Question #340147

The distribution function of a discrete random variable X is given by

Fx(k) = {0,k<10.1,1k<20.3,2k<30.7,3k<40.84k<51k5}\begin{Bmatrix} 0, & k < 1 \\ 0.1, & 1 \leq k <2 \\ 0.3, & 2 \leq k <3\\ 0.7, & 3 \leq k <4 \\ 0.8 & 4 \leq k < 5\\ 1 & k \geq 5 \end{Bmatrix}



1) What are the possible values of X?

2) Find the probability mass function of X.



1
Expert's answer
2022-05-25T12:19:39-0400

1) The possible values of XX are 1,2,3,4,51,2,3,4,5

2)


F(1)=P(X1)=p(1)=0.1F(1)=P(X\le1)=p(1)=0.1

F(2)=P(X2)=p(1)+p(2)=0.3F(2)=P(X\le2)=p(1)+p(2)=0.3


p(2)=0.30.1=0.2p(2)=0.3-0.1=0.2

F(3)=P(X3)=p(1)+p(2)+p(3)=0.7F(3)=P(X\le3)=p(1)+p(2)+p(3)=0.7


p(3)=0.70.3=0.4p(3)=0.7-0.3=0.4

F(4)=P(X4)=p(1)+p(2)+p(3)F(4)=P(X\le4)=p(1)+p(2)+p(3)

+p(4)=0.8+p(4)=0.8


p(4)=0.80.7=0.1p(4)=0.8-0.7=0.1

F(5)=P(X5)=p(1)+p(2)+p(3)F(5)=P(X\le5)=p(1)+p(2)+p(3)

+p(4)+p(5)=1+p(4)+p(5)=1


p(5)=10.8=0.2p(5)=1-0.8=0.2


x12345f(x)0.10.20.40.10.2\def\arraystretch{1.5} \begin{array}{c:c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 0.1 & 0.2 & 0.4 & 0.1 & 0.2 \\ \end{array}


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