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).
kvnmurty:
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.
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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
{ (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
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[tex]P(3<= x <= 4)\ \ \ = \int\limits^4_3 {1} \, dx = (4-3) = 1 \\ \\
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. \\
[/tex]
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