Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
Universidad Nacional Mayor de San Marcos
acultad de Ciencias Administrativas Esc#ela Acad)mico Pro*esional de Administraci+n de Negocios Internacio “Año de la Consolidación del Mar de Grau”
GUÍA 1 Curso: Estadística Aplicada a los Negocios II
Profesor: Alumno:
Lic. Emma Perez Palacios Niel Andrew Cortez Gala
Código: 14!4""
Ciclo: Aula y turno:
C#arto $% & 'a(ana
Sign up to vote on this title
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Sign In
Upload
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
Universidad Nacional Mayor de San Marcos
U1. Distribuciones de
Tem 3 Probabilidad
Search document
scuela E scuela
Universitaria de Negocios Internacional
Estadística Aplicada a los NN. NN. II, I I Guía de Ejercicios – I Distribución normal 1.
Calcu Calcule le las las sigui siguient entes es prob probabi abililida dades des:: a) Si la la variable variable X tiene distri distribución bución normal con media media 40 y desviación desviación estándar estándar 8 ca i) ! "X# 4$) 2
2
X 1 N ( μ= 40, σ =8
)
p ( x x < 45 )
(
p x x <
45− 40 8
)
p ( x x < 6.25 ) ≅ 0.63 =0.73565 ii) ! "X % &8) 2
2
X 1 N ( μ= 40, σ =8
)
p ( x x > 38 ) 1− p ( x < 38 )
(
1− p x <
38− 40 8
)
1− p ( x <−0.25 ) 1−0.40129 =0.59871 Sign up to vote on this title
iii) ! "&0 # X # $0)
2
2
X 1 N ( μ= 40, σ =8
)
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
2
2
X 1 N ( μ= 40, σ =8
U1. Distribuciones de
Tem 3 Probabilidad
Search document
)
p ( x < 55 / x > 28 ) p ( 28 ≤ x ≤ 55 ) p ( x ≤ 55 )− p ( x ≤ 28 )
(
p x <
55 −40 8
) (
28− 40
− p x <
8
)
p ( x < 1.875 ) − p ( x <−1.5 ) 0.96995 − 0.06681 =0.90314
b)
Si una variable X tiene distribución normal con µ * 1$0 y σ * 1+ ,allar el valor de i) ! "X #-) * 0.( 2
2
X 1 N ( μ=150, σ = 16
)
p ( x < k ) =0.92 k −150 16
You're Reading a Preview
=1.41
Unlock full access with a free trial.
k =172.56
Download With Free Trial
ii) ! "X % -) * 0.0($ 2
2
X 1 N ( μ=150, σ = 16
)
p ( x > k ) =0.025 1− p ( x > k )= 0.025
p ( x < k ) =0.975 k −150
1.96
Sign up to vote on this title
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
k −150 16
k −130 16
−
of 21
130− 150 16
U1. Distribuciones de
Tem 3 Probabilidad
Search document
=−0.05
=−0.05
k =129.20 iv) ! "-# X #1/$) * 0.+0 2
μ=150, σ = 16 X 1 N ¿
2
p ( k < x < 175 )=0.60 p ( x < 175 ) − p ( x < k )=0.60 175 −150 16
−
k − 150 16
=0.25
175 −k 16
= 0.25
k =171.00 (.
You're Reading a Preview l tiempo necesario para terminar un eamen parcial en un determinado curso se d Unlock full access with a free trial. normalmente con tiempo medio de 80 minutos y una desviación estándar de 10 minuto a) 2Cuál es la probabilidad de ue un alumno termine el eamen en más de +0 Download With Free Trial pero en menos de /$ minutos3 2
2
X 1 N ( μ=80, σ =10
)
p ( 60 < x < 75 ) p ( x < 75 ) − p ( x < 60 )
(
p x <
75 −80 10
) (
− p x < 60−80
p ( x <−0.5 )− p ( x <−2 )
10
)
Sign up to vote on this title
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
(
1− p x <
of 21
90 −80 10
U1. Distribuciones de
Tem 3 Probabilidad
Search document
)
1− p ( x < 1 ) 1−0.84134 =0.15866 60 × 0.15866 = 9.5196 ≅ 9
5 de alumnos: &.
Se cree ue las ventas de un determinado detergente tienen una distribución norm media de 10000 bolsas y una desviación estándar de 1$00 bolsas por semana. a)
2Cuál es la probabilidad de vender más de 1(000 bolsas en una semana3 2
X 1 N ( μ=10000, σ =1500
2
)
p ( x > 12000 ) 1− p ( x < 12000 )
(
1− p x <
12000 − 10000 1500
)
1− p ( x < 1.33 ) 1−0.90824 =0.09176
b)
You're Reading a Preview !ara tener una probabilidad del /.$6 de ue la empresa cuente con su eistencias para cubrir la demanda Unlock fullsemanal access with a2cuántas free trial. bolsas debe producir3 2
2
) X 1 N ( μ=10000, σ =1500 Download With Free Trial p ( x ≥ k )=0.975 1− p ( x ≤ k ) =0.975
p ( x ≤ k )=0.025 k −10000 1500
=−1.96
k =¿ !060 c)
Sign up to vote on this title
Useful
Not useful
2Cuál es la probabilidad de ue la venta semanal de bolsas di7iera de
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
d)
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
Si en la siguiente semana se asegura vender más de 11000 bolsas 2c probabilidad de ue en esa semana se venda menos de 1($00 bolsas3 2
X 1 N ( μ=10000, σ =1500
2
)
p ( 11000 < x < 12500 ) p ( x < 12500 ) − p ( x < 11000 )
(
p x <
12500 −10000 1500
) (
− p x <
11000 −10000 1500
)
p ( x < 1.67 )− p ( x < 0.6 ) 0.95154 −0.74537 =0.20617
4.
n una empresa ue 7abrica artculos de plástico los ingresos mensuales en miles se distribuyen normalmente. Sabiendo ue el 1(6 de ellos son superiores a +0$.$ y están entre &&.$ y +0$.$ calcular: a) l ingreso promedio mensual y su desviación estándar 2
X 1 N ( μ= ? , σ =? ) i.)
p ( 393.5 < x < 605.5 )=0.76 You're Reading a Preview p ( x < 605.5 ) − p ( x < 393.5 ) =0.76
(
p x <
605.5− μ
σ
605.5 − μ
σ
−
σ =
)− ( < σ
=0.71
=0.71
212 0.71
σ =298.5915493 ii.)
)
393.5 − μ =0.76 p x σ With Free Trial Download
393.5 − μ
605.5 −393.5
σ
Unlock full access with a free trial.
p ( x > 605.5 ) =0.12
Sign up to vote on this title
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Sign In
Upload
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
605.5 − μ 298.59
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
=1.17
μ=256.1478873
b)
9a probabilidad de ue los ingresos sean mayores ue $00 2
X 1 N ( μ=256.15, σ = 298.59
2
)
p ( x > 500 ) 1− p ( x < 500 )
(
1− p x <
500−256.15 298.59
)
1− p ( x < 0.82 ) 1−0.79389 =0.20611
$.
l gerente de anca !ersonal de un banco estimó ue los depósitos desde ue as dirección están normalmente distribuidos con una media de 10000 dólares y una de estándar de 1$00 dólares. l gerente estudia la posibilidad de una tasa de pre7erencial a sus clientes incentivando al a,orro ue consiste en lo siguiente:
Límite de a,orro mínimo2
X 1 N ( μ=10000, σ =1500
2
)
p ( x ≤ k )=0.20
(
p x <
k −10000 1500
k −10000 1500
)=
=−0.84
0.20 Sign up to vote on this title
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
k =11005.00
b) +.
>ctor ?a@ tiene un depósito de 1(8$0 dólares. 2!odrá lograr >ctor una tasa d superior al $63.
n una compaAa distribuidora de productos umicos se observa ue el nBmero de c cierto medicamento ue se distribuye mensualmente a un establecimiento es una aleatoria X normal con una media ($/ caas y una desviación estándar de (0 caas. a) 2Cuál es la probabilidad de ue en un establecimiento la compaAa distribuya más caas de este tipo de medicamento3 2
2
X 1 N ( μ=257, σ =20
)
p ( x > 280 ) 1− p ( x < 280 )
(
1− p x <
280−257 20
)
1− p ( x < 1.15 ) 1− 0.87493 =0.12507
b)
You'redel Reading a Preview Calcule e interprete en t;rminos enunciado el valor - tal ue : !"X ≤ -) * 0.0 2
2
X 1 N ( μ=257, σ =20 p ( x ≤ k )=0.90 k −257 20
Unlock full access with a free trial.
)
Download With Free Trial
=1.28
k =282.60 c)
Cada caa de este producto distribuido a un establecimiento cuesta $ nuevos sol cartera de clientes de la compaAa distribuidora cuenta con +4 estable Sign up to vote on this title comerciales
Useful
Not useful
c.1) Dbtener la media y la desviación estándar del monto total de dinero mensualmente por la compaAa distribuidora.
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
(
1− p x <
of 21
80000 − 82240 6400
U1. Distribuciones de
Tem 3 Probabilidad
Search document
)
1− p ( x <−0.35 ) 1− 0.36317 =0.63683
c.&) Calcule el monto total máimo ue obtendra la compaAa distribuidora tal ocurra con probabilidad de $6. 2
X 1 N ( μ=82240, σ =6400
2
)
p ( x ≤ k )=0.95
(
p x <
k −82240 6400
k −82240 6400
)=
0.95
= 1.64
k =92736.00 /.
l tiempo de conservación en buen estado X en meses de un producto ue es en eperimentalmente es una variable aleatoria con distribución normal "( 1) a) 2Cuál es la probabilidad de ue uno de tales productos envasados permane@ca estado más de 4 meses3 You're Reading a Preview 2
X 1 N ( μ=2, σ =1
2
)
Unlock full access with a free trial.
Download With Free Trial
p ( x > 4 ) 1− p ( x < 4 )
(
1− p x <
4 −2 1
)
1− p ( x < 2 ) 1−0.97725 =0.02275
b)
Sign up to vote on this title
Not useful Usefules Si el costo de producción de uno de estos productos C * ( E "&0 F X) 2C valor esperado del costo3
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
1− p ( x < 7000 )
(
7000 −6100
1− p x <
√ 692500
)
1− p ( x < 1.082 ) 1− 0.85993 =0.14007
b)
2Iu; ingreso total máimo podrá obtener en un mes de modo ue esto ocu probabilidad 0.+3. μ (¿¿ 1 + μ2=6100, σ 21 +σ 22=692500 ) X 1 + X 2 ¿ p ( x ≤ k )=0.96
(
p x <
k −6100
√ 692500
k −6100
√ 692500
)=
0.96
=1.75
k =7556.29 .
You're Reading a Preview
Unlock fullvalor accesstiene with a free trial. l precio ue se 7ia para cierto tipo de distribución normal con media J * desviación σ * K$.00. 9os compradores desean pagar una cantidad ue tambi Trial distribución normal con media J Download * K4$.00 yWith K(.$0. 2Cuál es la probabilidad de u σ *Free lugar una transacción3.
μ
(¿¿ X =50, σ X 2 =52) X ¿ μ
(¿¿ Y = 45, σ 2Y =2.52 ) Y ¿
Sign up to vote on this title
μ
(
2
2
)
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
reali@an una despu;s de la otra. l tiempo ue se emplea para la primera tarea variable normal con µ * 10 minutos y σ * 1.$ minutos. !ara la segunda tarea se em tiempo normal con µ * 1$ minutos y σ * ( minutos a) 2Cuál es la probabilidad ue en la inspección se emplee más de M ,ora3 μ (¿¿ 1 + μ2=25, σ 21 +σ 22=6.25 ) X 1 + X 2 ¿ p ( x > 30 ) 1− p ( x < 30 )
(
1− p x <
30−25
√ 6.25
)
1− p ( x < 2 ) 1− 0.97725 =0.02275
b)
2n u; tiempo máimo se concluirá la inspección con una probabilidad de 0.$ μ (¿¿ 1 + μ2=25, σ 21 +σ 22=6.25 ) X 1 + X 2 ¿ You're Reading a Preview p ( x ≤ k )=0.995
(
p x < k −25
√ 6.25
k −25
√ 6.25
)=
0.995
Unlock full access with a free trial.
Download With Free Trial
= 2.58
k =31.45
11. Gna empresa agroindustrial tiene dos plantaciones de piAas en el valle de C,anc,am di7erentes motivos en la presente temporada las piAas detitle ambas plant Sign cosec,adas up to vote on this presentan di7erencias. 9as de una plantación "@ona baa) son más y su pes Useful Notgrandes useful aproimadamente una distribución normal con µ * 1+$0 gr. y σ * 100 gr. 9os pesos de plantación "@ona alta) tambi;n siguen una distribución normal con µ * 14$0 gr. y σ * 1$
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
p ( x < 0.33 )− p ( x <−0.33 ) 0.62930 − 0.37070 =0.2586
b)
Si se toma al a@ar cuatro piAas de la @ona baa y se las pesa untas 2c probabilidad de ue el peso total sea mayor a +/$0 gr.3. 2
X 1 N ( μ=1650, σ =100
2
) μ=1650 × 4 =6600
σ =100 × 4 = 400 p ( x > 6750 ) 1− p ( x < 6750 )
(
1− p x <
6750 −6600 400
)
1− p ( x < 0.375 ) 1− 0.64803 =0.35197
c)
Si se toma al a@ar una piAa de cada plantación y se las pesa untas= 2cuá probabilidad de ue el peso sea mayor a &&00 gr.3 You're Reading a Preview
μ (¿¿ 1 + μ2=3100, σ 21 + σ 22=32500 ) X 1 + X 2 ¿
Unlock full access with a free trial.
Download With Free Trial
p ( x > 3300 ) 1− p ( x < 3300 )
(
1− p x <
3300−3100
√ 32500
)
1− p ( x < 1.109 ) Sign up to vote on this title
1− 0.86650 =0.13350
12. Sea Y = X
X
X ! donde X
N"4! 3#$% X
Useful
Not useful
N"6! 4#$ y X
N"8! 5#$ son v
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
μ
(¿¿ Y =18, σ 2Y =50 ) Y = N ¿ b)
-vale /"Y 20$ μ (¿ ¿ Y =18, σ Y = √ 50 ) Y N ¿ p ( x < 20 )
(
p x <
20 −18
√ 50
)
p ( x < 0.283 )=0.61026 c)
Si = 2X 1 3X2 3X3! allar la )ro&a&ilidad de +e sea s+)erior a 15
R= 2 X 1 −3 X 2+ 3 X 3 μ R= 2 μ
(¿¿ 1−3 μ2 + 3 μ3 , σ R2 =2 σ 21−3 σ 22+3 σ 23) You're Reading a Preview R N ¿ Unlock full access with a free trial.
μ (¿¿ R=14, σ R2 =√ 45 ) R = N ¿
Download With Free Trial
μ (¿¿ R=14, σ R2 =√ 45 ) R = N ¿ p ( x > 15 ) Sign up to vote on this title
1− p ( x < 15 )
(
1− p x <
15−14
)
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
p ( x < 7.59 ) =0.0249269 b)
!"X % (/.$) 2
X X (17 ) p ( x > 27.59 ) 1− p ( x < 27.59 ) 1−0.95 = 0.05
c)
!"+.408 ≤ X ≤ (/.$) 2
X X (17 ) p ( 6.408< x < 27.59 ) p ( x < 27.59 ) − p ( x < 6.408 ) 0.95− 0.01=0.94
d)
l valor de c tal ue !"X % c) * 0.01 2
X X (17 )
You're Reading a Preview
p ( x > c ) =0.01 1− p ( x < c )=0.01
p ( x < c ) =0.99
Unlock full access with a free trial.
Download With Free Trial
c = 33.4 2
14. Si X O
χ " 24$
,allar los valores de a y b tal ue !"a#X#b) * 0.88 si se tiene ue !"X%b) 2
X X (24 )
Sign up to vote on this title
p ( x > b ) =0.02 1 p ( x < b )
0.02
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
a =15.7 1$. Si X es una variable ue tiene distribución t de Student con una varian@a $'4. Calcul
0arianzaV ( T ) ⇒
n n−2
=
5 4
4 n=5 n −10
n =10 a)
!"N1.81( ≤ X ≤ (.((8) X T ( 10) p (−1.812 < x < 2.228 ) p ( x < 2.228 ) − p ( x <−1.812 ) 0.975 − 0.05= 0.925
b)
l valor de c tal ue : !"Nc # X # c) * 0.+ X T ( 10)
You're Reading a Preview
x <¿ c ∨¿
Unlock full access with a free trial.
¿ p ¿
1− p ( x < c )=
Download With Free Trial 1 −0.96 2
1− p ( x < c )=0.02
p ( x < c ) =0.98 c =2.35931
Sign up to vote on this title
una C,iiNcuadrado 1+. 9as variables aleatorias G y > se distribuyen como con m libertad y una t de Student con n grados de libertad respectivamente. Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
News
Documents
Sheet Music
Area Bajo La Curva
1
Download
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
p ( U > a )= 0.05 1− p ( U < a )=0.05
p ( U < a )= 0.95
2
U X (9 ) p ( U < a )= 0.95 a =16.9
2
U X (25) p ( U < a )= 0.95 a =37.7 You're Reading a Preview
b)
!"G#((.&/) con m * 1&
Unlock full access with a free trial.
Download With Free Trial 2
U X (13) p ( U < 22.37 )= 0.95
Sign up to vote on this title
c)
9os valores de b para ue !">%b)*00$ con n * (0 Useful y n * 40 Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
News
Documents
Sheet Music
Area Bajo La Curva
1
Download
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
V T (20) p ( V < b )= 0.95 b =1.725
V T (20) p ( V < b )= 0.95 b =1.684
1/. Si X es una v.a. ue tiene distribución Q con m y n grados de libertad.
b)
p ( x < 14.66 ) =0.99 !"X % /.&+) si m * $ y n * &.You're Reading a Preview F ( 5,3)
Unlock full access with a free trial.
p ( x > 7.36 )
Download With Free Trial
1− p ( x < 7.36 ) 1−0.934457 =0.065563
c)
!"X # &.$4) si m * + y n * . F ( 6,9) p ( x < 3.54 )=0.956059
d)
Sign up to vote on this title
Useful
Not useful
9os valores de a y b tales ue !"a # X # b) * 0.0 y !"X # b) * 0.$ si m * 8 y n
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
a =0.279 18. 9as variables X R son independientes con las siguientes distribuciones: X → N"µ = 50! σ= 8$
Y
→ ' "15$
2
→ χ "10$
1Tesponder lo siguiente: / "X − 50$
a)
2
)
⇒
A =
2
> c = 0.05
X − μ X −50 = = N ( 0,1) 8 σ
( A )2= A2 X 2(1)
(
X −50 8
)=
(
2
( X −50 )2 64
2
p X ( 1) > c
(
)=0.05
2
1− p X ( 1) < c
(
2
p X (1) < c 64
= X 2(1)
c 64
)=0.05
)=
0.95
=3.84
c =245.76
b) Calcular :
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial
/" Y >1.60$
Y t (15) Sign up to vote on this title
p (|Y |> 1.60 )
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
c)
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
X −µ 2 " σ $ / > & = 0.08 510
2
X X (1 ) , W X ( 10) ⇒ A =
1
W
F ( 1,10)
10
A F (1,10 ) p ( A > b )= 0.08 1− p ( A < b )=0.08
p ( A < b )= 0.92 b =3.79509 d)
2
W X (10)
= 0.10
You're Reading a Preview Unlock full access with a free trial.
p ( k < W ) =0.10
Download With Free Trial
1− p ( W < k )=0.05
p ( U < a )= 0.95 1. Sean X1 X ( X& y X 4 variables aleatorias independientes cada una con distribución est+ndar calcular las siguientes probabilidades:
a)
!U +.$ #R # 1(.8V si se sabe ue 2 1
2 2
2 3
2 4
Y = X + X + X + X
R
2 2 = X12 + X22 + XSign Xto 3 +up 4 vote on this title
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Upload
Sign In
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
Area Bajo La Curva
1
Download
News
Documents
Sheet Music
X
a)
of 21
(
→ χ"10)
R → t" (0)
U1. Distribuciones de
Tem 3 Probabilidad
Search document
(
S → χ"1$)
.
X X (10 ) p ( X > k ) =0.015 1− p ( X < k )=0.015
p ( X < k ) =0.985 k =22.0206 p ( c < X < k ) =0.94 p ( X < k ) − p ( X < c )= 0.94 0.985− p ( X < c ) =0.94
p ( X < c )=0.045 c = 3.82248
b)
!" R < (.04)
You're Reading a Preview Unlock full access with a free trial.
Y t (20)
Download With Free Trial
p (−2.04 < Y < 2.04 ) p ( Y < 2.04 )− p ( Y <−2.04 ) 0.972611 −0.0273891 =0.9452219 Sign up to vote on this title
c)
l valor de , de modo ue: !", # X E # &4.&81+) * 0.+$ Not useful Useful
S 1$ ! > ( 8$
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
68 views
Sign In
Upload
Join
RELATED TITLES
0
Guia 1 Unmsm Eap II- 2016 II Uploaded by OlórteguiRamosLuisNazir
Books
Audiobooks
Magazines
asdf
Save
Embed
Share
Print
News
Documents
Sheet Music
Area Bajo La Curva
1
Download
of 21
U1. Distribuciones de
Tem 3 Probabilidad
Search document
1−0.95 = 0.05
. Supongamos ue la variable aleatoria tiene una distribución W de Student con
de libertad se de7ine la siguiente variable aleatoria / 0 . ?etermine la distribució variable aleatoria /
X t ( m) 2
X (m ) 2
Y = X =(
N ( 0,1 )
(√ ) 2
X ( m) m
2
)=
[ N ( 0,1 ) ]2 2
X m
=
1 2
X (m )
= F ( 1, m )
m
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial
Sign up to vote on this title
Useful
Not useful
Home
Saved
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join