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Suppose that the distribution function of X given by

F(b)=0 â¶Ä…â¶Ä…â¶Ä…b<0b4 â¶Ä…â¶Ä…â¶Ä…0≤b<112+b−14 â¶Ä…â¶Ä…â¶Ä…1≤b<21112 â¶Ä…â¶Ä…â¶Ä…2≤b<31 â¶Ä…â¶Ä…â¶Ä…3≤b

(a) Find P{X=i},i=1,2,3.

(b) Find P12<X<32.

Short Answer

Expert verified

(a) The value for the P{x=i} if i=1,2,3 are

P{1}=14

P{2}=16

P{3}=112

(b) The value for theP12<X<32is12.

Step by step solution

01

Given information (Part a)

The distribution function of X given as

F(b)=0 â¶Ä…â¶Ä…â¶Ä…b<0b4 â¶Ä…â¶Ä…â¶Ä…0≤b<112+b−14 â¶Ä…â¶Ä…â¶Ä…1≤b<21112 â¶Ä…â¶Ä…â¶Ä…2≤b<31 â¶Ä…â¶Ä…â¶Ä…3≤b

02

Solution (Part a)

The calculation is given below,

P(X=1)=F(1+)−F(1−)

=12+1−14−14

=14

P(X=2)=F(2+)−F(2−)

=1112−12−2−14

=1112−12−14

=11−6−312

=16

Similarly,

P(X=3)=1−1112

=112

03

Final answer (Part a)

The value for the P{x=i} if i=1,2,3 are

P{1}=14

P{2}=16

P{3}=112

04

Given information (Part b)

Given in the question that

F(b)=0 â¶Ä…â¶Ä…â¶Ä…b<0b4 â¶Ä…â¶Ä…â¶Ä…0≤b<112+b−14 â¶Ä…â¶Ä…â¶Ä…1≤b<21112 â¶Ä…â¶Ä…â¶Ä…2≤b<31 â¶Ä…â¶Ä…â¶Ä…3≤b

05

Solution (Part b)

The calculation is given below,

P12<X<32=F(3/2)−F(1/2)

=12+32−14−124

=12+12×14−12×14

=12

06

Final answer (Part b)

The value for theP12<X<32is12.

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