capitulo6
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capitulo6

Disciplina:Introdução à Probabilidade e A Estatística II202 materiais1.638 seguidores
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\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

=\u2212×
1

0

1
)1|(

11
1

0

1
)1|( \u2212\u2212+

\u2212

=

\u2211 \uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

=\u2212× xnx
n

x

qp
x

n
xnXEp

11
1

0

1 1)( \u2212\u2212+
\u2212

=

\u2212 \u2211 \uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

=+× xnx
n

x

n qp
x

n
qpp

Portanto,

[ ]11)´()1|( 11
1

0

1 +\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

=+×+\u2212× \u2212\u2212+
\u2212

=

\u2212 \u2211 xqpxnqppnXEp xnx
n

x

n

Então:

( ) Aqp
x

n
x

qp
x

n
xqppnXEpnXEq

xnx
n

x

xnx
n

x

n

=\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

++

+\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

=+×++×+\u2212×

\u2212\u2212+
\u2212

=

\u2212

\u2212

=

\u2212

\u2211
\u2211

11
1

0

1

0

1

1
1

1
)´()1|()1|(

Separando o primeiro termo da primeira somatória e o último do segundo, tem-se:

( ) nxnx
n

x

xnx
n

x

n pnqp
x

n
xqp

x
n

xqA ×+\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

++\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

+×= \u2212\u2212+
\u2212

=

\u2212

\u2212

=

\u2211\u2211 112
0

1

0

1
1

1
0

O coeficiente de knkqp \u2212 , para k=1, 2, ..., n-1, será a soma do coeficiente da primeira somatória
quando x=k e o da segunda somatória quando x+1=k, ou seja, x=k-1, logo é igual a:

- cap.6 \u2013 pág. 16 --

bussab&morettin estatística básica

[ ] \uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

×=
\u2212

×=
\u2212

\u2212

×=+\u2212
\u2212

\u2212

×=

=\uf8fa\uf8fb
\uf8f9\uf8ef\uf8f0

\uf8ee
\u2212\u2212

\u2212

+
\u2212

\u2212

×=\uf8fa\uf8fb
\uf8f9\uf8ef\uf8f0

\uf8ee
\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u2212

\u2212

+\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

×=\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u2212

\u2212

×+\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

×

k
n

k
knk

nk
knk
nnkkkn

knk
nk

knk
n

knk
nk

k
n

k
n

k
k
n

k
k

n
k

)!(!
!

)!(!
)!1(

)!(!
)!1(

)!()!1(
)!1(

)!(!
)!1(

1
11

1
11

Substituindo em A, vem:

)|(0
0

1

1

nXEqkppnqp
k
n

kqA
n

k

knknknk
n

k

n
==×+\uf8f7\uf8f7\uf8f8

\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb

+×= \u2211\u2211
=

\u2212\u2212

\u2212

=

Como queríamos provar.

Problema 42.

(a) 705,0)285,0()285,0()135,0()2( =++=\u2264XP

(b) 236,0)154,0()068,0()014,0()2( =++=\u2264XP

(c) 933,0)179,0()377,0()377,0()2( =++=\u2264XP

Problema 43.

kn

n

k

knk

n

knkk

n

knk

n

knk

n

nnn
k

nk

n
n

n
k

nk

n
n

n
knnn

k

n
n

nkkn
npp

k
n

\u2212

\u221e\u2192

\u2212

\u221e\u2192

\u2212

\u221e\u2192

\u2212

\u221e\u2192

\u2212

\u221e\u2192

\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed

\uf8eb
\u2212×\uf8f7\uf8f8

\uf8f6\uf8ec\uf8ed
\uf8eb

\u2212×\uf8fa\uf8fb
\uf8f9\uf8ef\uf8f0

\uf8ee \uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed

\uf8eb \u2212
\u2212\uf8f7\uf8f8

\uf8f6\uf8ec\uf8ed
\uf8eb

\u2212××=

=\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed

\uf8eb \u2212
×\uf8fa\uf8fb

\uf8f9\uf8ef\uf8f0
\uf8ee \uf8f7\uf8f8

\uf8f6\uf8ec\uf8ed
\uf8eb \u2212

\u2212\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed

\uf8eb
\u2212××=

=\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed

\uf8eb \u2212
×\uf8f7\uf8f8

\uf8f6\uf8ec\uf8ed
\uf8eb

×+\u2212××\u2212××=

=\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed

\uf8eb \u2212
×\uf8f7\uf8f8

\uf8f6\uf8ec\uf8ed
\uf8eb

×
\u2212

=\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u3bb\u3bb\u3bb

\u3bb\u3bb

\u3bb\u3bb

\u3bb\u3bb

1111....111lim
!

11....111
!

lim

1)1(....)1(
!

lim

!)!(
!lim)1(lim

Como 11 \u2192\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed

\uf8eb
\u2212

\u2212 k

n
\u3bb ; 111....11 \u2192\uf8f7\uf8f8

\uf8f6\uf8ec\uf8ed
\uf8eb \u2212

\u2212\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed

\uf8eb
\u2212

n
k

n
 e \u3bb\u3bb \u2212\u2192\uf8f7\uf8f8

\uf8f6\uf8ec\uf8ed
\uf8eb

\u2212 e
n

n

1

Então:

!
)1(

k
epp

k
n kknk \u3bb\u3bb ×\u2192\u2212××\uf8f7\uf8f7\uf8f8

\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb \u2212

\u2212 quando \u221e\u2192n

Problema 44.
Usando a propriedade da soma de infinitos termos de uma P.G. de razão menor que 1.

(a)
3
1

411
41

4
1

2
1) (

11
2

=

\u2212

=== \u2211\u2211 \u221e
=

\u221e

= k
k

k
kparXP

(b)
8
7

2
1)3(

3

1

==< \u2211
=k

kXP

(c) 10
11

11 2
1

211
21

2
1)10( =

\u2212

==> \u2211\u221e
=k

kXP

- cap.6 \u2013 pág. 17 --

bussab&morettin estatística básica

Problema 45.

\u2211 \u2211 \u2211 \u2211 \u2211 +=+=+=+=+ bXaExpbxxpaxbpxaxpxpbaXbaXE )()()()()()()()(
( )[ ] [ ] [ ]( )

)()(2)(2

)()()()()(
2

2222222

XVaraXbEXbE

bbXEXEabaXEbaXEbaXVar

=\u2212+

+\u2212+\u2212=+\u2212+=+

[ ] [ ] [ ]22222 )()()()()()( \u2211 \u2211\u2211 \u2211 \u2211 \u2212=\u2212=\u2212= xxpxpxxxpxxpxpxXEXEXVar
Problema 46.

\u3bb\u3bb\u3bb\u3bb\u3bb \u3bb\u3bb\u3bb
\u3bb

======
\u2212

\u221e

=

\u221e

=

\u221e

=

\u2212

\u2212\u2211 \u2211\u2211 eekekekkXkPXE k k
k

k

k

0 11 !!
)()(

( )

\u2211\u2211\u2211
\u2211 \u2211\u2211

\u221e

=

\u2212

\u221e

=

\u2212

\u221e

=

+\u2212

\u221e

=

\u221e

=

\u2212

\u221e

=

\u2212

+=+=+

\u2212==

\u2212

====

0

2

00

1

0 11

222

!!!
)1(

1
)!1(!

)()(

j

j

j

j

j

j
k k

k

k

k

j
e

j
ej

j
ej

kj
k
ek

k
ekkXPkXE

\u3bb\u3bb\u3bb\u3bb\u3bb\u3bb\u3bb

\u3bb\u3bb

\u3bb\u3bb\u3bb

\u3bb\u3bb

[ ] \u3bb\u3bb\u3bb\u3bb =\u2212+=\u2212= 2222 )()()( XEXEXVar
Problema 47.
Para justificar a expressão, considere-se que a probabilidade de se extrair uma amostra com k
elementos marcados é dada pelo quociente entre o número de amostras em que existem k
elementos marcados e o número total de amostras de tamanho n, obtidas, sem reposição, de uma
população de tamanho N.
O número total de amostras de tamanho n, obtidas, sem reposição, de uma população de tamanho

N é dado por \uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb
n
N

.

Para o numerador da expressão a ser provada, deve-se raciocinar da seguinte maneira: é
necessário obter k elementos dentre os r que possuem o tributo e n-k dentre os N-r elementos

restantes. Portanto, justifica-se o valor \uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb
k
r

x \uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u2212

\u2212

kn
rN

e a probabilidade em questão é dada por:

\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u2212

\u2212

×\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

=

n
N

kn
rN

k
r

pk

Problema 48.
Cada resposta é um ensaio de Bernoulli com probabilidade de sucesso 0,50. Desse modo, o
número de respostas corretas, X, tem distribuição binomial com n=50 e p=0,50. Acertar 80% das
questões significa X = 40. Portanto:

61040 109)50,0()50,0(
40
50

)40( \u2212×=××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

==XP

Problema 49.

- cap.6 \u2013 pág. 18 --

bussab&morettin estatística básica

No caso de alternativas por questão, a variável aleatória X segue distribuição binomial com n=50
e p = 0,20. Desse modo,

191040 1021,1)80,0()20,0(
40
50

)40( \u2212×=××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

==XP

Problema 50.

20,012)1(3
3
3

12)1(
2
3

)3(12)2( 3232 =\u21d2=\u2212\u21d2\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

×=\u2212\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u21d2=×== pppppppXPXP Proble

ma 51.
Seja X o número de componentes que funcionam.Tem-se que X ~b (10; p).

(a) 10)10()( pXPfuncionarP ===

(b) 101)10() ( pXPfuncionarnãoP \u2212=<=

(c) 8282 )1(45)1(
2
10

)2( ppppXP \u2212××=\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

==

(d) \u2211
=

\u2212

\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

=\u2265
10

5

10)1(
10

)5(
k

kk pp
k

XP

Problema 52.

),;(
)1)(1(

)(

)1(
)1)(1(

)()1(
1!)!()!1(

!)(

)1(
)!1()!1(

!)1(
1

),;1( 1111

pnkb
pk

pkn

pp
k
n

pk
pknp

p
pp

kknk
nkn

pp
kkn

npp
k
n

pnkb

knkknk

knkknk

\u2212+

\u2212

=

\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

×
\u2212+

\u2212

=\u2212×
\u2212

××
\u2212+

\u2212

=

\u2212××
+\u2212\u2212

=\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

+
=+

\u2212\u2212

\u2212\u2212+\u2212\u2212+

Problema 53.
Para a variável Z, a mediana é qualquer valor pertencente a (1, 2), de acordo com a definição.
Nestes casos costuma-se indicar o ponto médio da classe que é 1,5.

Problema 54.
)25,0(q = qualquer valor entre (0, 1)

2)60,0( =q ,porque 60,075,0)2())60,0(( \u2265=\u2264=\u2264 XPqXP e 40,050,0)2( \u2265=\u2265XP
3)80,0( =q , pois 80,000,1)3( >=\u2264XP e 20,025,0)3( >=\u2265XP

Problema 55.

(e) \u2211\u221e
=

\u2212

=

\u2212\u2212

×=\u2212×
1

1 1
)1(1

1)1(
j

j

p
ppp

(f) \u2211\u2211 \u221e
=

\u221e

=

\u2212 ×=\u2212×=
11

1)1()(
j

j

j

j q
dq
dpppXE , onde 1- p =q.

- cap.6 \u2013 pág. 19 --

bussab&morettin estatística básica

Mas, q
q

dq
dq

dq
d

j

j

\u2212

=\u2211\u221e
=

11
, pois a série \u2211\u221e

= 1j

jq é convergente

Logo,

pq
pXE 1

)1(
1)( 2 =

\u2212

×=

Mesmo raciocínio para a Var(X).

(g)
( ) )()1(

)1(
1

)1(

)1(

)(
)()|( 1

1

1

1 tXPp
p
p

pp

pp

sXP
tsXPsXtsXP ts

ts

sj

j

tsj

j

\u2265=\u2212=
\u2212

\u2212

=

×\u2212

×\u2212

=

>

+>
=>+>

+

++

\u221e

+=

\u221e

++=

\u2211
\u2211

Problema 56.
Considere:
C: custo do exp.
X: nº de provas para sucesso.

)1(3001000 \u2212+= XXC
Portanto,.

6200300
2,0

11300300)(1300)( =\u2212×=\u2212= XECE

Problema 57.

{ }\u2211
=

\u2212====+
n

k
knYkXPnYXP

0

,)( , pois o evento { nYX =+ } pode ser escrito como a união

de eventos disjuntos { }knYkX \u2212== , , n=0,.....
{ } { } { }

\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb +

=\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u2212

×\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb +

=\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u2212

×\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

=

=\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

\u2212

×\u2212××\uf8f7\uf8f7\uf8f8
\uf8f6

\uf8ec\uf8ec\uf8ed
\uf8eb

=

=\u2212=×==\u2212====+

\u2211\u2211
\u2211

\u2211\u2211

==

=

+\u2212\u2212\u2212

==

m
nm

kn
m

k
n

pp
m
nm

pp
kn

m
k
n

pp
kn

m
pp

k
n

knYPkXPknYkXPnYXP

n

k

mn
n

k

mn

n

k

knmknknk

n

k

n

k

00

0

00

 pois ,)1()1(

)1()1(

,)(

- cap.6 \u2013 pág. 20 --