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: $P(X>6)$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{17}{100}$
We know that $Σp_i = 1$
$⇒ 0+ k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k=1$
$⇒ 10k^2 + 9k-1=0⇒ (10k-1) (k + 1) = 0$
$⇒ k=\frac{1}{10},-1$ but $k$ cannot be negative
$⇒ k=\frac{1}{10}$
$P(X > 6) = P(7) = 7k^2 + k = 7.\frac{1}{100}+\frac{1}{10}=\frac{17}{100}$