Capítulo CD 6 _(1 - 20_)
-
Upload
elmer-antonio-guerrero-saucedo -
Category
Documents
-
view
12 -
download
2
description
Transcript of Capítulo CD 6 _(1 - 20_)
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
6 Distribuciones de probabilidad
Distribucin binomial de Poisson Hipergeomtrica y normal
EJERCICIOS RESUELTOS
Se presenta el desarrollo de los 210 ejercicios que tiene este captulo
1. Solucin:
( ) %5,37375,0166
21
21 2424
22 ===
=
=CP x
22121
4
=
=
=
=
Xqpn
( ) %5,372 ==xP
(exactamente dos caras)
2. Solucin:
( )13
433 2
121
==
CP x
( ) %2525,0164
1614
21
21
!1!3!4
3 ===
=
=
=xP
32121
4
=
=
=
=
X
qpn
( ) %0,253 ==xP
(exactamente 3 caras)
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
2
3. Solucin:
( )
=
== 36
25361
!2!2!4
65
61 224
22 CP x
( ) %57,111157,0296.1150
296.1256
296.125
234
2 ===
=
=
=xP
26561
4
=
=
=
=
X
qpn
( ) %57,112 ==xP
(exactamente dos cincos)
4. Solucin:
a) 8=n ( )ganarP 8,0= 2,0=q 2=X ( ) ?2 ==xP
( ) ( ) ( ) ( ) %1146,0001146,02,08,0 62822 ====xP ( ) %1146,02 ==xP
b) 8=n ( )perderP 2,0= 8,0=q 2=X ( ) ?2 ==xP
( ) ( ) ( ) ( ) %36,292936,08,02,0 62822 ====xP ( ) %36,292 ==xP
c) 8=n ( )perderP 2,0= 8,0=q
8,7,6,5,4,3,2)2( ydosmnimox == ( ) ?2 =xP
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]
( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]7181808021087654322
8,02,08,02,01
1
+=
+=++++++=
x
x
P
PPPPPPPPPP
( ) [ ] %67,494967,05033,013355,01678,012 ===+=xP ( ) %67,492 =xP
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
3
d) 8=n ( )ganarP 8,0= 2,0=q 6,5,4,3,2,1,0 yX = ( ) ?6 =xP
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]088817876
8765432106
2,08,02,08,01
1
++=
+=++++++=
x
x
P
PPPPPPPPPP
( ) [ ] %67,494967,05033,011678,03355,016 ===+=xP ( ) %67,496 =xP
e) 8=n ( )perderp 2,0= 8,0=q 6=X ( )6=xP
( ) ( ) ( ) ( ) %1147,0001147,08,02,0 26866 ====xP Observemos que decir: seis pierdan es lo mismo que dos ganen
8=n ( )ganarp 8,0= 2,0=q 2=X ( )2=xP
( ) ( ) ( ) ( ) %1147,0001147,02,08,0 62822 ====xP ( ) %1147,02 ==xP
5. Solucin:
xnxnx qpCP
= 5,021
==p 5,021
==q 6=n
a) ( )
=
== 4
1161
!4!2!6
21
21 246
44 CP x
( ) %44,232344,06415
64115
641
256
4 ===
=
=
=xP ( ) %44,234 ==xP
(exactamente 4 caras)
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
4
b) Como mximo 4 caras
( )24
64
3363
4262
5161
60604 2
121
21
21
21
21
21
21
21
21
+
+
+
+
= CCCCCP x
( ) ( ) ++++= 41
161158
1812016
1411532
121664
1114xP
( ) %06,898906,06457
6415
6420
6415
646
641
4 ===++++=xP ( ) %06,894 =xP
Tambin se puede resolver de la siguiente forma:
( )
+
=
0666
15654 2
121
21
211 CCPx
( ) %06,898906,06457
647
6464
646
64114 ====
+=xP ( ) %06,894 =xP
(mximo 4 caras)
6. Solucin:
Aparicin de un cinco, la probabilidad es 1/6; Aparicin de un seis, la probabilidad es 1/6
31
62
61
61
==+=p 32
31
331 === pq
a) ( ) %80,121280,0187.2280
187.2835
278
811
!3!4!7
32
31 34
474 ===
=
=
==xP
(cuatro xitos) ( ) %80,124 ==xP
b) ( )34
74
6171
70704 3
231
..............32
31
32
31
+
+
= CCCP x
( ) ( ) ++++= 278
8113581
1627135243
329121729
64317187.2
128114xP
( ) ==++++= 187.2088.2
187.2280
187.2560
187.2672
187.2448
187.2128
4xP
%47,959547,0 == (mximo 4 xitos) ( ) %47,954 =xP
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
5
Tambin puede resolverse as:
( )
+
+
=
071625
4 32
317
732
317
632
317
51xP
( ) ( )
+
+
= 1187.2113
2729179
424312114xP
( ) 0453,01187.2991187.2
1187.214
187.28414 ==
++=xP
%47,959547,0 == ( ) %47,954 =xP
7. Solucin:
4=n 10,0=p 90,0=q
a) ( ) ( ) ( ) ( ) ( ) %61,656561,06561,0119,01,0 40400 ===== CP x ( ) %61,650 ==xP
b) ( ) ( ) ( ) ( ) ( ) %16,292916,0729,01,049,01,0 31411 ===== CP x ( ) %16,291 ==xP
c) ( ) ( ) ( ) ( ) ( ) %86,40486,081,001,069,01,0 22422 ===== CP x ( ) %86,42 ==xP
d) ( ) ( ) ( ) ( ) ( ) ( ) ( )2242314140402 9,01,09,01,09,01,0 CCCP x ++=
( ) %63,999963,00486,02916,06561,02 ==++=xP ( ) %36,992 =xP (no ms de dos defectuosos)
8. Solucin:
a) 40,0=p 60,0=q 5=n 2=X
( ) ( ) ( ) ( ) ( ) %56,343456,0216,016,0106,04,0 32522 ===== CP x ( ) %56,342 ==xP
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
6
b) ( ) ( ) ( ) ( ) ( )415150501 6,04,06,04,0 CCP x +=
( ) ( ) ( ) ( ) ( )1296,04,0!4!1!507776,01!5!0
!51 +=xP
( ) ( ) ( ) ( ) ( ) 2592,007776,01296,04,0507776,0111 +=+=xP
%69,333369,0 == (menos de 2 golpes) ( ) %69,331 =xP
9. Solucin:
8=n 5,0=p 5,0=q ,5,4,3,2,1,0=X
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )53836282718180805 5,05,05,05,05,05,05,05,0 CCCCP x +++=
( ) ( ) ( ) ( ) %54,8585543,05,05,05,05,0 35854484 ==++ CC ( ) %54,855 =xP
Es posible resolverlos de la siguiente forma:
( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]0888178726865 5,05,05,05,05,05,01 CCCP x ++=
( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]100396,015,000781,0825,0015625,02815 ++=xP
( ) [ ] %54,8585543,014457,0100396,003124,010937,015 ===++=xP ( ) %54,855 =xP (menos de 6 caras)
10. Solucin:
05,0=p 95,0=q 6=n ,2,1,0=X
( ) ( ) ( ) ( ) ( ) ( ) ( )4262516160602 95,005,095,005,095,005,0 CCCP x ++=
( ) ( ) ( ) ( ) ( ) ( ) ( )814506,00025,015773780,005,06735091,0112 ++=xP
( ) %78,99997768,0030543,0232134,0735091,02 ==++=xP ( ) %78,992 =xP
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
7
11. Solucin:
10,0=p 90,0=q 5=n 0=X
a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) %05,595905,05905,0111,09,09,01,0 055550500 ====== CCP x ( ) %05,590 ==xP
b) ( ) ( ) ( ) ( ) ( ) ( ) ( )0555145423533 9,01,09,01,09,01,0 CCCP x ++=
00856,000001,000045,000810,0 =++= ( ) %856,03 =xP
c) ( ) ( ) ( ) %81,000810,09,01,0 23533 ==== CP x ( ) %81,03 ==xP (exactamente 3 mueran)
12. Solucin:
2,0=p 8,0=q 4=n
a) ( ) ( ) ( ) ( ) ( ) %96,404096,0512,02,048,02,0 31411 ===== CP x ( ) %96,401 ==xP
b) ( ) ( ) ( ) ( ) ( ) %96,404096,04096,0118,02,0 40400 ===== CP x ( ) %96,400 ==xP
c) ( ) ( ) ( ) ( ) ( ) ( ) ( )2242314140402 8,02,08,02,08,02,0 CCCP x ++=
( ) %28,979728,01536,04096,04096,02 ==++=xP ( ) %28,972 =xP (no ms de dos cerrojos sean defectuosos)
13. Solucin:
4,0=p 6,0=q 5=n
a) Que ninguno se gradu:
( ) ( ) ( ) %78,70778,06,04,0 50500 ==== CP x ( ) %78,70 ==xP
b) Que se gradu uno:
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
8
( ) ( ) ( ) %92,252592,06,04,0 41511 ==== CP x ( ) %92,251 ==xP
c) Que se grade al menos uno:
( ) ( ) ( ) %22,929222,00778,016,04,01 50501 ==== CP x ( ) %22,991 =xP
14. Solucin:
61=p 65=q 5=n
a) ( ) %19,404019,0776.7125.3
296.1625
615
65
61 415
11 ===
=
=
=CP x ( ) %19,401 ==xP
b) ( ) %08,161608,0776.7250.1
216125
361106
561 325
22 ===
=
==
CP x ( ) %08,162 ==xP
c) ( ) %21,30321,0776.7250
3625
2161106
561 235
33 ===
=
==
CP x ( ) %21,33 ==xP
d) ( ) %32,00032,0776.725
65
296.1156
561 145
44 ===
=
==
CP x ( ) %32,04 ==xP
e) ( ) ( ) %19,404019,0776.7 125.311656150
500 ==
=
==
CP x (ninguna vez) ( ) %19,400 ==xP
15. Solucin:
10,0=p 90,0=q 4=n
a) ( ) ( ) ( ) %61,656561,09,01,0 40400 ==== CP x ( ) %61,650 ==xP
b) ( ) ( ) ( ) %39,343439,09,01,01 40401 === CP x ( ) %39,341 =xP
c) ( ) ( ) ( ) ( ) ( )314140401 9,01,09,01,0 CCP x +=
%77,949477,02916,06561,0 ==+= ( ) %77,941 =xP 16. Solucin:
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
9
2,0=p 8,0=q 10=n
a) ( ) ( ) ( ) %2,303020,08,02,0 821022 ==== CP x ( ) %2,302 ==xP
b) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]82102911011001003 8,02,08,02,08,02,01 CCCP x ++=
( ) [ ] %22,323222,06778,013020,02684,01074,013 ===++=xP ( ) %22,323 =xP
c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0101010191092810837107461066 8,02,08,02,08,02,08,02,08,02,0 CCCCCP x ++++=
0063,00000,00000,00008,00055,0 =+++= ( ) %63,06 =xP (Se us la tabla para el clculo)
d) ( ) ( ) ( ) %74,101074,08,02,0 1001000 ==== CP x ( ) %74,100 ==xP
17. Solucin:
5,0=p 5,0=q 10=n 0,1,2,3=X
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1001009110182102731033 5,05,05,05,05,05,05,05,0 CCCCP x +++=
%19,171719,00010,00098,00439,01172,0 ==+++= ( ) %19,173 =xP
npE = ( ) 100181719,0100 depersonasE =
18. Solucin:
5,0=p 5,0=q 10=n 10 9 ,8 ,7 yX =
( ) ( ) ( ) ( ) ( ) ( ) ( ) 01010101910928108371077 )5,0()5,0(5,05,05,05,05,05,0 CCCCP x +++=
( ) %19,171719,100010,00098,00439,01172,07 ==+++=xP ( ) %19,177 =xP
19. Solucin:
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
10
15=n 10,0=p 90,0=q
a) ( ) ( ) ( ) %05,10105,09,01,0 1051555 ==== CP x ( ) %05,15 ==xP
b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) +++= 31215124111511510151010 9,01,09,01,09,01,0 CCCP x
( ) ( ) ( ) ( ) ( ) ( ) 0000,09,01,09,01,09,01,0 015151511415142131513 =++ CCC ( ) 010 =xP
(Como se trabaja con cuatro decimales, aproximamos a cero) (Se utiliz la tabla)
A partir de x > 8 la probabilidad obtenida es demasiado pequea, casi cero.
c) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ +++= 1321521411511501505 9,01,09,01,09,01,01 CCCP x
( ) ( ) ( ) ( ) ]114154123153 9,01,09,01,0 CC +
Utilizando la tabla se tiene:
( ) [ ]9873,00428,01285,02669.03432,02059,015 =++++=xP ( ) %27,15 =xP
( ) %27,10127,09873,015 ===xP
20. Solucin:
20=n 25,0=p 75,0=q
a) ( ) ( ) ( ) 0...............0000,075,025,0 515201515 ==== CP x (ver tabla) ( ) 015 ==xP
b) ( ) ( ) ( ) ( ) ( ) ( ) ( )1642041912012002004 75,025,0...........75,025,075,025,0 CCCP x ++=
%48,414148,01897,01339,00669,00211,00032,0 ==++++= ( ) %48,414 =xP
c) ( ) ( ) ( ) ( ) ( ) ( ) ( )02020201192091282088 75,025,0...........75,025,075,025,0 CCCP x ++=
Es ms fcil resolverlo de la siguiente forma:
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
11
( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]1372071912012002008 75,025,0............75,025,075,025,01 CCCP x ++=
[ ] =+++++++= 1124,01686,02023,01897,01339,00669,00211,00032,01
%19,108981,01 == (por lo menos 8 defectuosas) ( ) %19,108 =xP
21. Solucin:
5,0=p 5,0=q 4=n
a) ( ) ( ) ( ) 9375,00625,015,05,01 40401 === CP x ( ) %75,931 =xP
( ) 875.19375,0000.2 ==E familias
b) ( ) ( ) ( ) 3750,05,05,0 22422 === CP x ( ) %50,372 ==xP
( ) familiasE 7503750,0000.2 ==
c) ( ) ( ) ( ) 0625,05,05,0 40400 === CP x ( ) %25,60 ==xP
( ) familiasE 1250625,0000.2 ==
(Se utilizaron las tablas)
22. Solucin:
( ) ( ) ( ) ( ) ( ) ( ) ( )1321521411511501502 95,005,095,005,095,005,0 CCCP x ++=
9639,01348,03658,04633,0 =++= = 96,39% ( ) %39,962 =xP
(Se utiliz la tabla)
23. Solucin:
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
12
40,0=p 20=n
( ) ( ) ( ) ( ) ( ) ( ) ( )02020208122012911201111 6,04,0........6,040,06,04,0 CCCP x +++=
Utilizando la tabla se tendr que:
( ) =+++++++++= 00000003,00013,00049,00146,00355,00710,011xP
%76,121276,0 == (mitad ms uno) ( ) %76,1211 =xP
(Se utiliz la tabla para el clculo)
24. Solucin:
20,0=p 80,0=q 18=n 8=X
( ) ( ) ( ) %20,10120,080,020,0 1081888 ==== CP x ( ) %20,18 ==xP
25. Solucin:
( ) ( ) ( ) ( ) ( ) ( ) ( )37107461065510575 5,05,05,05,05,05,0 CCCP x ++=
( ) %84,565684,01172,02051,02461,075 ==++= xP ( ) %84,5675 = xP
26. Solucin:
5=n ( )3xP 5,4,3=X 5,0=p 5,0=q
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
13
( ) ( ) ( ) ( )5433 === ++= xxxx PPPP
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )055514542353 5,05,05,05,05,05,0 ++=
%505000,003125,015625,03125,0 ==++= ( ) %503 =xP
27. Solucin:
cariescon 90,0109
= %1010,0cariessin == 5=n
a) Cuatro tengan caries 5=n 90,0=p 4=X
( ) ( ) ( ) ( ) %81,3232805,01,09,0 14544 ====xP ( ) %81,324 ==xP
b) Por lo menos dos tengan caries 90,0=p 5,4,3,2=X
( ) ( ) ( ) ( ) ( )54322 ==== +++= xxxxx PPPPP
( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ]41515050
10
1,09,01,09,01
1
+=
+=== xx PP
[ ] %95,999995,000045,000001,01 =+= ( ) %95,992 =xP
c) Por lo menos 2 no tengan caries: 10,0=p 5,4,3,2=X
( ) ( ) ( ) ( ) ( )54322 ==== +++= xxxxx PPPPP
( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ]41515050
10
9,01,09,01,01
1
+=
+=== xx PP
[ ] %15,89185,0132805,059049,01 ==+= ( ) %15,82 =xP
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
14
d) Por lo menos una tenga caries 90,0=p 5,4,3,2,1=X
( ) ( )01 1 = = xx PP
( ) ( ) ( ) %10099999,000001,011,09,01 5050 ==== ( ) %1001 =xP
28. Solucin:
20% pierden el 1 ao pierden lo no 80% 6=n
a) :aprueben 2 Mximo 210 , , X = 800,p =
( ) ( ) ( ) ( )2102 === ++= xxxx PPPP
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4262516160602 2,08,02,08,02,08,0 ++=xP
%70,101696,001536,0001536,0000064,0 ==++= ( ) %70,12 =xP
b) Todos aprueben: 800, p = 6=X
( ) ( ) ( ) ( ) %21,262621,02,08,0 06666 ====xP ( ) %21,266 ==xP
c) Ninguno apruebe 800, p = 0=X
( ) ( ) ( ) ( ) %0064,0000064,02,08,0 60600 ====xP ( ) %0064,00 ==xP
29. Solucin:
0,7000068004
=
.
.
Transporte pblico 30% 0,30 = otro servicio
a) No ms de 2 utilicen transporte pblico 70,p = 2,1,0=X 8=n
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
15
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )6282718180802 3,07,03,07,03,07,0 ++=xP
%13,101129,001000,00012247,00000656,0 ==+== ( ) %13,12 =xP
b) Por lo menos 3 no lo utilicen 30,0=p 8,7,6,5,4=X
( ) ( ) ( ) ( ) ( ) ( ) ( )8765433 ====== +++++= xxxxxxx PPPPPPP
( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]628271818080
210
7,03,07,03,07,03,01
1
++=
++==== xxx PPP
[ ] %82,,444482,02965,01977,00576,01 ==++= ( ) %82,443 =xP
c) Exactamente 2 no lo utilicen 30,0=p 2=X
( ) ( ) ( ) ( ) %65,292965,07,03,0 62822 ====xP ( ) %65,292 ==xP
d) Exactamente 2 lo utilicen 70,0=p 2=X
( ) ( ) ( ) ( ) %10100,03,07,0 62822 ====xP ( ) %12 ==xP
30. Solucin:
60% = 0,60 asisten 0,40 = 40% no asisten n = 8
a) asistan 7 menos loPor 6,0=p 8,7=X
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
16
( ) ( ) ( )877 == += xxx PPP
( ) ( ) ( ) ( ) ( ) ( )08881787 4,06,04,06,0 +=
%64,101064,00168,00896,0 ==+= ( ) %64,107 =xP
b) Por lo menos 2 no asistan 8=n 40,0=p 8,7,6,5,4,3,2=X
( ) ( ) ( ) ( )8322 .................... === +++= xxxx PPPP
( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ]71818080
10
6,04,06,04,01
1
+=
+=== xx PP
[ ] %36,898936,01064,010896,00168,01 ===+= ( ) %36,892 =xP
31. Solucin:
gafasusan 4,02000800
= gafasusan no0,6 = 5n =
a) gafasusan 2 menos loPor 40,0=p 5,4,3,2=X
( ) ( ) ( ) ( ) ( )54322 ==== +++= xxxxx PPPPP
( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ]41515050
10
6,04,06,04,01
1
+=
+=== xx PP
[ ] %3,666630,03370,012592,00778,01 ===+= ( ) %30,662 =xP
b) gafasusan no 2 menos loPor 60,0=p 5,4,3,2=X
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
17
( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %30,9191296,008704,010768,001024,01
4,06,04,06,01
1
4151
5050
102
===+=
+=
+=== xxx PPP
( ) %30,912 =xP
c) ( ) gafasusen no espera se alumnos,200.160,02000 === EnpE
32. Solucin:
repitentesson 33,031 = repitentes no0,67 = 4n =
a) repitentessean dos de mas No 33,0=p 2,1,0=X
( ) ( ) ( ) ( )2102 === ++= xxxx PPPP
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )%18,898918,02933,03970,02015,0
67,033,067,033,067,033,0 2242314
1404
0
==++=
++=
( ) %18,882 =xP 32y31con trabajamosS:Nota ( ) %89,882 =xP
b) repitente sea no 1 menos Al 67,0=p 4,3,2,1=X
( ) ( ) ( ) ( ) ( )43211 ==== +++= xxxxx PPPPP
32y31con os trabajamS :Nota ( ) %77,981 =xP
( ) ( )01 1 = = xx PP
( ) ( ) ( ) %81,989881,00119,0133,067,01 4040 ==== ( ) %81,981 =xP
33. Solucin:
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
18
16=n 6,0=p 16,15,14,13,12,11,10=X
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )313161341216125111611610161010 4,06,04,06,004,06,04,06,0 +++=xP
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) =+++ 016161611516152141614 4,06,04,06,04,06,0
0150,00468,01014,01623,01983,0 ++++=
%71,520003,00030,0 =++ ( ) %71,5210 =xP (diez o ms acontecimientos desfavorables)
34. Solucin:
accidentan se 25% accidentan se no 75%
accidentan se 3 menos loPor 7=n
( ) ( ) ( ) ( ) ( ) ( )765433 ===== ++++= xxxxxx PPPPPP
( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]527261717070
210
75,025,075,025,075,025,01
1
++=
++==== xxx PPP
[ ] %35,242435,07565,013115,03115,01335,01 ===++= ( ) %35,243 =xP
35. Solucin:
3% son defectuosos 97% Buenos n = 7
a) Por lo menos 3 sean buenos
( ) ( ) ( ) ( )7433 ....... === +++= xxxx PPPP
( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]527261717070
210
03,097,003,097,003,097,01
1
++=
++==== xxx PPP
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
19
[ ] %1001 a aproxima se0001 ==++= ( ) %1003 =xP
b) Por lo menos 3 sean defectuosos
( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %09,0%0009,09991,010162,01749,08080,01
97,003,097,003,097,003,01
1
5272
6171
7070
2103
===++=
++=
++==== xxxx PPPP
( ) %09,03 =xP
36. Solucin:
enferman01,0 ==p 5=n enferman no99,0 ==q
a) enfermos2=X
( ) ( ) ( ) ( ) %097,000097,099,001,0 32522 ====xP ( ) %097,02 ==xP
b) enfermo uno menos loPor 5432,1 , , , X =
( ) ( ) ( ) ( ) ( ) %9,4049,09510,0199,001,011 505001 ===== = xx PP ( ) %9,41 =xP
c) Por lo menos 2 no enfermen 5432 , , , X =
( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %1001 a aproxima se001
01,099,001,099,01
1
4151
5050
102
==+=
+=
+=== xxx PPP
( ) %1002 =xP
37. Solucin:
-
Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007
20
20% de mortalidad 80% de sobrevivir 5=n
a) Ninguno sobreviva 0=X ( )mueran todos,5aequivale =x
( ) ( ) ( ) ( ) ( ) ( ) ( ) %032,000032,08,02,02,08,0 055550500 =====xP ( ) %032,00 ==xP
b) Todos sobrevivan
( ) ( ) ( ) ( ) %77,323277,02,08,0 05555 ====xP ( ) %77,325 ==xP
c) Al menos 1 sobrevivan 54321 , , , , X =
( ) ( ) ( ) ( ) ( ) %97,99%968,9900032,012,08,011 505001 ===== = xx PP ( ) %97,991 =xP
d) Al menos 1 no sobrevivan 54321 , , , , X =
( ) ( ) ( ) ( ) ( ) %23,6767232,032768,018,02,011 505001 ===== = xx PP ( ) %23,671 =xP
38. Solucin:
scientfico 20%0,20255
== cientfico no 80%2520
= 4=n
a) Por lo menos 1 sea cientfica 4,3,2,1=X
( ) ( ) ( ) ( ) ( ) %04,595904,04096,018,02,011 404001 ===== = xx PP ( ) %04,591 =xP