A random variable X has the following probability distribution
|
$X$ |
2 |
3 |
4 |
5 |
|
$P(X)$ |
$5/k$ |
$7/k$ |
$9/k$ |
$11/k$ |
Then the value of $k$ is
Answer & explanation
Correct answer: option 3
The correct answer is Option (1) → 8
Given probability distribution:
$X:$ 2 3 4 5
$P(X): \frac{5}{k}, \frac{7}{k}, \frac{9}{k}, \frac{11}{k}$
Since total probability = 1,
$\frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1$
$\frac{(5 + 7 + 9 + 11)}{k} = 1$
$\frac{32}{k} = 1$
$k = 32$
Therefore, the value of $k$ is 32.