Solucion compendio 4
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SOLUCIÓN COMPENDIO 4 Taller de Aplicación: Basándose en los anteriores procedimientos construir intervalos y gráficos para los siguientes datos que corresponden a la edad de 50 microempresarios de la ciudad de Villavicencio 48 39 35 29 30 38 42 37 40 38 22 37 34 55 48 35 50 36 48 42 53 35 38 38 35 40 50 23 32 45 35 42 59 28 38 34 38 44 46 23 40 48 34 30 35 43 32 36 32 46 Códigos en R Resultados Ingresando datos: datos=c(48,38,22,35,53,40 ,35,34,40,43,39,42,37,50, 35,50,42,38,48,32,35,37,3 4,36,38,23,59,44,34,36,29 ,40,55,48,38,32,28,46,30, 32,30,38,48,42,35,45,38,2 3,35,46) [1] 48 38 22 35 53 40 35 34 40 43 39 42 37 50 35 50 42 38 48 32 35 37 34 36 38 [26] 23 59 44 34 36 29 40 55 48 38 32 28 46 30 32 30 38 48 42 35 45 38 23 35 46 Calculando el rango: Rang= max(datos)- min(datos) > Rang [1] 37
Transcript of Solucion compendio 4
SOLUCIÓN COMPENDIO 4
Taller de Aplicación:
Basándose en los anteriores procedimientos construir intervalos y gráficos para los siguientes
datos que corresponden a la edad de 50 microempresarios de la ciudad de Villavicencio
48 39 35 29 30
38 42 37 40 38
22 37 34 55 48
35 50 36 48 42
53 35 38 38 35
40 50 23 32 45
35 42 59 28 38
34 38 44 46 23
40 48 34 30 35
43 32 36 32 46
Códigos en R Resultados
Ingresando datos:
datos=c(48,38,22,35,53,40
,35,34,40,43,39,42,37,50,
35,50,42,38,48,32,35,37,3
4,36,38,23,59,44,34,36,29
,40,55,48,38,32,28,46,30,
32,30,38,48,42,35,45,38,2
3,35,46)
[1] 48 38 22 35 53 40 35 34 40 43
39 42 37 50 35 50 42 38 48 32 35
37 34 36 38
[26] 23 59 44 34 36 29 40 55 48 38
32 28 46 30 32 30 38 48 42 35 45
38 23 35 46
Calculando el rango:
Rang= max(datos)-
min(datos)
> Rang
[1] 37
Calculando el número de
intervalos
m=round(1+3.3*log10(50))
La función Round, redondea
al entero más cercano.
> m
[1] 7
Longitud del intervalo:
C=Rang/m
> C
[1] 5.285714
Este resultado se redondea al entero
más cercano, por exceso en este caso 6.
Redefinir=42-37=5
2 Xmin-2=20
Xmax +3=62
Ahora le damos forma a los
intervalos
intervalos=cut(datos,
breaks=c(20,26,32,38,44,5
0,56,62))
Intervalos
[1] (44,50] (32,38] (20,26]
(32,38] (50,56] (38,44] (32,38]
(32,38] (38,44]
[10] (38,44] (38,44] (38,44]
(32,38] (44,50] (32,38] (44,50]
(38,44] (32,38]
[19] (44,50] (26,32] (32,38]
(32,38] (32,38] (32,38] (32,38]
(20,26] (56,62]
[28] (38,44] (32,38] (32,38]
(26,32] (38,44] (50,56] (44,50]
(32,38] (26,32]
[37] (26,32] (44,50] (26,32]
(26,32] (26,32] (32,38] (44,50]
(38,44] (32,38]
[46] (44,50] (32,38] (20,26]
(32,38] (44,50]
Levels: (20,26] (26,32] (32,38]
(38,44] (44,50] (50,56] (56,62]
Ahora se forma las
frecuencias absolutas
f=table(intervalos)
f
intervalos
(20,26] (26,32] (32,38] (38,44]
(44,50] (50,56] (56,62]
3 7 19 9
9 2 1
Calculando el número de
elementos de la muestra
n=sum(f)
> n
[1] 50
Construimos las frecuencias
absolutas
h
h=f/n
h
intervalos
(20,26] (26,32] (32,38] (38,44]
(44,50] (50,56] (56,62]
0.06 0.14 0.38 0.18
0.18 0.04 0.02
Construyendo frecuencias
absolutas acumuladas
F=cumsum(f)
F
(20,26] (26,32] (32,38] (38,44]
(44,50] (50,56] (56,62]
3 10 29 38
47 49 50
Construyendo las frecuencias
relativas acumuladas.
H=cumsum(h)
H
(20,26] (26,32] (32,38] (38,44]
(44,50] (50,56] (56,62]
0.06 0.20 0.58 0.76
0.94 0.98 1.00
Ahora se arman la tabla de
frecuencias
cbind(f,h,F,H)
f h F H
(20,26] 3 0.06 3 0.06
(26,32] 7 0.14 10 0.20
(32,38] 19 0.38 29 0.58
(38,44] 9 0.18 38 0.76
(44,50] 9 0.18 47 0.94
(50,56] 2 0.04 49 0.98
(56,62] 1 0.02 50 1.00
Construyendo marcas de clase
LimSup=c(26,32,38,44,50,56,62) LimInf=c(20,26,32,38,44,50,56)
Marca= (LimSup+LimInf)/2
Marca
[1] 23 29 35 41 47 53 59
La tabla con las frecuencias y
la marca de clase
f Marca h F H
(20,26] 3 23 0.06 3 0.06
(26,32] 7 29 0.14 10 0.20
tabla=cbind(f,Marca,h,F,H) (32,38] 19 35 0.38 29 0.58
(38,44] 9 41 0.18 38 0.76
(44,50] 9 47 0.18 47 0.94
(50,56] 2 53 0.04 49 0.98
(56,62] 1 59 0.02 50 1.00
Gráfico de un histograma
hist(datos,
breaks=c(20,26,32,38,44,50,56,6
2), col = "green", border = 1,
main = "MICROEMPRESARIOS",xlab
= "EDAD" , ylab = "FRECUENCIA")
