A random variable X has the following probability mass function.
x
0
1
2
3
4
5
6
P(X=x)
k
3k
5k
7k
9k
11k
13k
(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. 6∑x=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=1⇒k=149 (b) P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3) =149+349+549+749=1649 P(X≥5)=P(X=5)+P(X=6)=1149+1349=2449 P(3<x≤6)=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(X≤0)=149<12;P(X≤1)=449<12 P(X≤2)=949<12;P(X≤3)=1649<12 P(X≤4)=2549>12 ∴ the smallest value of x for which P(X≤x)>12 is 4.