A random variable X has the following probability distribution.
X
0
1
2
3
4
5
6
7
P (X)
0
k
2k
2k
3k
k 2
2k 2
7k 2 + k
Determine
(i) k
(ii) P (X < 3)
(iii) P (X > 6)
(iv) P (0 < X < 3)
Attachments:
Answers
Answered by
2
Answer:
As we know, for any probablity distribution,
∑P(X)=1
So, for given distribution,
0+k+2k+2k+3k+k2+2k2+7k2+k=1
10k2+9k−1=0
10k2+10k−k−1=0
10k(k+1)−1(k+1)=0
(10k−1)(k+1)=0
As, k can not be negative, k=110
(ii) P(X<3)=P(X=0)+P(X=1)+P(X=2)
=3k=310
(iii) P(X>6)=P(X=7)
=7k2+k=7100+110=17100
(iv)P(0<X<3)=P(X=1)+P(X=2)
=3k=310
Similar questions