Solution Found!
Suppose that the distribution function of X is given by
Chapter 4, Problem 17P(choose chapter or problem)
Suppose that the distribution function of X is given by
\(F(b)=\left\{\begin{array}{ll}
0 & b<0 \\
\frac{b}{4} & 0 \leq b<1 \\
\frac{1}{2}+\frac{b-1}{4} & 1 \leq b<2 \\
\frac{11}{12} & 2 \leq b<3 \\
1 & 3 \leq b
\end{array}\right.\)
(a) Find \(P\{X=i\}, i=1,2,3\).
(b) Find \(P\left\{\frac{1}{2}<X<\frac{3}{2}\right\}\).
Questions & Answers
QUESTION:
Suppose that the distribution function of X is given by
\(F(b)=\left\{\begin{array}{ll}
0 & b<0 \\
\frac{b}{4} & 0 \leq b<1 \\
\frac{1}{2}+\frac{b-1}{4} & 1 \leq b<2 \\
\frac{11}{12} & 2 \leq b<3 \\
1 & 3 \leq b
\end{array}\right.\)
(a) Find \(P\{X=i\}, i=1,2,3\).
(b) Find \(P\left\{\frac{1}{2}<X<\frac{3}{2}\right\}\).
ANSWER:Step 1 of 2
Given the distribution function of X is
\(F(b)=\left\{\begin{array}{ll}
0 & b<0 \\
\frac{b}{4} & 0 \leq b<1 \\
\frac{1}{2}+\frac{b-1}{4} & 1 \leq b<2 \\
\frac{11}{12} & 2 \leq b<3 \\
1 & 3 \leq b
\end{array}\right.\)
a). We have to find \(P\{X=i\}, i=1,2,3\).
We have \(P(X=i)=P(X \leq i)-P(X<i)\)
\(\begin{aligned}
P(X=1)= & P(X \leq 1)-P(X<1) \\
& =F(1)-F(1-) \\
& =\frac{1}{2}+\frac{1-1}{4}+\frac{1}{4} \\
& =\frac{1}{4} \\
P(X=2) & =P(X \leq 2)-P(X<2) \\
& =F(2)-F(2-) \\
& =\frac{11}{12}-\left(\frac{1}{2}+\frac{2-1}{4}\right) \\
& =\frac{1}{6} \\
P(X=3) & =P(X \leq 3)-P(X<3) \\
& =F(3)-F(3-) \\
& =1-\frac{11}{12} \\
& =\frac{1}{12}
\end{aligned}\)