A random variable X has the following probability distribution:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 2k | k | 4k | 6k | 3k |
The mean of the distribution is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{39}{16}$
Sum of all probabilities is 1.
$⇒2k+k+4k+6k+3k=1$
$⇒16k=1$
$⇒k=\frac{1}{16}$
Mean of the random variable X.
$E(X)=∑XP(X)$
$E(X)=(0×2k)+(1×k)+(2×4k)+(3×6k)+(4×3k)$
$=39k=39×\frac{1}{16}=\frac{39}{16}$