compendio cuatro jenny urrea.docx

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")