The distribution function of a random variable is given by
F(x)= {0 , for x<-1
{(x+1)/2 , for -1<x<1
{1 , for x≥1.
Find P(3<x<4).

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is such a probability distribution possible ? probability seems to be not bounded. is it a text book question? if P(3<x<4) = 1, that means x is only between 3 & 4. actually integral F(x) dx from - infinity to + infinity will be 1.

Answers

2014-08-23T14:37:22+05:30
F(x)  =  {  0                for x<-1
            {  (x+1)/2       for -1<x<1
            {  1               for  x≥1
P(3<x<4) = integration from 3 to 4 of  F(x) dx
P(3<x<4) = integration from 3 to 4 of 1 dx
P(3<x<4) = limit from 3 to 4 of x
P(3<x<4) = (4-3) = 1
hence ,
P(3<x,4) = 1


0
2014-08-24T09:25:46+05:30

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P(3<= x <= 4)\ \ \ =  \int\limits^4_3 {1} \, dx  = (4-3) = 1 \\ \\ &#10;Actually\ \int\limits^a_b {F(x)} \, dx \ should\ be\ equal\ to \ 1\ when\ b = -infinity\ \ and\ a\ =\ +infinity. \\ Here\ in\ the\ question\ given,\ the\ integral\ is\ summing\ to\ infinity. \\ &#10;

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