capitulo6

\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 --