capítulo 6 cd (21 45-)

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

21

b) Por lo menos 2 no sean científicos 4,3,2=X

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %28,979728,00272,010256,00016,01

2,08,02,08,01

1

3141

4040

102

==−=+−=

+−=

+−= ==≥ xxx PPP

( ) %28,972 =≥xP

c) Una sea científica 1=X ( ) ( ) ( ) ( ) %96,404096,08,02,0 314

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

39. Solución:

clientes posibles 30% clientessean no 70% 8=n a) Tres o menos sean clientes 0,1,2,3=X ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )538

3628

2718

1808

03 7,03,07,03,07,03,07,03,0 +++=≤xP

%59,808059,02541,02965,01977,00576,0 ==+++= ( ) %59,803 =≤xP

b) Tres o más no sean clientes 8,7,6,5,4,3=X

( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ][ ] 0113,0101000,000122,000006,01

3,07,03,07,03,07,01

1

6282

7181

8080

2103

−=++−=

++−=

++−= ===≥ xxxx PPPP

%87,989887,0 == ( ) %87,983 =≥xP


22

40. Solución:

AaApoyan 40%0,40 52 == Aaapoyan No 60%60,0 =

a) Exactamente 5 apoyen a A 5=x ( ) ( ) ( ) ( ) %74,70774,06,04,0 257

55 ====xP ( ) %74,75 ==xP

b) Por lo menos 2 apoyen a A 7,6,5,4,3,2=x

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ]6171

7070

102

6,04,06,04,01

1

+−=

+−= ==≥ xxx PPP

[ ] %14,848414,01586,011306,00280,01 ==−=+−= ( ) %14,842 =≥xP

c) Por lo menos dos no apoyen a A

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ]6171

7070

102

4,06,04,06,01

1

+−=

+−= ==≥ xxx PPP 7,6,5,4,3,2=x

[ ] %12,989812,001884,010172,000164,01 ==−=+−= ( ) %12,982 =≥xP

41. Solución:

Verdadero0,5021 = falso50,0 = 16=n

a) A lo más dos preguntas correctas 210 , , X = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )14216

2

151161

1601602 5,05,05,05,05,05,0 ++=≤xP

%21,0%0021,000183,000024,0000015,0 ==++= ( ) %21,02 =≤xP


23

b) Por lo menos 2 sean verdaderas 16......,4,3,2=X

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ]151161

160160

102

5,05,05,05,01

1

+−=

+−= ==≥ xxx PPP

[ ] %97,999997,0000255,0100024,0000015,01 ==−=+−= ( ) %97,992 =≥xP

c) Por lo menos 2 no sean verdaderas

( ) ( ) ( )[ ]102 1 ==≥ +−= xxx PPP

( ) ( ) ( ) ( ) ( ) ( )[ ] %97,995,05,05,05,01 15116

116016

0 =+−= ( ) %97,992 =≥xP

42. Solución:

0,15203 = Defectuosos buenos85,0 = 8=n

a) Por lo menos dos defectuosos 8,7,6,5,432 , , X =

( ) ( ) ( )[ ]

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

80802

102

85,015,085,015,01

1

+−=

+−=

≥

==≥

x

xxx

P

PPP

[ ] %28,343428,06572,013847,02725,01 ==−=+−= ( ) %28,342 =≥xP

b) Por lo menos 2 no sean defectuosos 8,7,6,5,4,3,2=X

( ) ( ) ( )[ ]

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

8080

102

15,085,015,085,01

1

+−=

+−= ==≥ xxx PPP

[ ] %10099,9999999,000001,001 ≅==+−= ( ) %1002 =≥xP


24

c) ( ) ( ) ( ) ( ) %76,232376,085,015,0 62822 ====xP 2=x

( ) Artículos4752376,0000.2 ==E npE =

43. Solución:

fumadores%70 fumadoresno%30 = 16=x 18=n

( ) ( ) ( ) ( ) %58,404576,03,07,0 216181616 ====xP ( ) %58,416 ==xP

44. Solución:

responden%20 responden no%80 10=n a) La mayoría responden 10,9,8,7,6=X (mitad más uno) ( ) ( ) ( ) ( ) ( ) ( )1098766 =====≥ ++++= xxxxxx PPPPPP

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ++++= 1910

92810

83710

74610

6 8,02,08,02,08,02,08,02,0 ( ) ( ) ( ) =++++= 00000074,000079,00055,08,02,0 01010

10 %64,000636,0 == ( ) %64,06 =≥xP

b) Menos del 30% no respondan ( ) 3100,30 = 2,1,0=X ( ) ( ) ( ) ( )2102 ===≤ ++= xxxx PPPP

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )8210

29110

110010

0 2,08,02,08,02,08,0 ++= 000078,00000737,00000041,00000001,0 =++= %0078,0= ( ) %0078,02 =≤xP

c) Nadie responde 0=X ( ) ( ) ( ) ( ) %00001,00000001,08,02,0 10010

00 ====xP ( ) %010 ≅=xP


25

45. Solución:

10=n defectuoso%17 buenos%83 a) Ninguno defectuoso 0=X ( ) ( ) ( ) ( ) %52,151552,083,017,0 10010

00 ====xP ( ) %52,150 ==xP

b) Por lo menos 2 no sean defectuosos 10.......4,3,2=X

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %1001 a aproxima se001

17,083,017,083,01

1

91101

100100

102

==+−=

+−=

+−= ==≥ xxx PPP

( ) %1002 =≤xP

c) Como máximo dos defectuosos 21,0,=X ( ) ( ) ( ) ( )2102 ===≤ ++= xxxx PPPP

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )8210

29110

110010

0 83,017,083,017,083,017,0 ++= %59,767659,02929,03178,01552,0 ==++= ( ) %59,762 =≤xP

46. Solución: 80% adecuadamente 20% no adecuado 4=n a) Todos adecuadamente 4=X ( ) ( ) ( ) ( ) %96,404096,02,08,0 044

44 ====xP ( ) %96,404 ==xP

b) Falla uno (no adecuado) 1=X ( ) ( ) ( ) ( ) %96,404096,08,02,0 314

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


26

c) Uno o más fallan 4,3,2,1=X ( ) ( )01 1 =≥ −= xx PP

( ) ( ) ( ) %04,595904,04096,018,02,01 404

0 ==−=−= ( ) %04,591 =≥xP

47. Solución:

%6060,0106 == Detectados 40% no detectados 8=n

a) Por lo menos 5 veces sea detectado 8,7,6,5=X ( ) ( ) ( ) ( ) ( )87655 ====≥ +++= xxxxx PPPPP

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

8178

7268

6358

55 4,06,04,06,04,06,040,060,0 +++=≥xP

%41,595941,00168,00896,02090,02787,0 ==+++= ( ) %41,595 =≥xP

b) Exactamente 2 no sea detectado ( ) ( ) ( ) ( ) %90,202090,06,04,0 628

22 ====xP ( ) %90,202 ==xP

48. Solución:

naranja de jugotoman %9,9 90,1% tomanlo no 0,901;0,099 ==

a) Por lo menos 2 toman jugo 5,4,3,2=X

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %80800,09200,013262,05938,01

901,0099,0901,0099,01

1

4151

50502

102

==−=+−=

+−=

+−=

≥

==≥

x

xxx

P

PPP

( ) %82 =≥xP


27

b) Como máximo 3 no lo toman 3,2,1,0=X

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %80800,09200,015938,03262,01

099,0901,0099,0901,01

1

0555

1454

5143

==−=+−=

+−=

+−= ==≤ xxx PPP

( ) %83 =≤xP

49. Solución:

vende25,041 = vendeno 0,75= 5=n

Por lo menos 3 compren 5,4,3=X

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

%35,101035,000098,00146,00879,0

75,025,075,025,075,025,0 0555

1454

2353

5433

==++=

++=

++= ===≥ xxxx PPPP

( ) %35,103 =≥xP

50. Solución:

pierden lo16,0 pierden lo no 0,84= 10=n No mayor a 5 ni menor a tres no lo pierdan 5,4,3=X

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

%30,10130,00110,00018,000019,0

16,084,016,084,016,084,0 55105

64104

73103

54353

==++=

++=

++= ===≤≤ xxxx PPPP

( ) %30,153 =≤≤xP


28

51. Solución:

tardanse%20 tardanse no 80%= 5=n a) Dos veces se retardan 2=X ( ) ( ) ( ) ( ) %48,202048,08,02,0 325

22 ====xP ( ) %84,202 ==xP

b) Por lo menos 2 no se retardan 5,4,3,2=X

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %33,9900672,010064,000032,01

2,08,02,08,01

1

4151

5050

102

=−=+−=

+−=

+−= ==≥ xxx PPP

( ) %33,992 =≥xP

52. Solución:

años 65 de mas%36 65 de menores 64% = Quince o más tengan más de 65 años

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) =+++=≥01818

1811718

1721618

1631518

1515 64,036,064,036,064,036,064,036,0xP

0000519,0000000049,0000047,0 =+++= %00519,0= ( ) %00519,015 =≥xP

53. Solución: 30% se retiran n = 12 70% siguen a) Por lo menos 9 sigan ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )01212

1211112

1121012

103912

99 3,07,03,07,03,07,03,07,0 +++=≥xP

%25,494925,00138,00712,01678,02397,0 ==+++= ( ) %25,499 =≥xP


29

b) Como mínimo tres se retiren 1243 ...., ........, X =

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %72,742528,011678,00712,00138,01

7,03,07,03,07,03,01 102122

111121

1201203

=−=++−=

++−=≥xP

( ) %72,743 =≥xP

54. Solución:

éxitos 0,10101 = éxitos de nada 0,90 6=n

Por lo menos 2 sean éxitos financieros 6,5,4,3,2=x

( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %43,118857,013543,05314,01

9,01,09,01,01

1

5161

6060

102

=−=+−=

+−=

+−= ==≥ xxx PPP

( ) %43,112 =≥xP

55. Solución:

0,20255 = Técnicos 0,80 No técnicos 4=n

a) Por lo menos 1 sea técnico 432 ,1 , , X = ( ) ( ) ( ) ( ) ( ) %04,594096,018,02,011 404

001 =−=−=−= =≥ xx PP ( ) %04,591 =≥xP

b) Por lo menos 2 no sean técnicos 432 , , X =


30

( ) ( ) ( )[ ]

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

[ ] %28,970272,010256,00016,01

2,08,02,08,01

1

3141

4040

102

=−=+−=

+−=

+−= ==≥ xxx PPP

( ) %28,972 =≥xP

DISTRIBUCIÓN DE POISSON 56. Solución:

10,0=p 90,0=q 10=n 2=X a) Distribución binomial ( ) ( ) ( ) ( ) ( ) %37,191937,04305,001,0459,01,0 8210

22 ===== CP x ( ) %37,192 ==xP

b) Distribución Poisson

!X

ep

x λλ −

= 2=X ( ) 11,010 === pnλ

( )( ) ( )

%39,1818394,0236788,01

!21 12

2 ====−

=e

P x ( ) %39,182 ==xP

57. Solución:

001,0=p np=λ 999,0=q ( ) 2001,0000.2 ==λ 000.2=n

a) ( )( )

%04,1818045,06

13534,08!3

2!

23

3 =====−−

=e

xe

Px

x

λλ ( ) %04,183 ==xP

b) ( ) 000.2................................,5,4,3?;2 ==> XxP

( )( ) ( ) ( )

++−=> !2

13534,02!1

13534,02!0

13534,021

210

2xP


31

( ) ( ) =++−=≥ 27068,027068,013534,013xP

%33,323233,067670,01 ==−= ( ) %33,323 =≥xP

58. Solución:

03,0=p 97,0=q ( ) 310003,0 ==λ 100=n pn=λ

a) ( )( )

%98,404979,01

04979,01!0

3 30

0 ====−

=e

P x ( ) %98,40 ==xP

b) ( )( )

%94,1414937,0! 1

04979,031

1 ====xP ( ) %94,141 ==xP

c) ( )( )

%40,2222404,0!2

04979,032

2 ====xP ( ) %40,222 ==xP

d) ( )( )

%40,2222404,0!3

04979,033

3 ====xP ( ) %40,223 ==xP

e) ( )( )

%80,1616803,0!4

04979,034

4 ====xP ( ) %80,164 ==xP

f) ( )( )

%08,1010082,0!5

04979,035

5 ====xP ( ) %08,105 ==xP

59. Solución:

00003,0=p ( ) 6000.20000003,0 ==λ 000.200=n

a) ( )( )

%25,0002479,0 1

002479,01

!0

6 60

0 ====−

=

eP x ( ) %25,00 ==xP

b) ( )( )

%46,4044622,0!2

002479,062

2 ====xP ( ) %46,42 ==xP


32

c) ( )( ) ( ) ==== 720

002479,0656.46!6

002479,066

6xP

%06,161606,0720

66,115 === ( ) %06,166 ==xP

d) ( )( ) ( ) ==== 320.40

002479,0616.679.1!8

002479,068

8xP

%33,1010326,0320.40

76,163.4 === ( ) %33,108 ==xP

e) 8,7,6,5,4=X

( )( ) ( ) ( ) ( ) ( )

!8002479,06

!7002479,06

!6002479,06

!5002479,06

!4002479,06 87654

84 ++++=≤≤ xP

( )( ) ( ) ( ) +++=≤≤ 720

002479,0656.46120

002479,0776.724002479,0296.1

84 xP

( ) ( ) =++320.40

002479,0616.679.1040.5

002479,0936.29

( ) =++++=≤≤ 320.4079,163.4

040.596,693

72066,115

12028,19

2421,3

84 xP

( ) =++++=≤≤ 1033,01377,01606,01606,01338,084 xP

( )84 ≤≤ xP %60,696960,0 == ( ) %60,6984 =≤≤ xP

f) ( )( ) ( ) ( ) =++=≤ !2

002479,06!1

002479,06!0

002479,06 210

2xP

( ) =++=≤ 044622,0014874,0002479,02xP

( )2≤xP %20,6061975,0 == ( ) %20,62 =≤xP

60. Solución:


33

0066,01501 ==p ( ) ( ) 9,9500.10066,0 ==λ 500.1=n

a) 3=X

( )( )

%81,0008085,06

04851,06

00005,029,970!3

9,9 9,93

3 =====−

=e

P x ( ) %81,03 ==xP

Con calculadora programable el resultado es = 0,0081141 = 0,81% b) 500.1..............7,6,5,4=X

( )

+++−=

−−−−

≥ !39,9

!29,9

!19,9

!09,9

19,939,929,919,90

4eeee

P x

( )( ) ( ) ( )

+++−=≥ 008085,0200005,00,98

100005,09,9

100005,01

14xP

( ) [ ]00808,000245,000049,000005,014 +++−=≥xP

( ) [ ] %89,9898893,001107,014 ==−=≥xP ( ) %89,984 =≥xP

NOTA: ¿Menos de 5 vuelos se retrazaran más de una hora?

c) ( ) !49,9

!39,9

!29,9

!19,9

!09,9 9,949,939,929,919,90

4

−−−−−

≤ ++++= eeeeeP x

( ) 24480298,000808,000245,000049,000005,04 ++++=≤xP

( ) =++++=≤ 02001,000808,000245,000049,000005,04xP

( )4≤xP %11,303108,0 == ( ) %11,34 =≤xP

61. Solución:

0001,0=p 000.10=n ( ) 1000.100001,0 ==λ 5=X


34

( )( )

%31,0003065,0120

36788,01

!5

1 15

5 ====−

=e

P x ( ) %31,05 ==xP

62. Solución:

001,0=p 999,0=q ( ) 2000.2001,0 ==λ 000.2=n

a) ( ) ?3 ==xP !X

xx eP

−

= λ

( )( )

%04,1818045,06

0872,16

13534,08!3

2 23

3 =====−

=e

P x ( ) %04,183 ==xP

b) ( ) ?3 =≥xP 000.2.............,6,5,4,3=X También se puede plantear así:

( )

++−=

−−−

≥ !22

!12

!02

1222120

3eee

P x

( )( ) ( ) ( )

++−=≥ 213534,04

113534,02

113534,01

13xP

( ) [ ] =++−=≥ 27067,027067,013534,013xP

( ) =≥3xP %33,323233,06767,01 ==− ( ) %33,323 =≥xP

63. Solución:

a) ( )( )

%68,484868,0!04868,072,0 0

0 ====xP ( ) %68,480 ==xP

b) ( )( )

%05,3535049,0!14868,072,0 1

1 ====xP ( ) %05,351 ==xP

c) ( )( ) ( ) ==== 2

4868,05184,0

!2

4868,072,0 2

2xP

( ) ==2xP ( ) %62,12126178,04868,02592,0 == ( ) ==2xP %62,12


35

d) ( )( ) ( ) ==== 6

4868,03732,0

!3

4868,072,0 3

3xP

( ) ==3xP %03,3030278,06

181673,0 == ( ) %03,33 ==xP

64. Solución:

5,2== npλ !X

eP

x λλ −

=

a) ( )( )

%21,808208,01

08208,01!0

5,2 5,20

0 ====−

=e

P x ( ) %21,80 ==xP

b) ( )( )

%52,202052,0108208,05,2

!15,2 5,21

1 ====−

=e

P x ( ) %52,201 ==xP

c) ( )( )

%65,252565,0208208,05,2

!25,2 25,22

2 ====−

=e

P x ( ) %65,252 ==xP

d) ( )( )

%37,212137,0608208,05,2

!35,2 35,23

3 ====−

=e

P x ( ) %37,213 ==xP

65. Solución:

0004,0000.52 ==p ( ) 4,00004,0000.1 === npλ

Por lo menos 2 tengan problemas .........., ........,X 32 = 1.000


36

( ) ( ) ( )[ ]

[ ] %16,60616,09384,012681,06703,01

!14,0

!04,0

1

1

4,014,00

102

==−=+−=

+−=

+−=

−−

==≥

ee

PPP xxx

( ) %16,62 =≥xP

66. Solución:

0005,0000.21 ==p ( ) 30005,0000.6 ==λ

a) Más de 3 se incendien ..............,7,654 ,,X =

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

[ ]

%28,353528,06472,01

2240,02240,01494,00498,01

!33

!23

!13

!03

1

1

33323130

32104

==−=

+++−=

+++−=

+++−=

−−−−

====≥

eeee

PPPPP xxxxx

( ) %28,354 =≥xP

b) ( ) %40,222240,0!2

3 32

2 ===−

=e

P x ( ) %40,222 ==xP

67. Solución:

2=λ a) No más de 3 atracos


37

( ) ( ) ( ) ( ) ( )

%71,858571,01804,02707,02707,01353,0

!32

!22

!12

!02 23222120

32103

==+++=

+++=

+++=

−−−−

====≤

eeee

PPPPP xxxxx

( ) %71,853 =≤xP

b) A lo más 2 atracos

( ) ( ) ( ) ( )

!2

2

!1

2

!0

2 222120

2102

−−−

===≤

++=

++=

eee

PPPP xxxx

%67,676767,02707,02707,01353,0 ==++= ( ) %67,672 =≤xP

68. Solución:

0001,0%01,0 ==p ( ) 10001,0000.10 ==λ Máximo 3 se accidentan

( ) ( ) ( ) ( ) ( )

!3

1

!2

1

!1

1

!0

1 13121110

32103

−−−−

====≤

+++=

+++=

eeee

PPPPP xxxxx

%10,989810,00613,01839,03679,03679,0 ==+++= ( ) %10,983 =≤xP

69. Solución:

0024,0%24,0 ==p ( ) 6,30024,0500.1 ==λ


38

a) Dos o menos defectuosos

( ) ( ) ( ) ( )

!2

6,3

!1

6,3

!0

6,3 6,326,316,30

2102

−−−

===≤

++=

++=

eee

PPPP xxxx

%28,303028,01771,00984,00273,0 ==++= ( ) %28,302 =≤xP

b) Más de 2 defectuosos

( ) ( ) ( ) ( )[ ]

++−=

++−=

−−−

===≥

!2

6,3

!1

6,3

!0

6,31

1

6,326,316,30

2103

eee

PPPP xxxx

[ ] %72,696972,03028,011771,00984,00273,01 ==−=++−= ( ) %72,693 =≥xP

70. Solución:

5semestreun en 10 =⇒= λλ en un trimestre

( ) ( ) ( ) ( )

%46,121246,00842,00337,00067,0

!25

!15

!05 525150

2102

==++=

++=

++=

−−−

===≤

eee

PPPP xxxx

( ) %46,122 =≤xP

71. Solución:


39

3=λ

a) Ninguna demanda 0=X

( ) %98,40498,0!0

3 30

0 ===−

=ePx ( ) %98,40 ==xP

b) Por lo menos 2 demandas ..........,5,4,3,2=X

( ) ( ) ( )[ ]

[ ] %08,808008,01992,011494,00498,01

!13

!03

1

1

3130

102

==−=+−=

+−=

+−=

−−

==≥

ee

PPP xxx

( ) %08,802 =≥xP

72. Solución:

0005,0=p ( ) 4,00005,0800 ==λ a) Mínimo 3 equivocaciones ..........6,5,4,3=X

( ) ( ) ( ) ( )[ ]

[ ] %80,00080,09920,010536,02681,06703,01

1 2103

==−=++−=

++−= ===≥ xxxx PPPP

( ) %80,02 =>xP

b) Máximo 2 equivocaciones 2,1,0=X ( ) ( ) ( ) ( ) %20,992102 =++= ===≤ xxxx PPPP ( ) %20,992 =≤xP

(Ver ejercicio anterior)


40

73. Solución:

003,0000.13 = ( ) 5,1003,0500 ==λ

a) Más de 2 mueran ...,5,4,3=X

( ) ( ) ( ) ( )[ ]

[ ] %12,198088,012510,03347,02231,01

!25,1

!15,1

!05,1

1

1

5,125,115,10

2103

=−=++−=

++−=

++−=

−−−

===≥

eee

PPPP xxxx

( ) %12,193 =≥xP

b) Como máximo dos mueran 2,1,0=X ( ) ( ) ( ) ( ) %88,808088,02102 ==++= ===≤ xxxx PPPP ( ) %88,802 =≤xP

74. Solución:

horapor 12=λ minutos diezen 2=λ a) Por lo menos 2 se acerquen ....4,3,2=X


41

( ) ( ) ( )[ ]

[ ]

%40,595940,0

4060,012707,01353,01!1

2

!0

21

1

2120

102

==

=−=+−=

+−=

+−=

−−

==≥

ee

PPP xxx

( ) %40,592 =≥xP

b) No más de dos se acerquen al especialista

( ) ( ) ( ) ( )

%67,676767,02707,02707,01353,0

!22

!12

!02 222120

2102

==++=

++=

++=

−−−

===≤

eee

PPPP xxxx

( ) %67,672 =≤xP

75. Solución:

0001,0000.101 = 3)0001,0(000.30 ==λ

a) Por lo menos uno sufra reacción ,....4,3,2,1=X

( ) ( ) %02,959502,00498,01!0

311

30

01 ==−=

−=−=

−

=≥e

PP xx ( ) %02,951 =≥xP

b) Más de una sufra reacción ,....4,3,2=X


42

( ) ( ) ( )[ ]

[ ]

%08,808008,0

1992,011494,00498,01

!13

!03

1

1

3130

102

==

−=+−=

+−=

+−=

−−

==≥

ee

PPP xxx

( ) %08,802 =≥xP

76. Solución:

minutos 2 cada llamadas20=λ a) Exactamente 4 llamadas en 30 segundos 5=λ

( ) %55,171755,0!4

5 54

4 ===−

=e

P x ( ) %55,174 ==xP

b) Como máximo dos en 15 segundos 5,2=λ 2,1,0=X

( ) ( ) ( ) ( )

%38,545438,02565,02052,00821,0

!25,2

!15,2

!05,2 5,225,215,20

2102

==++=

++=

++=

−−−

===≤

eee

PPPP xxxx

( ) %38,542 =≤xP

77. Solución:

horapor clientes8,6=λ a) Por lo menos uno en la primera media hora ....4,3,2,1=X


43

( ) ( )

%67,969667,00334,01

!0

4,31

1

4,30

01

==−=

−=

−=

−

=≥

e

PP xx

( ) %67,961 =≥xP

b) Ninguno en el primer cuarto de hora 71,λ = 0=X

( ) %27,181827,0!0

7,1 7,10

0 ===−

=e

P x

( ) %27,180 ==xP

c) Más de uno, en cualquier hora 8,6=λ ....4,3,2=X

( ) ( ) ( )[ ]

[ ] %13,999913,00087,010076,00011,01

!18,6

!08,6

1

1

8,618,60

102

==−=+−=

+−=

+−=

−−

==≥

ee

PPP xxx

( ) %13,992 =≥xP

78. Solución:

minutos 30en 9=λ a) Por lo menos 4 en la primera media hora ....6,5,4=X


44

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

[ ]

%88,979788,00212,01

0150,00050,00011,00001,01

!39

!29

!19

!09

1

1

93929190

32104

==−=

+++−=

+++−=

+++−=

−−−−

====≥

eeee

PPPPP xxxxx

( ) %88,974 =≥xP

b) Ninguno en los 10 primeros minutos 0=X 3=λ

( ) 98,40498,0!0

3 30

0 ===−

=e

P x % ( ) %98,40 ==xP

79. Solución:

año7,5=λ a) Ninguno en los 4 meses 037,5 == Xλ

( ) %96,141496,0!0

9,1 9,10

0 ===−

=e

P x ( ) %96,140 ==xP

b) Por lo menos 1 en el semestre 85,2=λ ...,4,3,2,1=X

( ) ( ) %22,940578,01!0

85,211

85,20

01 =−=−=−=−

=≥e

PP xx ( ) %22,941 =≥xP


45

HIPERGEOMETRICA 80. Solución: a) 15=N 6=A 5=n 2=X

( )( )( )

( ) %96,414196,0155

61525

62

2 ===−−

=xP ( ) %96,412 ==xP

b) 15=N 9=A 5=n 2=X

( )( )( )

( ) %98,232398,0155

91525

92

2 ===−−

=xP ( ) %98,232 ==xP

81. Solución:

12=N 4=A 3=n 0=X

( )( )( )

( ) %45,252545,0123

41203

40

0 ===−−

=xP ( ) %45,250 ==xP

82. Solución:

15=N 10=A 5=n a) A dos les guste 2=X

( )( )( )

( ) %99,141499,0155

101525

102

2 ===−−

=xP ( ) %99,142 ==xP

b) A dos no les guste 15=N 5=A 5=n 2=X

( )( )( )

( ) %96,393996,0155

51525

52

2 ===−−

=xP ( ) %96,392 ==xP


46

83. Solución:

25=N 6=A 4=n 2=X

( )( )( )

( ) %28,202028,0254

62524

62

2 ===−−

=xP ( ) %88,202 ==xP

(dos que no requieren ser ajustadas) 84. Solución:

40=N 35=A 5=n ....,4,3,2,1=X

( ) ( )( )( )

( )

%100199999,0

0000015,0111405

354005

350

01

=≅=

−=

−=−=

−−

=≥ xx PP

(por lo menos uno es economista) ( ) %1001 =≥xP

capítulo 6 cd (21 45-)

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