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Licenciatura en Economía
Cuarto semestre. Probabilidad y Estadística
Unidad II. Distribuciones de Probabilidad
Prof. Jorge Mendoza
Actividad 4
1.
Supón que una distribución normal dada tiene una media de 20 y una
desviación estándar de 4. Calcula las probabilidades siguientes usando la regla
empírica:
a)
P(X>24)
z
=
x
−
μ
σ
=
z
=
24
−
20
4
=
1
(
0.3413
+
0.5
)
−
1
=
0.1587
=
15.87
La probabilidad de que sea mayor de 24 es del 15.87%
b) P(16<X<24)
z
=
x
−
μ
σ
=
z
=
16
−
20
4
=−
1
z
=
24
−
20
4
=
1
(
0.3413
+
0.3413
)
=
0.6826
=
68.26
La probabilidad de que este entre 16 y 24 es de 68.26%
c)
P(12<X<24)
z
=
x
−
μ
σ
=
z
=
12
−
20
4
=−
2
z
=
24
−
20
4
=
1
(
0.4772
+
0.3413
)
=
0.8185
=
81.85
La probabilidad de que este entre 12 y 24 es de 81.85%
d) P(X>28)
z
=
x
−
μ
σ
=
z
=
28
−
20
4
=
2
(
0.4772
+
0.5
)
−
1
=
0.0228
La probabilidad de que sea mayor que 28 es de 2.28%
VIDEO Tabla de “Z”
http://www.est.uc3m.es/esp/nueva_docencia/comp_col_leg/ing_tec_inf_gestion/estadistica/
Documentacion/Tablas/tablas2caras.pdf
VIDEO DISTRIBUCIÓN ESTÁNDAR
https://www.youtube.com/watch?v=sDQgKLaQ5UM
https://www.youtube.com/watch?v=sDQgKLaQ5UM
1
Licenciatura en Economía
Cuarto semestre. Probabilidad y Estadística
Unidad II. Distribuciones de Probabilidad
Prof. Jorge Mendoza
https://www.youtube.com/watch?v=PKgk6wmM2FI
2.
Si una variable aleatoria
x
se distribuye normalmente con una media de 70 y
una desviación estándar de 5, encuentra la probabilidad de que
x
(usa la tabla de
distribución
z
):
a)
Esté entre 60 y 80
z
=
x
−
μ
σ
=
z
=
68
−
7 0
5
=−
0.4
z
=
80
−
70
5
=
2
(
0.1554
+
0.4772
)
=
0.6326
=
63.26
La probabilidad de que este entre 60 y 80 es de 63.26%
b)
Sea menor que 65
z
=
x
−
μ
σ
=
z
=
65
−
70
5
−
1
(
0.3413
+
0.5
)
−
1
=
0.1587
=
15.87
La probabilidad de que sea menor que 65 es de 15.87%
c)
Sea menor que 81
z
=
x
−
μ
σ
=
z
=
81
−
70
5
=
2.2
(
0.4861
+
0.5
)
=
0.9861
=
98.61
La probabilidad de que sea menor que 81 es de 98.61%
d)
Sea mayor que 67.5
z
=
x
−
μ
σ
=
z
=
67.5
−
70
5
=
0.5
(
0.1915
+
0.5
)
=
0.6915
=
69.15
La probabilidad de que sea mayor que 67.5 es de 69.15%
e)
Sea menor que 67 o mayor que 77
z
=
x
−
μ
σ
=
z
=
67
−
7 0
5
=−
0.6
z
=
77
−
70
5
=
1.4
(
0.2257
+
0.5
)
−
1
=
0.2743,
(
0.4192
+
0.5
)
−
1
=
0.0808,0.2743
+
0.0808
=
0.3551
La probabilidad de que sea menor que 67 y mayor que 77es de 35.51%
2
Licenciatura en Economía
Cuarto semestre. Probabilidad y Estadística
Unidad II. Distribuciones de Probabilidad
Prof. Jorge Mendoza
3.
Una máquina produce pernos cuyos diámetros se distribuyen normalmente con
un promedio de 0.25 pulgadas y una desviación estándar de 0.02 pulgadas. ¿Cuál es la
probabilidad de que un perno tenga un diámetro?
a)
Mayor de 0.3 pulgadas
z
=
x
−
μ
σ
=
z
=
0.3
−
0.25
0.02
=
2.5
(
0.4938
+
0.5
)
−
1
=
0.0062
=
0.62
La probabilidad de que sea mayor que 0.3 pulgadas es de 0.62%
b)
Entre 0.2 y 0.3 pulgadas
z
=
x
−
μ
σ
=
z
=
0.2
−
0.25
0.02
=−
2.5
z
=
0.3
−
0.25
0.02
=
2.5
(
0.4938
+
0.4938
)
=
0.9876
=
98.76
La probabilidad de que este entre 0.2 y 0.3 pulgadas es de 98.76
c)
Menor que 0.19 pulgadas
z
=
x
−
μ
σ
=
z
=
0.19
−
0.25
0.02
=−
3
(
0.4987
+
0.5
)
−
1
=
0.0013
=
0.13
La probabilidad de que sea menor que 0.19 pulgadas es de 0.13%
4.
Supón que
x
se distribuye exponencialmente con parámetro λ = 6.4. Determina
su media, su varianza y su desviación estándar.
Respuesta:
Media= 6.4
Varianza= 6.4
Desviación Estándar= 2.5298
3
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