Capítulo 6 cd (1 20-)

Estadística y muestreo, 12ª.ed. (Segunda reimpresión) – CD Cap.6 Distribuciones de probabilidad Ciro Martínez Bencardino – Ecoe Ediciones Actualizado en diciembre de 2007

6 Distribuciones de probabilidad

Distribución binomial – de Poisson – Hipergeométrica y normal

EJERCICIOS RESUELTOS

Se presenta el desarrollo de los 210 ejercicios que tiene este capítulo 1. Solución:

( ) %5,37375,0166

21

21

242422 ===

=

−

= CPx

22121

4

====

X

qpn

( ) %5,372 ==xP

(exactamente dos caras) 2. Solución:

( )13

433 2

121

== CPx

( ) %2525,016

4

16

14

2

1

2

1

!1!3

!43 ===

=

==xP

3

21

21

4

====

X

q

p

n

( ) %0,253 ==xP

(exactamente 3 caras)


2

3. Solución:

( )

=

== 36

25361

!2!2!4

65

61

22422 CPx

( ) %57,111157,0296.1

150

296.1

256

296.1

25

2

342 ===

=

⋅==xP

2

65

61

4

====

X

q

p

n

( ) %57,112 ==xP

(exactamente dos cincos) 4. Solución: 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

( ) ( ) ( ) ( ) %36,292936,08,02,0 628

22 ====xP ( ) %36,292 ==xP

c) 8=n ( )perderP 2,0= 8,0=q 8,7,6,5,4,3,2)2( ydosmínimox == ( ) ?2 =≥xP

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]7181

80802

1087654322

8,02,08,02,01

1

+−=

+−=++++++=

≥

≥

x

x

P

PPPPPPPPPP

( ) [ ] %67,494967,05033,013355,01678,012 ==−=+−=≥xP ( ) %67,492 =≥xP


3

d) 8=n ( )ganarP 8,0= 2,0=q 6,5,4,3,2,1,0 yX = ( ) ?6 =≤xP

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]0888

17876

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

( ) ( ) ( ) ( ) %1147,0001147,08,02,0 268

66 ====xP

Observemos que decir: seis pierdan es lo mismo que dos ganen 8=n ( )ganarp 8,0= 2,0=q 2=X ( )2=xP

( ) ( ) ( ) ( ) %1147,0001147,02,08,0 628

22 ====xP ( ) %1147,02 ==xP

5. Solución:

xnxnx qpCP −= 5,0

21 ==p 5,0

21 ==q 6=n

a) ( )

=

== 4

1161

!4!2!6

21

21

24644 CPx

( ) %44,232344,064

15

64

115

64

1

2

564 ===

=

×==xP ( ) %44,234 ==xP

(exactamente 4 caras)


4

b) Como máximo 4 caras

( )24

64

3363

4262

5161

60604 2

121

21

21

21

21

21

21

21

21

+

+

+

+

=≤ CCCCCPx

( ) ( )

+

+

+

+

=≤ 4

116115

81

8120

161

4115

321

216

641114xP

( ) %06,898906,064

57

64

15

64

20

64

15

64

6

64

14 ===++++=≤xP ( ) %06,894 =≤xP

También se puede resolver de la siguiente forma:

( )

+

−=≤

0666

15654 2

121

21

211 CCPx

( ) %06,898906,064

57

64

7

64

64

64

6

64

114 ===−=

+−=≤xP ( ) %06,894 =≤xP

(máximo 4 caras) 6. Solución: Aparición de un cinco, la probabilidad es 1/6; Aparición de un seis, la probabilidad es 1/6

31

62

61

61 ==+=p

32

31

331 =−=−= pq

a) ( ) %80,121280,0187.2

280187.2835

278

811

!3!4!7

32

31 34

474 ===

=

=

==xP

(cuatro éxitos) ( ) %80,124 ==xP

b) ( )34

74

6171

70704 3

231..............

32

31

32

31

+

+

=≤ CCCPx

( ) ( )

+

+

+

+

=≤ 27

881135

8116

27135

24332

9121

72964

317

187.2128114xP

( ) ==++++=≤ 187.2088.2

187.2280

187.2560

187.2672

187.2448

187.2128

4xP

%47,959547,0 == (máximo 4 éxitos) ( ) %47,954 =≤xP


5

También puede resolverse así:

( )

+

+

−=≤

071625

4 3

2

3

1773

2

3

1763

2

3

175

1xP

( ) ( )

+

+

−=≤ 1

187.211

32

72917

94

24312114xP

( ) 0453,01187.2991

187.21

187.214

187.28414 −=−=

++−=≤xP

%47,959547,0 == ( ) %47,954 =≤xP

7. Solución:

4=n 10,0=p 90,0=q a) ( ) ( ) ( ) ( ) ( ) %61,656561,06561,0119,01,0 404

00 ===== CPx ( ) %61,650 ==xP

b) ( ) ( ) ( ) ( ) ( ) %16,292916,0729,01,049,01,0 314

11 ===== CPx ( ) %16,291 ==xP

c) ( ) ( ) ( ) ( ) ( ) %86,40486,081,001,069,01,0 224

22 ===== CPx ( ) %86,42 ==xP

d) ( ) ( ) ( ) ( ) ( ) ( ) ( )224

2314

1404

02 9,01,09,01,09,01,0 CCCPx ++=≤

( ) %63,999963,00486,02916,06561,02 ==++=≤xP ( ) %36,992 =≤xP (no más de dos defectuosos) 8. Solución: a) 40,0=p 60,0=q 5=n 2=X ( ) ( ) ( ) ( ) ( ) %56,343456,0216,016,0106,04,0 325

22 ===== CPx ( ) %56,342 ==xP


6

b) ( ) ( ) ( ) ( ) ( )415

1505

01 6,04,06,04,0 CCPx +=≤

( ) ( ) ( ) ( ) ( )1296,04,0!4!1

!507776,01

!5!0!5

1 +=≤xP

( ) ( ) ( ) ( ) ( ) 2592,007776,01296,04,0507776,0111 +=+=≤xP

%69,333369,0 == (menos de 2 golpes) ( ) %69,331 =≤xP

9. Solución:

8=n 5,0=p 5,0=q ,5,4,3,2,1,0=X

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )5383

6282

7181

80805 5,05,05,05,05,05,05,05,0 CCCCPx +++=≤

( ) ( ) ( ) ( ) %54,8585543,05,05,05,05,0 358

5448

4 ==++ CC ( ) %54,855 =≤xP

Es posible resolverlos de la siguiente forma:

( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]0888

1787

26865 5,05,05,05,05,05,01 CCCPx ++−=≤

( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]100396,015,000781,0825,0015625,02815 ++−=≤xP

( ) [ ] %54,8585543,014457,0100396,003124,010937,015 ==−=++−=≤xP ( ) %54,855 =≤xP (menos de 6 caras) 10. Solución:

05,0=p 95,0=q 6=n ,2,1,0=X

( ) ( ) ( ) ( ) ( ) ( ) ( )4262

5161

60602 95,005,095,005,095,005,0 CCCPx ++=≤

( ) ( ) ( ) ( ) ( ) ( ) ( )814506,00025,015773780,005,06735091,0112 ++=≤xP

( ) %78,99997768,0030543,0232134,0735091,02 ==++=≤xP ( ) %78,992 =≤xP


7

11. Solución:

10,0=p 90,0=q 5=n 0=X a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) %05,595905,05905,0111,09,09,01,0 055

5505

00 ====== CCPx ( ) %05,590 ==xP

b) ( ) ( ) ( ) ( ) ( ) ( ) ( )055

5145

4235

33 9,01,09,01,09,01,0 CCCPx ++=≥

00856,000001,000045,000810,0 =++= ( ) %856,03 =≥xP

c) ( ) ( ) ( ) %81,000810,09,01,0 235

33 ==== CPx ( ) %81,03 ==xP (exactamente 3 mueran) 12. Solución:

2,0=p 8,0=q 4=n a) ( ) ( ) ( ) ( ) ( ) %96,404096,0512,02,048,02,0 314

11 ===== CPx ( ) %96,401 ==xP

b) ( ) ( ) ( ) ( ) ( ) %96,404096,04096,0118,02,0 404

00 ===== CPx ( ) %96,400 ==xP

c) ( ) ( ) ( ) ( ) ( ) ( ) ( )224

2314

1404

02 8,02,08,02,08,02,0 CCCPx ++=≤

( ) %28,979728,01536,04096,04096,02 ==++=≤xP ( ) %28,972 =≤xP (no más de dos cerrojos sean defectuosos) 13. Solución:

4,0=p 6,0=q 5=n a) Que ninguno se gradué: ( ) ( ) ( ) %78,70778,06,04,0 505

00 ==== CPx ( ) %78,70 ==xP

b) Que se gradué uno:


8

( ) ( ) ( ) %92,252592,06,04,0 41511 ==== CPx ( ) %92,251 ==xP

c) Que se gradúe al menos uno: ( ) ( ) ( ) %22,929222,00778,016,04,01 505

01 ==−=−=≥ CPx ( ) %22,991 =≥xP

14. Solución:

61=p 65=q 5=n

a) ( ) %19,404019,0776.7

125.3

296.1

625

6

15

6

5

6

141

511 ===

=

== CPx ( ) %19,401 ==xP

b) ( ) %08,161608,0776.7250.1

216125

36110

65

61

32522 ===

=

== CPx ( ) %08,162 ==xP

c) ( ) %21,30321,0776.7

2503625

216110

65

61

23533 ===

=

== CPx ( ) %21,33 ==xP

d) ( ) %32,00032,0776.725

65

296.115

65

61

14544 ===

=

== CPx ( ) %32,04 ==xP

e) ( ) ( ) %19,404019,0776.7125.311

65

61

50500 ==

=

== CPx (ninguna vez) ( ) %19,400 ==xP

15. Solución:

10,0=p 90,0=q 4=n a) ( ) ( ) ( ) %61,656561,09,01,0 404

00 ==== CPx ( ) %61,650 ==xP

b) ( ) ( ) ( ) %39,343439,09,01,01 404

01 ==−=≥ CPx ( ) %39,341 =≥xP

c) ( ) ( ) ( ) ( ) ( )314

1404

01 9,01,09,01,0 CCPx +=≤

%77,949477,02916,06561,0 ==+= ( ) %77,941 =≤xP

16. Solución:


9

2,0=p 8,0=q 10=n a) ( ) ( ) ( ) %2,303020,08,02,0 8210

22 ==== CPx ( ) %2,302 ==xP

b) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]8210

29110

110010

03 8,02,08,02,08,02,01 CCCPx ++−=≥

( ) [ ] %22,323222,06778,013020,02684,01074,013 ==−=++−=≥xP ( ) %22,323 =≥xP

c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )01010

101910

92810

83710

74610

66 8,02,08,02,08,02,08,02,08,02,0 CCCCCPx ++++=≥

0063,00000,00000,00008,00055,0 =+++= ( ) %63,06 =≥xP (Se usó la tabla para el cálculo) d) ( ) ( ) ( ) %74,101074,08,02,0 10010

00 ==== CPx ( ) %74,100 ==xP

17. Solución:

5,0=p 5,0=q 10=n 0,1,2,3=X

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )100100

91101

82102

731033 5,05,05,05,05,05,05,05,0 CCCCPx +++=≤

%19,171719,00010,00098,00439,01172,0 ==+++= ( ) %19,173 =≤xP

npE = ( ) 100181719,0100 depersonasE ≅= 18. Solución:

5,0=p 5,0=q 10=n 10 9 ,8 ,7 yX =

( ) ( ) ( ) ( ) ( ) ( ) ( ) 0101010

19109

28108

371077 )5,0()5,0(5,05,05,05,05,05,0 CCCCPx +++=≥

( ) %19,171719,100010,00098,00439,01172,07 ==+++=≥xP ( ) %19,177 =≥xP

19. Solución:


10

15=n 10,0=p 90,0=q a) ( ) ( ) ( ) %05,10105,09,01,0 10515

55 ==== CPx ( ) %05,15 ==xP

b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) +++=≥

3121512

4111511

510151010 9,01,09,01,09,01,0 CCCPx

( ) ( ) ( ) ( ) ( ) ( ) 0000,09,01,09,01,09,01,0 01515

1511415

1421315

13 =++ CCC ( ) 010 =≥xP

(Como se trabaja con cuatro decimales, aproximamos a cero) (Se utilizó la tabla) A partir de x > 8 la probabilidad obtenida es demasiado pequeña, casi cero.

c) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ +++−=≥13215

2

141151

1501505 9,01,09,01,09,01,01 CCCPx

( ) ( ) ( ) ( ) ]11415

412315

3 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. Solución:

20=n 25,0=p 75,0=q a) ( ) ( ) ( ) 0...............0000,075,025,0 51520

1515 ==== CPx (ver tabla) ( ) 015 ==xP

b) ( ) ( ) ( ) ( ) ( ) ( ) ( )16420

419120

120020

04 75,025,0...........75,025,075,025,0 CCCPx ++=≤

%48,414148,01897,01339,00669,00211,00032,0 ==++++= ( ) %48,414 =≤xP

c) ( ) ( ) ( ) ( ) ( ) ( ) ( )02020

2011920

912820

88 75,025,0...........75,025,075,025,0 CCCPx ++=≥

Es más fácil resolverlo de la siguiente forma:


11

( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]13720

719120

120020

08 75,025,0............75,025,075,025,01 CCCPx ++−=≥

[ ] =+++++++−= 1124,01686,02023,01897,01339,00669,00211,00032,01 %19,108981,01 =−= (por lo menos 8 defectuosas) ( ) %19,108 =≥xP

21. Solución:

5,0=p 5,0=q 4=n a) ( ) ( ) ( ) 9375,00625,015,05,01 404

01 =−=−=≥ CPx ( ) %75,931 =≥xP

( ) 875.19375,0000.2 ==E familias b) ( ) ( ) ( ) 3750,05,05,0 224

22 === CPx ( ) %50,372 ==xP

( ) familiasE 7503750,0000.2 == c) ( ) ( ) ( ) 0625,05,05,0 404

00 === CPx ( ) %25,60 ==xP

( ) familiasE 1250625,0000.2 == (Se utilizaron las tablas) 22. Solución:

( ) ( ) ( ) ( ) ( ) ( ) ( )132152

141151

1501502 95,005,095,005,095,005,0 CCCPx ++=≤

9639,01348,03658,04633,0 =++= = 96,39% ( ) %39,962 =≤xP

(Se utilizó la tabla) 23. Solución:


12

40,0=p 20=n

( ) ( ) ( ) ( ) ( ) ( ) ( )0202020

8122012

911201111 6,04,0........6,040,06,04,0 CCCPx +++=≥

Utilizando la tabla se tendrá que:

( ) =+++++++++=≥ 00000003,00013,00049,00146,00355,00710,011xP

%76,121276,0 == (mitad más uno) ( ) %76,1211 =≤xP

(Se utilizó la tabla para el cálculo) 24. Solución:

20,0=p 80,0=q 18=n 8=X

( ) ( ) ( ) %20,10120,080,020,0 1081888 ==== CPx ( ) %20,18 ==xP

25. Solución:

( ) ( ) ( ) ( ) ( ) ( ) ( )37107

46106

5510575 5,05,05,05,05,05,0 CCCP x ++=≤≤

( ) %84,565684,01172,02051,02461,075 ==++=≤≤ xP ( ) %84,5675 =≤≤ xP

26. Solución:

5=n ( )3≥xP 5,4,3=X 5,0=p 5,0=q


13

( ) ( ) ( ) ( )5433 ===≥ ++= xxxx PPPP

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )055

5145

4235

3 5,05,05,05,05,05,0 ++= %505000,003125,015625,03125,0 ==++= ( ) %503 =≥xP

27. Solución:

cariescon 90,0109 = %1010,0cariessin == 5=n

a) Cuatro tengan caries 5=n 90,0=p 4=X ( ) ( ) ( ) ( ) %81,3232805,01,09,0 145

44 ====xP ( ) %81,324 ==xP

b) Por lo menos dos tengan caries 90,0=p 5,4,3,2=X ( ) ( ) ( ) ( ) ( )54322 ====≥ +++= xxxxx PPPPP

( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ]4151

5050

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

( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ]4151

5050

10

9,01,09,01,01

1

+−=

+−= == xx PP

[ ] %15,89185,0132805,059049,01 =−=+−= ( ) %15,82 =≥xP


14

d) Por lo menos una tenga caries 90,0=p 5,4,3,2,1=X ( ) ( )01 1 =≥ −= xx PP

( ) ( ) ( ) %10099999,000001,011,09,01 505

0 ==−=−= ( ) %1001 =≥xP

28. Solución: 20% pierden el 1ª año pierden lo no 80% 6=n a) :aprueben 2 Máximo 210 , , X = 800,p = ( ) ( ) ( ) ( )2102 ===≤ ++= xxxx PPPP

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )426

2516

1606

02 2,08,02,08,02,08,0 ++=≤xP

%70,101696,001536,0001536,0000064,0 ==++= ( ) %70,12 =≤xP

b) Todos aprueben: 800, p = 6=X ( ) ( ) ( ) ( ) %21,262621,02,08,0 066

66 ====xP ( ) %21,266 ==xP

c) Ninguno apruebe 800, p = 0=X ( ) ( ) ( ) ( ) %0064,0000064,02,08,0 606

00 ====xP ( ) %0064,00 ==xP

29. Solución:

0,7000068004 =.. Transporte público 30% 0,30 = otro servicio

a) No más de 2 utilicen transporte público 70,p = 2,1,0=X 8=n


15

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )628

2718

1808

02 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

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]6282

7181

8080

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 ( ) ( ) ( ) ( ) %65,292965,07,03,0 628

22 ====xP ( ) %65,292 ==xP

d) Exactamente 2 lo utilicen 70,0=p 2=X ( ) ( ) ( ) ( ) %10100,03,07,0 628

22 ====xP ( ) %12 ==xP

30. Solución: 60% = 0,60 asisten 0,40 = 40% no asisten n = 8 a) asistan 7 menos loPor 6,0=p 8,7=X


16

( ) ( ) ( )877 ==≥ += xxx PPP

( ) ( ) ( ) ( ) ( ) ( )088

8178

7 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

( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ]7181

8080

10

6,04,06,04,01

1

+−=

+−= == xx PP

[ ] %36,898936,01064,010896,00168,01 ==−=+−= ( ) %36,892 =≥xP

31. Solución:

gafasusan 4,02000800 = gafasusan no0,6 = 5n =

a) gafasusan 2 menos loPor 40,0=p 5,4,3,2=X ( ) ( ) ( ) ( ) ( )54322 ====≥ +++= xxxxx PPPPP

( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ]4151

5050

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


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. Solución:

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 2242

3141

4040

==++=

++=

( ) %18,882 =≤xP

32y31con trabajamosSí:Nota ( ) %89,882 =≤xP

b) repitente sea no 1 menos Al 67,0=p 4,3,2,1=X ( ) ( ) ( ) ( ) ( )43211 ====≥ +++= xxxxx PPPPP

32y31con os trabajamSí :Nota ( ) %77,981 =≥xP

( ) ( )01 1 =≥ −= xx PP

( ) ( ) ( ) %81,989881,00119,0133,067,01 404

0 ==−=−= ( ) %81,981 =≥xP

33. Solución:


18

16=n 6,0=p 16,15,14,13,12,11,10=X

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )3131613

4121612

5111611

610161010 4,06,04,06,004,06,04,06,0 +++=≥xP

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) =+++ 01616

16

1151615

2141614 4,06,04,06,04,06,0

0150,00468,01014,01623,01983,0 ++++= %71,520003,00030,0 =++ ( ) %71,5210 =≥xP (diez o más acontecimientos desfavorables) 34. Solución:

accidentan se 25% accidentan se no 75%

accidentan se 3 menos loPor 7=n

( ) ( ) ( ) ( ) ( ) ( )765433 =====≥ ++++= xxxxxx PPPPPP

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]5272

6171

7070

210

75,025,075,025,075,025,01

1

++−=

++−= === xxx PPP

[ ] %35,242435,07565,013115,03115,01335,01 ==−=++−= ( ) %35,243 =≥xP

35. Solución: 3% son defectuosos 97% Buenos n = 7 a) Por lo menos 3 sean buenos ( ) ( ) ( ) ( )7433 ....... ===≥ +++= xxxx PPPP

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]5272

6171

7070

210

03,097,003,097,003,097,01

1

++−=

++−= === xxx PPP


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. Solución:

enferman01,0 ==p 5=n enferman no99,0 ==q a) enfermos2=X ( ) ( ) ( ) ( ) %097,000097,099,001,0 325

22 ====xP ( ) %097,02 ==xP

b) enfermo uno menos loPor 5432,1 , , , X = ( ) ( ) ( ) ( ) ( ) %9,4049,09510,0199,001,011 505

001 ==−=−=−= =≥ 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. Solución:


20

20% de mortalidad 80% de sobrevivir 5=n a) Ninguno sobreviva 0=X ( )mueran todos,5aequivale =x ( ) ( ) ( ) ( ) ( ) ( ) ( ) %032,000032,08,02,02,08,0 055

5505

00 =====xP ( ) %032,00 ==xP

b) Todos sobrevivan ( ) ( ) ( ) ( ) %77,323277,02,08,0 055

55 ====xP ( ) %77,325 ==xP

c) Al menos 1 sobrevivan 54321 , , , , X = ( ) ( ) ( ) ( ) ( ) %97,99%968,9900032,012,08,011 505

001 ==−=−=−= =≥ xx PP ( ) %97,991 =≥xP

d) Al menos 1 no sobrevivan 54321 , , , , X = ( ) ( ) ( ) ( ) ( ) %23,6767232,032768,018,02,011 505

001 ==−=−=−= =≥ xx PP ( ) %23,671 =≥xP

38. Solución:

scientífico 20%0,20255 == científico no 80%

2520 = 4=n

a) Por lo menos 1 sea científica 4,3,2,1=X ( ) ( ) ( ) ( ) ( ) %04,595904,04096,018,02,011 404

001 ==−=−=−= =≥ xx PP ( ) %04,591 =≥xP

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