Question
A random variable X has the following probability distribution:
|
X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
1 |
|
P(X) |
0 |
k |
2k |
2k |
3k |
k2 |
2k2 |
7k2 + k |
Determine (i) k (ii) P(X < 3) (iii) P(X > 6) (iv) P(0 < X < 3)
Solution
(i) Since Σ P(X) = 1
∴ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1
⇒ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7 k2 + k = 1
But k cannot be negative as probability is always non-negative.
∴ probability distribution of X is
