A random variable 'X' has the following probability
x 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k^2 2k^2 7k^2 + k

Question

A random variable 'X' has the following probability 
x          0   1   2      3      4       5        6          7
P(X)     0    k   2k    2k    3k   k^2   2k^2    7k^2 + k

anonymous · Answer

We know that sum of all probabilits is equal to 1.....

anonymous · Answer

(i) find k (ii) Evaluate P(X<6) ,P(x>=6) (iii) if P(X<=C) find the minimum value of C (iv)P[1.52]

amistre64 · Answer

i found k, its up there in the question box .....

anonymous · Answer

then  0 +k  +2k   +2k  + 3k + k^2 +2k^2 +7k^2 + k=1

anonymous · Answer

by adding them then you can find k eassily....

anonymous · Answer

yeah!!

amistre64 · Answer

.. why did i think that was going to infinity :/  good eye

anonymous · Answer

P(x < 6)	 
=	P(x=0) + P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5)
           (Or)
 =	1 − P(x ≥ 6)
=	1 − [P(x=6) + P(x=7)]

anonymous · Answer

if P(X<=C)  is nothing but 
for C what value minimum we can take...