U2A6
Mahonri VALENZUELA
12/5/2021
U2A6 Introduccion a la probabilidad frecuentista.
Interpretacion frecuntista a la probabilidad
- Veamos el ejemplo del lanzamiento de una moneda
Lanzamientos_10 <-sample(c("A", "S") ,10, replace = TRUE )
Lanzamientos_10## [1] "S" "S" "A" "S" "A" "A" "A" "A" "S" "S"Ahora calcularemos la seciencia de frecuencias relativas del aguila
cumsum(Lanzamientos_10 == "A")## [1] 0 0 1 1 2 3 4 5 5 5Dividiendo
round(cumsum(Lanzamientos_10 == "A")/1:10, 2)## [1] 0.00 0.00 0.33 0.25 0.40 0.50 0.57 0.62 0.56 0.50- Funciones de #
Distribucion normal
Si x es una variable alertoria con distribucion normal de media 3, y su desviacion estandar es de 0.5
- La probabilidad de que x sea menor de 3.5 se calcula de la siguiente forma:
pnorm(3.5, mean = 3, sd=0.5)## [1] 0.8413447- Ahora generemos numeros alertorios con distribucion normal Vamos a generar 100 nueros con media 10 y desviacion estandar de 1
x <- rnorm(100, mean =10, sd=1)
x## [1] 8.091489 8.771203 9.503781 10.121784 9.982801 9.953070 9.442160
## [8] 10.148671 9.370480 10.765881 9.148562 11.246485 9.497014 10.484135
## [15] 10.773744 8.529666 8.186518 9.349004 9.100886 10.724972 9.729650
## [22] 10.133220 11.839416 9.787668 10.457823 9.526529 9.390244 9.260358
## [29] 9.158464 9.344678 9.278009 11.082810 10.934880 10.472940 9.130336
## [36] 8.314051 10.583565 9.280034 8.941975 8.703059 10.503686 12.036867
## [43] 10.447422 8.297895 10.201410 10.356025 8.778816 9.585895 11.043409
## [50] 8.668621 8.517240 12.524504 11.517344 9.808466 9.396607 9.640591
## [57] 10.057088 10.086321 10.456577 10.526752 8.077326 9.578387 9.963051
## [64] 10.579040 11.060030 10.203068 10.779884 10.024696 9.826049 9.804742
## [71] 10.920272 12.278870 9.586146 11.854899 10.142741 9.075645 7.965633
## [78] 7.871611 9.568005 10.101146 9.908047 8.181278 11.399700 12.065958
## [85] 10.722922 10.928465 8.997187 9.456160 9.072977 10.146165 7.205129
## [92] 9.361789 10.055742 11.595194 8.509943 8.838014 9.117298 8.478679
## [99] 11.278066 11.031561Ahora calculamos el promedio de estos numeros
promedio <- mean(x)
promedio## [1] 9.866051hist(x)
boxplot(x)
Histograma de frecuencia con la curva normalizada
hist(x, freq = FALSE)
curve(dnorm(x, mean = 10, sd=1),from = 7, to = 13, add=TRUE)
## Distribucion binomal
Generado 20 valores de extios (1) versus fracasos (0) con una probabilidad de 0.5
x <- rbinom(20,1,0.5)
x## [1] 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 1 1 1Contemos exitos versus fracasos
table(x)## x
## 0 1
## 13 7Probabilidad binomal de obtener un 1
P <- 13/20
P## [1] 0.65Ejercicio distribucion binomial Hoy 12 preguntas de seleccion multiple en un examen
Cada pregunta tiene 5 alternativas y solo 1 es correcta
- calcule la probabilidad de obtener al menos 4 respuestas correctas si contestamos enteramente al azar.
dbinom(0,size = 12, prob = 0.2) + dbinom(1,size = 12, prob = 0.2) +
dbinom(2,size = 12, prob = 0.2) +
dbinom(3,size = 12, prob = 0.2) +
dbinom(4,size = 12, prob = 0.2) ## [1] 0.9274445Tarea :
1- Calcular la probabilidad de obtener unn valor menor a 9 si tenemos media de 8 y desviacion estandar de 2, usando la distribucion normal
pnorm(9, mean = 8, sd=2)## [1] 0.69146252- Generar 150 numeros alertorios de media 5 y desviacion de 0.5 usando la distribucion normal
x <- rnorm(150, mean =5, sd=0.5)
x## [1] 4.718762 4.550818 5.197500 5.194996 5.357916 5.564499 4.847459 5.400481
## [9] 5.029514 4.325216 4.763021 4.851740 4.824983 4.441766 4.871353 5.743269
## [17] 5.036042 5.359821 4.800466 5.559450 4.487127 4.002037 4.935519 5.806808
## [25] 4.680308 4.507313 4.430342 5.402804 5.275418 5.058193 4.809358 4.956804
## [33] 4.761777 5.195851 5.767652 4.965905 4.569738 5.857131 5.137060 4.333248
## [41] 5.184409 5.186012 4.473652 5.386613 4.873230 5.034156 4.958277 4.656874
## [49] 4.921057 4.888300 4.743584 4.870393 4.491047 6.530680 5.579160 5.502931
## [57] 5.070619 4.626568 4.811678 4.700951 5.324637 4.781787 4.602252 4.812218
## [65] 4.408912 4.270978 5.146302 4.833642 5.072355 4.521698 5.364250 4.913395
## [73] 4.416199 5.035608 4.953920 5.612128 4.356683 5.915956 4.381844 4.600751
## [81] 4.885396 5.475305 4.661024 5.561768 4.806680 4.350545 4.780608 5.842600
## [89] 5.595953 4.638173 5.021432 5.550011 4.770930 5.363658 4.858490 5.269415
## [97] 5.718981 4.897810 5.252546 3.895291 4.397448 4.875865 3.995955 5.157935
## [105] 5.434956 5.420995 5.379956 5.498651 4.204143 5.086698 5.067900 4.753782
## [113] 4.401485 4.970190 5.552025 5.642207 4.526110 4.267315 5.404732 4.512691
## [121] 5.038963 4.919725 5.658641 6.333509 5.172755 5.011431 4.089777 5.570563
## [129] 5.082828 4.276819 4.760519 5.520737 5.718145 4.983795 5.447953 4.282002
## [137] 5.587231 5.170313 5.379146 3.892532 5.147085 5.574713 4.595288 4.609293
## [145] 4.749075 4.647105 5.766835 5.464001 5.437931 5.0311903- Obtener media, mediana, moda de los datos generados 150 y grafico de caja y bigote
Media
media <- mean(x)
media## [1] 4.999085Mediana
mediana <- median(x)
mediana## [1] 4.962091La moda
library(modeest)
mlv(x, method = "mfv")## [1] 3.892532 3.895291 3.995955 4.002037 4.089777 4.204143 4.267315 4.270978
## [9] 4.276819 4.282002 4.325216 4.333248 4.350545 4.356683 4.381844 4.397448
## [17] 4.401485 4.408912 4.416199 4.430342 4.441766 4.473652 4.487127 4.491047
## [25] 4.507313 4.512691 4.521698 4.526110 4.550818 4.569738 4.595288 4.600751
## [33] 4.602252 4.609293 4.626568 4.638173 4.647105 4.656874 4.661024 4.680308
## [41] 4.700951 4.718762 4.743584 4.749075 4.753782 4.760519 4.761777 4.763021
## [49] 4.770930 4.780608 4.781787 4.800466 4.806680 4.809358 4.811678 4.812218
## [57] 4.824983 4.833642 4.847459 4.851740 4.858490 4.870393 4.871353 4.873230
## [65] 4.875865 4.885396 4.888300 4.897810 4.913395 4.919725 4.921057 4.935519
## [73] 4.953920 4.956804 4.958277 4.965905 4.970190 4.983795 5.011431 5.021432
## [81] 5.029514 5.031190 5.034156 5.035608 5.036042 5.038963 5.058193 5.067900
## [89] 5.070619 5.072355 5.082828 5.086698 5.137060 5.146302 5.147085 5.157935
## [97] 5.170313 5.172755 5.184409 5.186012 5.194996 5.195851 5.197500 5.252546
## [105] 5.269415 5.275418 5.324637 5.357916 5.359821 5.363658 5.364250 5.379146
## [113] 5.379956 5.386613 5.400481 5.402804 5.404732 5.420995 5.434956 5.437931
## [121] 5.447953 5.464001 5.475305 5.498651 5.502931 5.520737 5.550011 5.552025
## [129] 5.559450 5.561768 5.564499 5.570563 5.574713 5.579160 5.587231 5.595953
## [137] 5.612128 5.642207 5.658641 5.718145 5.718981 5.743269 5.766835 5.767652
## [145] 5.806808 5.842600 5.857131 5.915956 6.333509 6.530680Grafico de caja y bigote
boxplot(x)
histograma de frecuencias absolutas
hist(x)
Histograma de frecuencias con la curva normalizada
hist(x, freq = FALSE)
curve(dnorm(x, mean = 8, sd=0.5), from = 7, to =13, add=TRUE )
*5- Genere un conteo y calculo de probabilidad de 30 numeros de forma binomial
x <- rbinom(30,5,0.40)
x## [1] 1 3 3 3 3 3 3 1 1 2 3 2 2 3 2 1 2 2 2 2 2 2 2 0 3 2 3 3 3 1table(x)## x
## 0 1 2 3
## 1 5 12 12A <- 1/30
A## [1] 0.03333333