A random variable X has the following probability distribution:
Determine: $P(X>6)$ |
$\frac{1}{100}$ $\frac{13}{100}$ $\frac{17}{100}$ $\frac{11}{100}$ |
$\frac{17}{100}$ |
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 $⇒ k=\frac{1}{10}$ $P(X > 6) = P(7) = 7k^2 + k = 7.\frac{1}{100}+\frac{1}{10}=\frac{17}{100}$ |