Answer to Question #261927 in Statistics and Probability for zaimsquah

Question

A random variable X has a CDF such that

F(x)= ( x/2 0<x≤1

( (x-1/2 1<x≤3/2

a)Graph F(x).

b/Graph the pdf f(x).

c)FindP[X≤1/2].

d)Find P[X  ≥1/2].

e)Find P[X≤ 1.25].

f)What is P[X= 1.25]?

Answer 1

(a)

b) Lets find pdf. $f(x)=F'(x)$ , where f(x) - pdf, F(x) - cdf. So,

f(x) = 0, x ≤ 0

f(x) = 0.5, 0 < x ≤ 1

f(x) = 1, 1 < x ≤ 1.5

f(x) = 0, x > 1.5

The graph of pdf is presented below

c) P[X ≤ 0.5] = F(0.5) = 0.25

d) P[X ≥ 0.5] = 1 - P[X ≤ 0.5] - P[X = 0.5] = 0.75 - 0 = 0.75

e) P[X ≤ 1.25] = F(1.25) = 1.25 - 0.5 = 0.75

f) Since X is a continuous random variable, then for any x P[X = x] = 0. So, P[X = 1.25] = 0