For a discrete random variable X, whose probability distribution is defined as :
\(P(x) = \begin{cases} 2k(x+1); & x = 0,\ 1 \\ 3kx; & x = 2 \\ k(5-x); & x = 3 \end{cases}\)
The value of mean will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{11}{7}\)
The probabilities are
\(P(0) = 2k(0+1) = 2k\)
\(P(1) = 2k(1+1) = 4k\)
\(P(2) = 3k \times 2 = 6k\)
\(P(3) = k(5-3) = 2k\)
The probabilities must add up to 1:
\(2k + 4k + 6k + 2k = 14k = 1\)
\( \displaystyle k = \frac{1}{14} \)
The mean is \(E(X) = \sum x\,P(x)\):
\(E(X) = 0 \times 2k + 1 \times 4k + 2 \times 6k + 3 \times 2k = 22k\)
\( \displaystyle E(X) = \frac{22}{14} = \frac{11}{7} \)