Cdf of a random variable X is given by Find pdf of X Using j

C.d.f of a random variable X is given by Find p.d.f of X. Using just the c.d.f, compute P(X = 4), P(X = 6) and P(3

FOR FINDING PDF YOU JUST NEED TO DO A SUBTRACTION

FOR EXAMPLE FOR THE FIRST INTERVAL IS 0 BECAUSE THAT DOESNT CHANGE BECAUSE HIS INTERVAL DONT HAVE OTHER INTERVAL BEFORE OF IT

FOR INTERVAL 1 < T < 3 WILL BE 0.3 - 0 = 0.3

FOR INTERVAL 3 < T 4 WILL BE 0.4 - 0.3 = 0.1

FOR INTERVAL 4 < T < 8 WILL BE 0.6 - 0.4 = 0.2

FOR INTERVAL T > 8 WILL BE 1 - 0.6 = 0.4

WE NEED TO SEE IF THERE IS A PMF SUM ALL OF PROBABILITIES AND SEE IF THE RESULT IS 1

0.3 + 0.1 + 0.2 + 0.4 = 1

SO THERE IS A PMF

P( x = 4 ) = F (4) - F(3) = 0.6 - 0.4 = 0.2

P(X = 6) = P(X=4) = 0.2 BECAUSE IS IN THE SAME INTERVAL

P( 3 < X < 8 ) = F(8) - F(2) = 1 - 0.3 = 0.7

IS F(2) BECAUSE THE SIGN IS ONLY < NOT EQUAL AND <