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UNIVERSIDAD NACIONAL DEL CENTRO DEL PERÚ FACULTAD DE INGENIERÍA QUÍMICA PRACICA CALIFICADA DE ESTADÍSTICA Y DISEÑO DE EXPERIMENTOS

Apellidos y Nombres: ……………………………………………………………….…………………………………...……… 1. De los valores obtenidos de la medición de la concentración de Fe de una solución acuosa son las siguientes 0,5 0,3 0,6 0,9 0,8 0,8 0,8 0,3 0,6 0,4 0,6 0,2 0,8 0,4 0,2 0,4 0,7 0,5 a) Desarrolle la distribución de frecuencia, los parámetros de medida de tendencia central, medidas de dispersión y realice la gráfica de barras. b) Determine los cuartiles, y el coeficiente de asimetría c) Determine qué % son menores al valor de 0,5 > concentracion_Fe=c(.5,.3,.6,.9,.8,.8,.8,.3,.6,.4,.6,.2,.8,.4,.2,.4,.7,.5) > length(concentracion_Fe) [1] 18 > tabfre tabfre concentracion_Fe Freq 1 0.2 2 2 0.3 2 3 0.4 3 4 0.5 2 5 0.6 3 6 0.7 1 7 0.8 4 8 0.9 1 > barplot(tabfre$Freq,names.arg=tabfre$concentracion_Fe,col=c(2,4,6,8,10,12,14,16,18,2 0) ) > transform(tabfre,F=cumsum(Freq)) concentracion_Fe Freq F 1 0.2 2 2 2 0.3 2 4 3 0.4 3 7 4 0.5 2 9 5 0.6 3 12 6 0.7 1 13 7 0.8 4 17 8 0.9 1 18 > transform(tabfre,F=cumsum(Freq),h=round(prop.table(Freq),2)) concentracion_Fe Freq F h 1 0.2 2 2 0.11 2 0.3 2 4 0.11 3 0.4 3 7 0.17 4 0.5 2 9 0.11 5 0.6 3 12 0.17 6 0.7 1 13 0.06 7 0.8 4 17 0.22 8 0.9 1 18 0.06 > transform(tabfre,F=cumsum(Freq),h=round(prop.table(Freq),2),H=round(cumsum(prop.tabl e(Freq)),2)) concentracion_Fe Freq F h H 1 0.2 2 2 0.11 0.11 2 0.3 2 4 0.11 0.22 3 0.4 3 7 0.17 0.39 4 0.5 2 9 0.11 0.50 5 0.6 3 12 0.17 0.67 6 0.7 1 13 0.06 0.72 7 0.8 4 17 0.22 0.94 8 0.9 1 18 0.06 1.00 > transform(tabfre,F=cumsum(Freq),h=round(prop.table(Freq),2),H=round(cumsum(prop.tabl e(Freq)),2),POR=round(prop.table(Freq),2)*100) concentracion_Fe Freq F h H POR 1 0.2 2 2 0.11 0.11 11 2 0.3 2 4 0.11 0.22 11 3 0.4 3 7 0.17 0.39 17 4 0.5 2 9 0.11 0.50 11

5 0.6 3 12 0.17 0.67 17 6 0.7 1 13 0.06 0.72 6 7 0.8 4 17 0.22 0.94 22 8 0.9 1 18 0.06 1.00 6 > media=mean(concentracion_Fe) > media [1] 0.5444444 > mediana=median(concentracion_Fe) > mediana [1] 0.55 > install.packages("modeest") WARNING: Rtools is required to build R packages but is not currently installed. Please download and install the appropriate version of Rtools before proceeding: https://cran.rstudio.com/bin/windows/Rtools/ Installing package into ‘C:/Users/user/Documents/R/win-library/4.0’ (as ‘lib’ is unspecified) probando la URL 'https://cran.rstudio.com/bin/windows/contrib/4.0/modeest_2.4.0.zip' Content type 'application/zip' length 142718 bytes (139 KB) downloaded 139 KB package ‘modeest’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\user\AppData\Local\Temp\Rtmp4YVYFq\downloaded_packages > library(modeest) > moda=mfv(concentracion_Fe) > moda [1] 0.8 > Q1=quantile(concentracion_Fe,0.25) > Q1 25% 0.4 > Q2=quantile(concentracion_Fe,0.5) > Q2 50% 0.55 > Q3=quantile(concentracion_Fe,0.75) > Q3 75% 0.775 > install.packages("psych") WARNING: Rtools is required to build R packages but is not currently installed. Please download and install the appropriate version of Rtools before proceeding: https://cran.rstudio.com/bin/windows/Rtools/ Installing package into ‘C:/Users/user/Documents/R/win-library/4.0’ (as ‘lib’ is unspecified) probando la URL 'https://cran.rstudio.com/bin/windows/contrib/4.0/psych_1.9.12.31.zip' Content type 'application/zip' length 3783807 bytes (3.6 MB) downloaded 3.6 MB package ‘psych’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\user\AppData\Local\Temp\Rtmp4YVYFq\downloaded_packages > library(psych) Registered S3 method overwritten by 'psych': method from plot.residuals rmutil > coe_sim=skew(concentracion_Fe) > coe_sim [1] -0.02093996 > kurtosis=kurtosi(concentracion_Fe) > kurtosis [1] -1.417979 Porc=((0.5-0.4)/17)*100 Porc [1] 38.8

2. De la base de datos de Excel con nombre metales responder las siguientes preguntas a. Desarrolle la distribución de frecuencia, los parámetros de medida de tendencia central, medidas de dispersión y realice la gráfica de barras. b. Determine los cuartiles, el percentil 80 y el coeficiente de asimetría

c. Determine qué % son valores mayores de 10 y menores que 15 file.choose() [1] "E:\\ESTADISTICA IV\\Metales.xlsx" > b_datos_metales="E:\\ESTADISTICA IV\\Metales.xlsx" > excel_sheets(b_datos_metales) [1] "ZINC" > B_datos B_datosa=data.frame(B_datos) > Conc_Zn=B_datosa[,1] > Conc_Zn [1] 13.7 14.4 8.5 12.4 12.4 11.3 16.3 14.6 19.0 13.4 11.7 11.2 10.6 15.3 14.1 13.6 17.1 13.1 7.4 9.4 [21] 12.8 15.5 18.1 18.2 13.0 14.1 13.3 13.5 16.6 14.3 9.1 13.4 12.9 13.0 19.5 14.1 16.4 14.9 14.2 9.4 [41] 14.6 13.1 12.9 10.0 15.9 14.9 12.5 16.8 19.0 10.3 12.6 12.9 15.7 13.6 8.9 12.4 13.8 9.2 15.8 9.5 > n=length(Conc_Zn) > k=round(1+3.3*log(n,10)) > rango=max(Conc_Zn)-min(Conc_Zn) > A=rango/k > histograma=hist(Conc_Zn,breaks=seq(min(Conc_Zn),max(Conc_Zn),by=A),col="blue",main="HIS TOGRAMA DE CONCENTRACION DE ZINC",xlab="CONCENTRACION",ylab="FRECUENCIA") > library(agricolae) > TF=(table.freq(histograma)) > TF Lower Upper Main Frequency Percentage CF CPF 1 7.400000 9.128571 8.264286 4 6.7 4 6.7 2 9.128571 10.857143 9.992857 7 11.7 11 18.3 3 10.857143 12.585714 11.721429 7 11.7 18 30.0 4 12.585714 14.314286 13.450000 22 36.7 40 66.7 5 14.314286 16.042857 15.178571 10 16.7 50 83.3 6 16.042857 17.771429 16.907143 5 8.3 55 91.7 7 17.771429 19.500000 18.635714 5 8.3 60 100.0 > TFC=round(TF,2) > TFC Lower Upper Main Frequency Percentage CF CPF 1 7.40 9.13 8.26 4 6.7 4 6.7 2 9.13 10.86 9.99 7 11.7 11 18.3 3 10.86 12.59 11.72 7 11.7 18 30.0 4 12.59 14.31 13.45 22 36.7 40 66.7 5 14.31 16.04 15.18 10 16.7 50 83.3 6 16.04 17.77 16.91 5 8.3 55 91.7 7 17.77 19.50 18.64 5 8.3 60 100.0 > GRA=TFC[,3:4] > GRA Main Frequency 1 8.26 4 2 9.99 7 3 11.72 7 4 13.45 22 5 15.18 10 6 16.91 5 7 18.64 5 > byd=barplot(GRA$Frequency,names.arg=GRA$Main,col=c("red","green","blue","yellow","orang e","blue","red")) > media=sum(TFC[,3]*TFC[,4])/n > media [1] 13.50767 > mediana=12.59+((n/2-18)/22)*A > mediana [1] 13.53286 > moda=12.59+((22-7)/((22-7)+(22-10)))*A > moda [1] 13.55032 > desmedia=sum(abs((TFC[,3]-media)*TFC[,4]))/n > desmedia [1] 1.979889 > varianza=sum((TFC[,3]-media)^2*TFC[,4])/3 > varianza [1] 145.588 > desestan=sqrt(varianza) > desestan [1] 12.06598 > desestan=sqrt(varianza) > coe_var=desestan/media

> coe_var [1] 0.8932693 > Eror_est=desestan/sqrt(n) > Eror_est [1] 1.557712 > Q1=10.86+A*(15-11)/(18-11) > Q1 [1] 11.84776 > Q2=12.59+A*(30-18)/(40-18) > Q2 [1] 13.53286 > Q3=14.31+A*(45-40)/(50-40) > Q3 [1] 15.17429 > fx=(19.5-10)*7/A > fx=(19.5-10)*11/A > fx=(19.50-10.00)*7/A

3. De os valores de pH responder las siguientes preguntas 5 5 6 4 6 6 7 8 6 5 7 8 4 8 6 9 5 5 6 4 6 6 7 8 6 5 7 8 4 8 6 9 5 5 6 4 6 6 7 8 a) Desarrolle la distribución de frecuencia, los parámetros de medida de tendencia central, medidas de dispersión y desarrolla la gráfica de barras. b) Determine los percentiles 30 y 50 c) Determina la cantidad de datos de pH menores a 5 y mayores a 7 pH=c(5,5,6,4,6,6,7,8,6,5,7,8,4,8,6,9,5,5,6,4,6,6,7,8,6,5,7,8,4,8,6,9,5,5,6,4,6,6,7,8 ) > n=length(pH) > K=1+3.3*log10(n) > K=round(K) > rango=max(pH)-min(pH) > A=rango/K > histograma=hist(pH,breaks=seq(min(pH),max(pH),by=A)) > Tab_Fre=(table.freq(histograma)) > Tab_Fre Lower Upper Main Frequency Percentage CF CPF 1 4.000000 4.833333 4.416667 5 12.5 5 12.5 2 4.833333 5.666667 5.250000 8 20.0 13 32.5

3 4 5 6 > > 1 2 3 4 5 6 > >

5.666667 6.500000 6.083333 13 32.5 6.500000 7.333333 6.916667 5 12.5 7.333333 8.166667 7.750000 7 17.5 8.166667 9.000000 8.583333 2 5.0 TFC=round(Tab_Fre,2) TFC Lower Upper Main Frequency Percentage CF CPF 4.00 4.83 4.42 5 12.5 5 12.5 4.83 5.67 5.25 8 20.0 13 32.5 5.67 6.50 6.08 13 32.5 26 65.0 6.50 7.33 6.92 5 12.5 31 77.5 7.33 8.17 7.75 7 17.5 38 95.0 8.17 9.00 8.58 2 5.0 40 100.0 GRA=TFC[,3:4] GRA Main Frequency 4.42 5 5.25 8 6.08 13 6.92 5 7.75 7 8.58 2

26 65.0 31 77.5 38 95.0 40 100.0

1 2 3 4 5 6 > byd=barplot(GRA$Frequency,names.arg=GRA$Main,col=c("red","green","blue","yellow","or ange","blue")) > media=sum(TFC[,3]*TFC[,4])/n > media [1] 6.22875 > mediana=5.67+(n/2-13)/13*A > mediana [1] 6.118718 > moda=5.67+((13-8)/((13-8)+(13-5)))*A > moda [1] 5.990513 > P30=quantile(pH,0.3) > P30 30% 5 > P50=quantile(pH,0.5) > P50 50% 6 > fx=(5.00-4.83)*8/A > fx [1] 1.632 > fx1=(5.67-5.00)*8/A > fx1 [1] 6.432 > FXA=fx1+fx > FXA [1] 8.064 > FXB=(9.00-7.00)*5/A > FXB [1] 12

MSc. Henrry Ochoa León