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The probability distribution of a discrete random variable is given as:
The value of E(X) is: |
$\frac{15}{11}$ $\frac{38}{11}$ $\frac{38}{13}$ $\frac{15}{13}$ |
$\frac{38}{11}$ |
The correct answer is Option (2) → $\frac{38}{11}$ $2k + 5k + k + 3k = 11k = 1 \Rightarrow k = \frac{1}{11}$ $E(X) = 2(2k) + 3(5k) + 4(k) + 5(3k)$ $= 4k + 15k + 4k + 15k = 38k$ $= \frac{38}{11}$ $E(X) = \frac{38}{11}$ |
