Problem 4a

n=6

k= 1:6

p = 1/n

dice <- dunif(k,0,p)


par(mfrow = c(1,2))

plot(dice, main = "PMF", ylab = "Density" , xlab = "X")

plot(ecdf(1:6), main="CDF" , xlab=" X" , ylab = "Probability")

Problem 4b

dice_2 <- sample(1:6 , 150 , replace = T) 


table(dice_2)
## dice_2
##  1  2  3  4  5  6 
## 19 28 19 27 28 29
hist(dice_2, breaks = seq(from=0, to=6 , by=1), probability = T , xaxt="n" , xlab = " X from 1 to 6" , ylab = "Probability", main = " Histogramm: Dice  n = 150")

# In the discrete case, if all outcomes have the same probability the mean equals also as the expected value.


Excpected_Value <- mean(dice_2)
Variance <- var(dice_2)


Excpected_Value
## [1] 3.693333
Variance
## [1] 2.898613

Problem 5

# a) on seperate Sheet , b&c) Did not understand

Problem 6

# X ~ N(2,16)  MEAN = 2  ,  SD = 4



# P(X< 4)

p1 <- pnorm(4, mean = 2, sd = 4)

p1
## [1] 0.6914625
# P(0 < X < 4)

p2 <- pnorm(4, mean = 2, sd = 4) - pnorm(0, mean = 2, sd = 4)

p2
## [1] 0.3829249
# P(X > q1)  =  0.95

q1 <- qnorm(0.95, mean = 2, sd = 4)

q1
## [1] 8.579415
# P(X < -q2) = -0.05

q2 <- qnorm(0.05, mean = 2, sd = 4, lower.tail = F)

q2
## [1] 8.579415

Problem 7a

?distribution

fx <- pexp(1:1000, rate=0.01)

plot(fx, type = "l", main = "PMF" , xlab = "X", ylab = "Probability")

Problem 7b

# f(x) -> antiderivative -> F(X)