The probability distribution of a random variable X is given as:
\( P(X=x)=\begin{cases}kx^2 & \mathit{for}\quad x=1,2,3\\ 2kx & \mathit{for}\quad x=4,5,6\\ 0 & \mathit{otherwise}\end{cases} \)
Where \( k \) is an arbitrary constant, then E(X) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{95}{22} \)
Since the probabilities add up to 1,
\( \displaystyle k(1+4+9)+2k(4+5+6)=1 \)
\( \displaystyle 14k+30k=1 \)
\( \displaystyle k=\frac{1}{44} \)
The expected value is \( E(X)=\sum x\,P(X=x) \):
\( \displaystyle E(X)=k(1+8+27)+2k(16+25+36) \)
\( \displaystyle =36k+154k=190k \)
\( \displaystyle =\frac{190}{44}=\frac{95}{22} \)