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2026-04-14
A: BIEN LIEN TUC
I: PHAN PHOI DEU UNIFORM- Hinh chu nhat, doan du lieu #Một biến ngẫu nhiên tuân theo phân phối đều trên đoạn [5,15] . Tính xác suất trong khoảng từ 7 đến 12
#1: punif: xac suat tich luy(x<=X voi xac suat bao nhieu)
#tinh tay:
x<-(12-7)/(15-5)
x## [1] 0.5#tinh bang punif: ham phan phoi xac suat tich luy
punif (12, min = 5, max = 15) - punif(7, min = 5, max = 15)## [1] 0.5#punif (12, min = 5, max = 15) = (12-5)/(15-5)
#punif(7, min = 5, max = 15) = (7-5)(15-5)
punif(6, min = 2, max = 8) - punif(3, min = 2, max = 8)## [1] 0.5#2: dunif: mat do xac suat( ko phai xac suat, la chieu cao do thi)
#chieu cao tai x=7:
dunif(7, min = 5, max = 15) #f(x)=1/(b-a)=1/15-5## [1] 0.1#3: qunif: tim gia tri theo xac suat
#50% xac suat la gia tri nao
qunif(0.5, min = 5, max = 15) #=a+p*(b-a)=5+0.5*(15-5)## [1] 10#4: runif: sinh so ngau nhien
#ko trung lap: dung set.seed(1)
set.seed(1)
x<-runif(5, min = 5, max = 15)
y<-unique(x)
y## [1] 7.655087 8.721239 10.728534 14.082078 7.016819II: PHAN PHOI CHUAN NORMAL - chieu cao, diem so #Chiều cao của sinh viên một trường đại học tuân theo phân phối chuẩn với trung bình 170 cm và độ lệch chuẩn 5 cm.
#1: pnorm - xac suat tich luy (x<=X voi xac suat bao nhieu)*
pnorm(175, 170, 5) ## [1] 0.8413447f <- function(t){
(1/(5*sqrt(2*pi))) * exp(-(t-170)^2/(2*5^2))
}
p<- integrate(f, -Inf, 175)$value
p## [1] 0.84134471-pnorm(175, 170, 5) #chieu cao >175## [1] 0.1586553#xac suat trong khoang 165-175
pnorm(175, 170, 5)-pnorm(165, 170, 5)## [1] 0.6826895#2: dnorm - mat do
dnorm(175, 170, 5) #(1/(sigma*sqrt(2*pi))) * exp(-(x-tb)^2/(2*sigma^2))## [1] 0.04839414#(1/(5*sqrt(2*pi))) * exp(-(175-170)^2/(2*5^2))
#3: qnorm - tim gia tri theo xac suat
qnorm(0.4, 170, 5)## [1] 168.7333#4: rnorm - sinh du lieu
rnorm(11, 170, 5)## [1] 176.3621 172.0732 162.3002 165.3572 168.5264 169.9712 182.0233 173.8180
## [9] 166.0050 164.2617 168.5527III: PHAN PHOI MU EXP - thoi gian cho giua cac su kien #Thời gian giữa hai lần khách hàng vào quán cà phê tuân theo phân phối mũ với trung bình 5 phút (mean=5 => rate=lambda=1/mean=1/5)
#1: pexp - xac suat (exp: tg cho)
pexp(3, 1/5) #tg cho <= 3p =1-exp(-lambda*x)=1-exp(-1/5*3)## [1] 0.45118841-pexp(3, 1/5) #tg cho >=3p ## [1] 0.5488116#2: dexp - mat do
dexp(3, 1/5) #mat do tg cho <=3 lambda*exp(-lambda*x)=1/5*exp(-1/5*3)## [1] 0.1097623#3: qexp - tim gia tri dua tren xac suat
qexp(0.45, 1/5) #gia tri co xac suat 45% la =(-1/lambda)*log(1-p)=(-1/(1/5)*log(1-0.45)## [1] 2.989185#4: rexp - sinh du lieu
rexp(11, 1/5)## [1] 6.1134950 1.1795597 0.9956583 4.1111169 2.9423986 11.8225763
## [7] 3.2094629 1.4706019 2.8293276 0.5303631 0.2971958A: BIEN ROI RAC I: PHAN PHOI POISSON - dem so su kien (cuoc goi, so khach hang, so loi, tb lambda) #Số cuộc gọi đến trong một phút tuân theo phân phối Poisson với λ = 3, xac suat co dung 5 cuoc goi
#1: ppois - xac suat <=
ppois(3, lambda = 3) #xac suat <=3 ## [1] 0.6472319#for(i in 0:k){exp(-lambda) * lambda^i / factorial(i)}
#p<-0
#for(i in 0:3){
# p<-p+exp(-3) * 3^i / factorial(i)
#}
#p
1- ppois(2, lambda = 3) #xac suat >=x: 1- ppois(x-1, lambda)## [1] 0.5768099# Tính xác suất có ít nhất 3 khách hàng đến trong một phút với λ = 2.
1-ppois(2, lambda = 2)## [1] 0.3233236#2: dpois - xac suat dung =
dpois(5, lambda = 3) #exp(-lambda) * lambda^k / factorial(k)## [1] 0.1008188#exp(-3)*3^5/factorial(5)
#3: qpois: tim gia tri
qpois(0.5, lambda = 3)## [1] 3#4: rpois: sinh du lieu
rpois(10, lambda = 3)## [1] 4 0 3 4 4 3 5 3 2 1II: PHAN PHOI NHI THUC BINOMIAL (xac suat thanh cong / that bai sau n lan lap) #Một nhà máy có xác suất sản phẩm lỗi là 𝑝=0.1, Kiểm tra 10 sản phẩm.
#1: pbinom - xac suat <=
pbinom(3, 10, 0.1) #Xác suất có nhiều nhất 3 sản phẩm lỗi## [1] 0.98720481-pbinom(3, 10, 0.1) #Xác suất có it nhất 3 sản phẩm lỗi## [1] 0.0127952#2: dbinom - xac suat dung =
dbinom(2, 10, 0.1) #Xác suất có đúng 2 sản phẩm lỗi## [1] 0.1937102#3: qbinom - tim gia tri dua tren xac suat
qbinom(0.23, 10, 0.1) ## [1] 0#4: rbinom - sinh du lieu
rbinom(4, 10, 0.1)## [1] 0 0 1 1C: BO DU LIEU BOSTON
library(MASS)
myd<-(Boston)
model<-lm(medv~nox+tax+rm, data = myd)
summary(model)##
## Call:
## lm(formula = medv ~ nox + tax + rm, data = myd)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.784 -3.219 -0.782 2.412 41.616
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -17.658150 3.235967 -5.457 7.63e-08 ***
## nox -7.176586 3.175910 -2.260 0.0243 *
## tax -0.012709 0.002176 -5.840 9.41e-09 ***
## rm 7.854108 0.407533 19.272 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.084 on 502 degrees of freedom
## Multiple R-squared: 0.565, Adjusted R-squared: 0.5624
## F-statistic: 217.3 on 3 and 502 DF, p-value: < 2.2e-16coef(model)## (Intercept) nox tax rm
## -17.65815000 -7.17658571 -0.01270934 7.85410819#Kiem tra da cong tuyen vif
library(car)## Warning: package 'car' was built under R version 4.5.3## Loading required package: carData## Warning: package 'carData' was built under R version 4.5.3vif(model) #=1->ko co, >5:co, >10: da cong tuyen manh## nox tax rm
## 1.847723 1.835551 1.118573#stepAIC chon cac bien quan trong
model_step <- stepAIC(model, direction = "both")## Start: AIC=1831.33
## medv ~ nox + tax + rm
##
## Df Sum of Sq RSS AIC
## <none> 18582 1831.3
## - nox 1 189.0 18771 1834.5
## - tax 1 1262.3 19844 1862.6
## - rm 1 13748.6 32331 2109.6summary(model_step)##
## Call:
## lm(formula = medv ~ nox + tax + rm, data = myd)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.784 -3.219 -0.782 2.412 41.616
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -17.658150 3.235967 -5.457 7.63e-08 ***
## nox -7.176586 3.175910 -2.260 0.0243 *
## tax -0.012709 0.002176 -5.840 9.41e-09 ***
## rm 7.854108 0.407533 19.272 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.084 on 502 degrees of freedom
## Multiple R-squared: 0.565, Adjusted R-squared: 0.5624
## F-statistic: 217.3 on 3 and 502 DF, p-value: < 2.2e-16#so sanh 2 mo hinh bang MSE
mse1 <- mean(model$residuals^2)
mse2 <- mean(model_step$residuals^2) #trung binh sai so binh phuong
mse1## [1] 36.72346mse2## [1] 36.72346#do thi residual vs fitted
plot(model$fitted.values, resid(model),
col = "lightseagreen",
pch = 16,
cex = 0.5,
xlab = "Fitted values",
ylab = "Residuals",
main = "Residuals vs Fitted")
abline(h = 0, col = "red")
axis(1, at = seq(0, 50, by = 10))
BO DU LIEU IRIS
#su dung bo du lieu iris
md<-(iris)
head(md)## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1 5.1 3.5 1.4 0.2 setosa
## 2 4.9 3.0 1.4 0.2 setosa
## 3 4.7 3.2 1.3 0.2 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 5 5.0 3.6 1.4 0.2 setosa
## 6 5.4 3.9 1.7 0.4 setosa#loc ko lay virsinica, chuyen ten loai thanh vecto 0, 1
loc<-subset(md, Species!="virginica") #loc<-md[md$Species!="virsinica", ]
#ma hoa nhan
loc$Species<-ifelse(loc$Species=="versicolor", 1, 0)
#mo hinh hoi quy ligistic
model<-glm(Species~Sepal.Length+Petal.Length+Petal.Width, data = loc, family = binomial)## Warning: glm.fit: algorithm did not converge## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(model)##
## Call:
## glm(formula = Species ~ Sepal.Length + Petal.Length + Petal.Width,
## family = binomial, data = loc)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 5.213 575458.883 0 1
## Sepal.Length -16.539 132345.998 0 1
## Petal.Length 23.860 76702.645 0 1
## Petal.Width 27.591 136533.255 0 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1.3863e+02 on 99 degrees of freedom
## Residual deviance: 1.4971e-09 on 96 degrees of freedom
## AIC: 8
##
## Number of Fisher Scoring iterations: 25#ma tran nnham lan
# Xác suất dự đoán
prob <- predict(model, type = "response")
# Phân lớp với ngưỡng 0.5
pred <- ifelse(prob > 0.5, 1, 0)
x<-table(Predicted = pred, Actual = loc$Species)
x## Actual
## Predicted 0 1
## 0 50 0
## 1 0 50#do chinh xac cua mo hinh
accuracy <- mean(pred == loc$Species)
accuracy## [1] 1TP <- x["1","1"]
TN <- x["0","0"]
FP <- x["1","0"]
FN <- x["0","1"]
# Accuracy
accuracy <- (TP + TN) / sum(x)
# Precision
precision <- TP / (TP + FP)
# Recall
recall <- TP / (TP + FN)
# F1-score
f1 <- 2 * precision * recall / (precision + recall)
accuracy## [1] 1precision## [1] 1recall## [1] 1f1 ## [1] 1#petal.length
# Tạo dải giá trị
x_seq <- seq(min(loc$Petal.Length), max(loc$Petal.Length), length=100)
# Giữ các biến khác cố định (lấy trung bình)
new_data <- data.frame(
Sepal.Length = mean(loc$Sepal.Length),
Petal.Length = x_seq,
Petal.Width = mean(loc$Petal.Width)
)
# Dự đoán xác suất
prob_curve <- predict(model, newdata = new_data, type = "response")
# Vẽ đồ thị
plot(x_seq, prob_curve, type="l", lwd=2,
xlab="Petal.Length", ylab="Probability",
main="Đường cong Sigmoid")
abline(h=0.5, col="red", lty=2) # ngưỡng 0.5
#petal.width
x_seq <- seq(min(loc$Petal.Width), max(loc$Petal.Width), length=100)
new_data <- data.frame(
Sepal.Length = mean(loc$Sepal.Length),
Petal.Length = mean(loc$Petal.Length),
Petal.Width = x_seq
)
prob_curve <- predict(model, newdata = new_data, type = "response")
plot(x_seq, prob_curve, type="l", lwd=2,
xlab="Petal.Width", ylab="Probability",
main="Sigmoid theo Petal.Width")
abline(h=0.5, col="red", lty=2)
#sepal.length
x_seq <- seq(min(loc$Sepal.Length), max(loc$Sepal.Length), length=100)
new_data <- data.frame(
Sepal.Length = x_seq,
Petal.Length = mean(loc$Petal.Length),
Petal.Width = mean(loc$Petal.Width)
)
prob_curve <- predict(model, newdata = new_data, type = "response")
plot(x_seq, prob_curve, type="l", lwd=2,
xlab="Sepal.Length", ylab="Probability",
main="Sigmoid theo Sepal.Length")
abline(h=0.5, col="red", lty=2)
#kiem dinh hosmer-lemeshow
library(ResourceSelection)## Warning: package 'ResourceSelection' was built under R version 4.5.3## ResourceSelection 0.3-6 2023-06-27hoslem.test(loc$Species, prob, g=10 )## Warning in hoslem.test(loc$Species, prob, g = 10): The data did not allow for
## the requested number of bins.##
## Hosmer and Lemeshow goodness of fit (GOF) test
##
## data: loc$Species, prob
## X-squared = 7.4857e-10, df = 1, p-value = 1