The random variable X has a probability distribution
\(P(X=x)=\begin{Bmatrix}5k, & x=2\\2k, & x=3\\3k, & x=1\\0, & otherwise\end{Bmatrix}\)
Then, the value of E(X) is:
Answer & explanation
Correct answer: option 3
The correct answer is option (3) → 1.9
Sum of probabilities must be equal to 1.
$⇒5k+2k+3k+0=1$
$⇒k=\frac{1}{10}$
Now,
$E(X)=∑x.P(X=x)$
$=\frac{5}{10}.2+\frac{2}{10}.3+\frac{3}{10}.1$
$=1+0.6+0.3$
$=1.9$