COMPENDIO..

1. 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

2. 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

3. 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

4. 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,
62), col = "green", border =
1, main =
"MICROEMPRESARIOS",xlab =
"EDAD" , ylab = "FRECUENCIA")