The probability distribution of a discrete random variable is given as:
|
X |
2 |
3 |
4 |
5 |
|
P(X) |
2k |
5k |
k |
3k |
The value of E(X) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{38}{11}$
$2k + 5k + k + 3k = 11k = 1 \Rightarrow k = \frac{1}{11}$
$E(X) = 2(2k) + 3(5k) + 4(k) + 5(3k)$
$= 4k + 15k + 4k + 15k = 38k$
$= \frac{38}{11}$
$E(X) = \frac{38}{11}$