Suppose that the distribution function of X is given by

Chapter 4, Problem 17P

(choose chapter or problem)

Get Unlimited 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 \(P\left\{\frac{1}{2}<X<\frac{3}{2}\right\}\).

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  \(P(X=i)=P(X \leq i)-P(X<i)\)

\(\begin{aligned}
P(X=1)= & P(X \leq 1)-P(X<1) \\
& =F(1)-F(1-) \\
& =\frac{1}{2}+\frac{1-1}{4}+\frac{1}{4} \\
& =\frac{1}{4} \\
P(X=2) & =P(X \leq 2)-P(X<2) \\
& =F(2)-F(2-) \\
& =\frac{11}{12}-\left(\frac{1}{2}+\frac{2-1}{4}\right) \\
& =\frac{1}{6} \\
P(X=3) & =P(X \leq 3)-P(X<3) \\
& =F(3)-F(3-) \\
& =1-\frac{11}{12} \\
& =\frac{1}{12}
\end{aligned}\)

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back