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2. Por lo que la función y la variable aleatoria continua tendrán
la misma representación pero con ligeras diferencias.
DISTRIBUCIÓN NORMAL
Es
Un modelo teórico
Puede
Aproximar satisfactoriamente
el valor de una variable
aleatoria continua a una
situación ideal.
Adapta
Una variable aleatoria
continua a una función que
depende de la media y
la desviación típica.
Gráfica de: m es el valor esperado y s es la
desviación estándar poblacional
FUNCIÓN DE PROBABILIDAD
que exp {} significa e elevado a lo
que hay dentro del paréntesis,
siendo e el número 2.71828...
p es el número
3.14159....
Los parámetros: m y s.
−
−
= 2
2
)
(
2
1
exp
2
1
)
(
s
m
s
x
x
p
3. Distribución
normal estándar
N (0, 1)
𝑍 =
ത
𝑋 − 𝜇
𝛿
Un parámetro muy
importante es la media
(µ) y siempre estará al
centro de la curva con
forma de campana.
una distribución normal con media igual a 8. la figura es:
La probabilidad de
la variable X
dependerá del área
del recinto.
Para calcularla
utilizaremos una
tabla.
Tipificación de la
variable
Para utilizar la
tabla tenemos
que transformar
2
variables
La X que sigue
una distribución
N(µ, σ)
Y la Z que sigue
la distribución
N(0, 1)
El calculo se
realiza mediante
la expresión:
4. DISTRIBUCIÓN NORMAL
ESTÁNDAR N (0,1)
Es
Una distribución normal
Toma en cuenta
la media(µ) y la
desviación estándar(σ).
Caracterizada por:
El área bajo
la curva es
igual a 1.
Es simétrica
respecto al
centro, o a la
media.
50% de los valores
son mayores que la
media, y 50% de los
valores son menores
que la media.
La media es
igual a la
mediana y a
la moda.
Tiene una
asíntota en
y = 0 (eje x).
Tiene una media igual a cero
Una Desviación estándar
igual a uno
Tiene
Esta función de densidad
normal estandarizada,
trabaja con la variable
estandarizada z en el eje
horizontal.
5. Por ejemplo, si se desea
encontrar la probabilidad de
que la variable
estandarizada z, tome un
valor entre 0 y 1,50; hay que
encontrar el área bajo la
curva entre z = 0 y z = 1,50.
Este valor se calcula en la tabla
9. En una ciudad se estima que la temperatura máxima en
el mes de junio sigue una distribución normal, con media
230C y desviación típica 50C
Calcular el número de días del mes en los
que se espera alcanzar máximas entre 210C
y 270C. Ver la figura. Para el desarrollo del
problema se aplica la formula.
𝑍 =
ത
𝑋 − 𝜇
𝛿
Para X= 210C se tiene
𝑍 =
21 − 23
5
=
−2
5
= −0,4 𝐸𝑠𝑡𝑒 𝑣𝑎𝑙𝑜𝑟 𝑒𝑛 𝑙𝑎 𝑡𝑎𝑏𝑙𝑎
P(Z) = 0,1554
0,1554
Para X= 270C se tiene
𝑍 =
27 − 23
5
=
4
5
= 0,8 𝐸𝑠𝑡𝑒 𝑣𝑎𝑙𝑜𝑟 𝑒𝑛 𝑙𝑎 𝑡𝑎𝑏𝑙𝑎
P(Z) = 0.2881
Luego la probabilidad de que se presenten
temperaturas entre 21 y 27 grados es de:
P(21<Z<27) = 0,1554 + 0.2881 = 0,444
El numero de días esperado seria igual:
ndias = 30*P= 0,444*30 = 13 días
Problema Solución
10. La media de los pesos de 500 estudiantes de un colegio es 70 kg y la
varianza de 81kg. Suponiendo que los pesos se distribuyen
normalmente, hallar cuántos estudiantes pesan: Más de 90 kg, Menos de 64 kg.
Mas de 90 Menos de 64
𝑍 =
90 − 70
9
=
20
9
= 2,22
𝑍 =
64 − 70
9
=
−6
9
= −0,666
P(z=2,222)=0.4783 luego p(z>90) = 0,5 – 0,4783 = 0,0217
0,4783
Cuantos estudiantes pesan mas de 90 Kg
n= 500*0,0217= 11 estudiantes
P(z=-0,666)=0.2454 luego p(z<64) = 0,5 – 0,2454 = 25,46%
0.2454
Cuantos estudiantes
pesan menos de 64 Kg
n= 500*0,2546= 127 estudiantes
Problema
Solución
11.
12.
13.
14.
15. 𝑍 =
3 − 4,2
1,25
= −0,96
𝜇=4,2horas 𝛿 = 1,25 horas p(x<3horas)
Datos:
Procedimiento, se calcula el valor de Z
p(x<3horas)=0,5-0,3315=0,17
0,3315
La probabilidad de que un
usuario permanezca
menos de 3 horas en las
redes sociales es del 17%
0,5
Se ha encontrado que el tiempo dedicado al Internet diariamente en las
redes sociales, se distribuye normalmente, con media 4,2 horas y
desviación estándar 1,25 horas. Si esto es así: ¿Cuál es la probabilidad de
que un usuario permanezca en las redes sociales menos de 3 horas al día?
Problema
Solución
16. Dependerá de qué tan asimétrica
sea la distribución poblacional
TEOREMA CENTRAL DEL LÍMITE
Para
El muestreo aleatorio con n “grande”
aprox. normal sin importar la forma
de la distribución poblacional
La distribución muestral de la
media muestral tiene aprox.
una distribución normal
n ≥ 30 es
suficiente
Es
¿Qué tan grande será n? Generalmente
Si la población sigue la distribución normal, la distribución muestral
de la media muestral seguirá también la distribución normal
Limite de la distribución
𝒁 =
𝑿 − 𝝁
𝝈
𝒏
𝒁 =
𝑿 − 𝝁
𝝈
𝒏
=
𝑿 − 𝝁 ∗ 𝒏
𝝈
Cuando n
tiende a infinito
El valor de z también se lee
en la tabla de distribución
normal. Recordar que Z
empieza en cero
17. La duración de la enfermedad del Covid 19 varía entre 1 a 16 días (desde los síntomas al
fallecimiento), el promedio de las personas que han fallecidos es de 70 años con una desviación
estándar de 40 años. Se seleccionan los expedientes de 25 pacientes ya fallecidos. Encuentre:
a) La probabilidad de que una persona menor de 50 años muera es de:
En 25 pacientes, la
probabilidad de que
una persona menor de
50 años muera es del
0,62%.
𝑍 =
50 − 70
40
25
=
−20 ∗ 25
40
=
−100
40
= −2,5
Usando la tabla A(Z) = 0,4938
Luego p(x<50)=0,5-0,4938=0,0062=0,62%
𝜇 = 70 𝜎 = 40 𝑛 = 25 𝑝 𝑥 < 50 =?
Ejemplo 1
Solución
18. Suponga que la media del precio de venta de un galón de gasolina en Colombia es de $9.600 Además, asuma que la
distribución está posiblemente inclinada, con una desviación estándar de $500. ¿Cuál es la probabilidad de
seleccionar una muestra de 35 estaciones de gasolina y encontrar una media muestral entre $9400 y 9800?
En ambos casos el problema propone estos datos:
𝑍1 =
9400−9600
500
35
=
−200 ∗ 35
500
= -2,36
𝜇=$9.600 𝜎 = 500 𝑛 = 35 P(9400<X<9800) =?
Caso a)
𝑍2 =
9800−9600
500
35
=
200 ∗ 35
500
= 2,36
Caso b)
A(z1)=0,4909
0,4909
0,4909
P(9400<X<9800) =0,4909*2=0,98
A(z2)=0,4909
La probabilidad de
encontrar precios en la
estación elegida entre 9600
y 9800 es del 98%
Ejemplo 2
Solución
Caso a: Las estaciones que
venden entre 9400 y 9600.
Caso b: Las estaciones que
venden entre 9600 y 9800.