A random variable X has the following probability distribution:
| \(X\) | 1 | 2 | 3 |
| \(P(X)\) | 0.2 | 0.5 | 0.3 |
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(E(X)\) | (I) | 0.49 |
| (B) | \(E(X^{2})\) | (II) | 2.1 |
| (C) | \(\mathrm{Var}(X)\) | (III) | 0.7 |
| (D) | Standard deviation of X | (IV) | 4.9 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(A) \(E(X)\)
This matches (II) 2.1. The mean is \(\sum x\,P(x)\):
\(E(X) = 1(0.2) + 2(0.5) + 3(0.3)\)
\(= 0.2 + 1.0 + 0.9 = 2.1\)
(B) \(E(X^{2})\)
This matches (IV) 4.9.
\(E(X^{2}) = 1(0.2) + 4(0.5) + 9(0.3)\)
\(= 0.2 + 2.0 + 2.7 = 4.9\)
(C) \(\mathrm{Var}(X)\)
This matches (I) 0.49. Using \(\mathrm{Var}(X) = E(X^{2}) - [E(X)]^{2}\):
\(\mathrm{Var}(X) = 4.9 - (2.1)^{2}\)
\(= 4.9 - 4.41 = 0.49\)
(D) Standard deviation of X
This matches (III) 0.7.
\(\sqrt{0.49} = 0.7\)