per-sonnes qui aiment Jay Leno), on a : x = 0, 1, 2, 3. P (0) = P
(DDD) = (0.48)3 =0.1106. P (1) = P (DJD) + P (DDJ) + P (JDD) =
3(0.52)2(0.48) = 0.3594.P (2) = P (DJJ) + P (JDJ) + P (JJD) =
3(0.52)(0.48) = 0.3894. On a P (3) =P (JJJ) = (0.52)3 = 0.1406
b)
c) P (1) = 0.3594d) = xp(x) = (0)(0.1106) + (1)(0.3594) +
(2)(0.3894) + (3)(0.1406) = 1.56 et
2 = (x)2p(x) =
(01.56)2(0.1106)+(11.56)2(3594)+(21.56)2(0.3894)+(3 1.56)2(0.1406)
= 0.7488
[ 4.8 #4.89] On a comme possibilites : E1 = {F1F2} E2 = {F1M1}
E3 = {F1M2} E4 = {F1M3}E5 = {F2F1} E6 = {F2M1} E7 = {F2M2} E8 =
{F2M3} E9 = {M1F1} E10 ={M1F2} E11 = {M1M2} E12 = {M1M3} E13 =
{M2F1} E14 = {M2F2} E15 ={M2M1} E16 = {M2M3} E17 = {M3F1} E18 =
{M3F2} E19 = {M3M1} E20 ={M3M2}
a) Pour p(x), on a p(0) = P (E11, E12, E15, E16, E19, E20) = 620
=310 , p(1) =
P (E2, ..., E10, E13, E14, E17, E18) = 1220 =35 et p(2) = P (E1,
E5) =
220 =
110 .
b)
[ 4.8 #4.90][ 4.8 #4.95] On a x1 = D sil ny a aucun vol. On a x2
= (1000D) sil y a vol. On a p(x1) =
0.99 et p(x2) = 0.01. On cherche =xp(x) = D(0.99)+(D50000)(0.01)
= 1000.
On trouve D 500 = 1000 et D = 1500.[ 4.8 #4.96][ 4.8 #4.98][ 4.8
#4.99][ 4.8 #4.102][ 4.8 #4.103]
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[ 4.8 #4.106][ 4.8 #4.107][ 4.8 #4.114][ 4.8 #4.121][ 4.8
#4.122][ 4.8 #4.130]
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