Question

. Suppose X ∼ Unif(−1, 3). Find the probabilities of the following events, both by hand calculation and with R’s punif function. (a) (X ≤ 2) (b) (X ≥ 1) (c) (−0.5 < X < 1.5) (d) (X = 0)

Answer 1

a)
`P(X \leq 2) = (2+1)/(3+1)= 0.75`

b)
`P(X \geq 1) = 1- P(X <1) =1-(1+1)/(3+1)= 1-0.5=0.5`

c)
`P(-0.5 <X <1.5)= P(X<1.5)- P(X<-0.5)= (1.5+1)/(4) -(-0.5+1)/(4)=0.625-0.125= 0.5`

d)
`f(x) = (1)/(3+1)= 0.25`

Explanation:

Let X the random variable of interest and we know that the distribution is given by:

`X \sim Unif (a= -1, b=3)`

And for this problem we can use the cumulative distribution function in order to solve the items:

`F(x) =(x-a)/(b-a), a\leq x \leq b`

Part a

We want to find this probability:

`P(X \leq 2) = (2+1)/(3+1)= 0.75`

Part b

`P(X \geq 1) = 1- P(X <1) =1-(1+1)/(3+1)= 1-0.5=0.5`

Part c

`P(-0.5 <X <1.5)`

And we can calculate the probability with this difference:

`P(-0.5 <X <1.5)= P(X<1.5)- P(X<-0.5)= (1.5+1)/(4) -(-0.5+1)/(4)=0.625-0.125= 0.5`

Part d

Since we have a continuous distribution the the probability for an unique value would be:

`f(x) = (1)/(3+1)= 0.25`