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

Answer 1

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

Answer 2

(i) find k
(ii) Evaluate P(X<6) ,P(x>=6)
(iii) if P(X<=C) find the minimum value of C
(iv)P[1.5<x<4.5 / x>2]

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

yeah!!

Answer 7

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

Answer 8

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)]

Answer 9

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