Let X be a r.v. has the probability function given by
Evaluate F(6),F(1.5),P(X>2.5)
If X<1 ,then F(x)=0If 1≤X<2 ,then F(1)=P(1)=3⁄7If 2≤X<3 ,then F(2)=P(1)+P(2)=3⁄7+2⁄7=5⁄7If 3≤X<4 ,then F(3)=P(1)+P(2)+P(3)=6⁄7If 4≤X ,then F(4)=1Thus:F(x)={( 0 X<1 3⁄7 1≤X<2 ( 5)⁄7 2≤X<3 6⁄7 3≤X<4 1 4≤X )┤F(6)=F(4)=1F(1.5)=3⁄7P(X>2.5)=1-P(X≤2.5)=1-5/7=2/7
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If X<1 ,then F(x)=0
need an explanation for this answer? contact us directly to get an explanation for this answerIf 1≤X<2 ,then F(1)=P(1)=3⁄7
If 2≤X<3 ,then F(2)=P(1)+P(2)=3⁄7+2⁄7=5⁄7
If 3≤X<4 ,then F(3)=P(1)+P(2)+P(3)=6⁄7
If 4≤X ,then F(4)=1
Thus:
F(x)={( 0 X<1 3⁄7 1≤X<2 ( 5)⁄7 2≤X<3 6⁄7 3≤X<4 1 4≤X )┤
F(6)=F(4)=1
F(1.5)=3⁄7
P(X>2.5)=1-P(X≤2.5)=1-5/7=2/7