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")