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Question

A random variable X has the following probability mass function.
x0123456
P(X=x)k3k5k7k9k11k13k
(a) Find k.
(b) Evaluate P(X<4),P(X 5) and P(3<X 6).
(c) What is the smallest value of x for which P(X x)> 12?

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Solution

(a) P(X=x) is a probability mass function.
6x=0P(X=x)=1
(i.e.,) P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=1.
k+3k+5k+7k+9k+11k+13k=1
49k=1k=149
(b) P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)
=149+349+549+749=1649
P(X5)=P(X=5)+P(X=6)=1149+1349=2449
P(3<x6)=P(X=4)+P(X=5)+P(X=6)
=949+1149+1349=3349
(c) The minimum value of x may be determined by trial and error method.
P(X0)=149<12;P(X1)=449<12
P(X2)=949<12;P(X3)=1649<12
P(X4)=2549>12
the smallest value of x for which P(Xx)>12 is 4.

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