Thread: Statistics - Discrete Random Variable

Hi, i'm completly at a loss as to how to approach this question and don't know how to answer it, can someone help me please.

The discrete random variable X has c.d.f. F(x) defined by

F(x) = (2+x)/7 ; x = 1, 2, 3, 4 and 5

a/ Find P(X<=3)
b/ Show that P(X=4) = 1/7
c/ Find the probability distribution for X.

Originally Posted by Tom G

Hi, i'm completly at a loss as to how to approach this question and don't know how to answer it, can someone help me please.

The discrete random variable X has c.d.f. F(x) defined by

F(x) = (2+x)/7 ; x = 1, 2, 3, 4 and 5

a/ Find P(X<=3)
b/ Show that P(X=4) = 1/7
c/ Find the probability distribution for X.

F is a cdf - cumulative distribution function, so:

P(X<=a)=F(a), a=1, 2, 3, 4, 5.

a) Hence P(X<=3)=F(3)=5/7.

b) P(X=4)=P(X<=4)-P(X<=3)=F(4)-F(3)=6/7-5/7=1/7

c) P(X=1)=P(X<=1)=F(1)=3/7,
P(X=2)=P(X<=2)-P(X<=1)=F(2)-F(1)=1/7

and similarly: P(X=3)=1/7, P(X=4)=1/7.

Finaly P(X=5)=1-P(X<=4)=1-6/7=1/7.

RonL