Posted by monea on Thursday, June 14, 2012 at 3:15pm.
Given probability distribution (pdf):
x 1 2 3 4 5 6 7
p(x) .04 .05 .08 .26 .39 .16 .02
Since ∑p(x) for x=1 to 7 =1.0
we conclude that:
p(0)=0
p(1)=0.04
...
p(7)=0.02
p(8+)=0
(a)
P(x=4)=0.26/1=0.26
(b)
P(x>4)
=P(5)+P(6)+P(7)+P(8+)
=0.39+0.16+0.02+0
=0.53
(c)
P(x≤5)=P(0-)+P(1)+P(2)+P(3)+P(4)+P(5)
=?
(d) P(x≥5)=P(x>4)
(e)
P(3≤x≤6)=P(3)+P(4)+P(5)+P(6)
P(3<x<6)=P(4)+P(5)
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