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COMPENDIO CUATRO JENNY ROCIO URREA BUSTAMANTE ROSA ELVIRA ARIZA LIC. JORGE OBANDO UNIVERSIDAD COOPERATIVA DE COLOMBIA VILLAVICENCIO – META CONTADURIA PÚBLICA ESTADISTICA DESCRIPTIVA GRUPO 502 2015

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JENNY ROCIO URREA BUSTAMANTEROSA ELVIRA ARIZA

COMPENDIO CUATRO

JENNY ROCIO URREA BUSTAMANTEROSA ELVIRA ARIZA

LIC. JORGE OBANDO

UNIVERSIDAD COOPERATIVA DE COLOMBIAVILLAVICENCIO METACONTADURIA PBLICAESTADISTICA DESCRIPTIVAGRUPO 5022015Taller de Aplicacin:

Basndose en los anteriores procedimientos construir intervalos y grficos para los siguientes datos que corresponden a la edad de 50 microempresarios de la ciudad de Villavicencio

4839352930

3842374038

2237345548

3550364842

5335383835

4050233245

3542592838

3438444623

4048343035

4332363246

Cdigos en RResultados

Ingresando datos:datos=c(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)> datos[1] 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[26] 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

Calculando el rango:Rang= max(datos)-min(datos)> Rang[1] 37

Calculando el nmero de intervalosm=round(1+3.3*log10(50)) La funcin Round, redondea al entero ms cercano.> m[1] 7

Longitud del intervalo:C=Rang/m> C[1] 5.285714Este resultado se redondea al entero ms cercano, por exceso en este caso 6.Redefinir=42-37=52 Xmin-2=22 Xmax +3=62

Ahora le damos forma a los intervalos>intervalos=cut(datos,breaks=c(20,26,32,38,44,50,56,62))> intervalos [1] (44,50] (38,44] (32,38] (26,32] (26,32] (32,38] (38,44] (32,38] (38,44][10] (32,38] (20,26] (32,38] (32,38] (50,56] (44,50] (32,38] (44,50] (32,38][19] (44,50] (38,44] (50,56] (32,38] (32,38] (32,38] (32,38] (38,44] (44,50][28] (20,26] (26,32] (44,50] (32,38] (38,44] (56,62] (26,32] (32,38] (32,38][37] (32,38] (38,44] (44,50] (20,26] (38,44] (44,50] (32,38] (26,32] (32,38][46] (38,44] (26,32] (32,38] (26,32] (44,50]Levels: (20,26] (26,32] (32,38] (38,44] (44,50] (50,56] (56,62]

Ahora se forma las frecuencias absolutasf=table(intervalos)intervalos(20,26] (26,32] (32,38] (38,44] (44,50] 3 7 19 9 9 (50,56] (56,62] 2 1

Calculando el nmero de elementos de la muestran=sum(f)>> n[1] 50

Construimos las frecuencias absolutash=f/n

intervalos(20,26] (26,32] (32,38] (38,44] (44,50] 0.06 0.14 0.38 0.18 0.18 (50,56] (56,62] 0.04 0.02

Construyendo frecuencias absolutas acumuladasF=cumsum(f)(20,26] (26,32] (32,38] (38,44] (44,50] 3 10 29 38 47(50,56] (56,62] 49 50

Construyendo las frecuencias relativas acumuladas.H=cumsum(h)(20,26] (26,32] (32,38] (38,44] (44,50] 0.06 0.20 0.58 0.76 0.94 (50,56] (56,62]0.98 1.00

Ahora se arman la tabla de frecuenciascbind(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> MarcaMarca[1] 23 29 35 41 47 53 59

La tabla con las frecuencias y la marca de clasetabla=cbind(f,Marca,h,F,H)f Marca H F(20,26] 3 23 0.06 3(26,32] 7 29 0.20 10(32,38] 19 35 0.58 29(38,44] 9 41 0.76 38(44,50] 9 47 0.94 47(50,56] 2 53 0.98 49(56,62] 1 59 1.00 50

Grfico de un histograma>hist(datos,breaks=c(20,26,32,38,44,50,56,62),col="pink",border=1,main="Microempresarios Villavicencio",xlab="EDAD",ylab="FRECUENCIA")