Question

For the following probability distribution of a random variable \(X\), the Expected value and Variance of \(X\) are respectively

\(X = x\)	\(1\)	\(2\)	\(3\)
\(P(X = x)\)	\(1/5\)	\(2/5\)	\(2/5\)

• A. \(\frac{27}{5},\frac{27}{25}\)
• B. \(\frac{11}{5},\frac{14}{25}\)
• C. \(\frac{4}{5},\frac{14}{25}\)
• D. \(\frac{7}{5},\frac{11}{25}\)

Hint:

Use E(X) = sum x p and Var(X) = E(X squared) - E(X) squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a discrete random variable the mean is \(E(X)=\sum xP(x)\), and the variance is \(E(X^2)-[E(X)]^2\).

Step 2: Mean
\[ E(X)=1\cdot\frac15+2\cdot\frac25+3\cdot\frac25=\frac{1+4+6}{5}=\frac{11}{5} \]

Step 3: Second moment
\[ E(X^2)=1\cdot\frac15+4\cdot\frac25+9\cdot\frac25=\frac{1+8+18}{5}=\frac{27}{5} \]

Step 4: Variance
\[ \text{Var}(X)=\frac{27}{5}-\left(\frac{11}{5}\right)^2=\frac{135}{25}-\frac{121}{25}=\frac{14}{25} \]

Step 5: Check
The pair \(\left(\frac{11}{5},\frac{14}{25}\right)\) is option (B). Option (A) gives the second moment as the mean.

Final Answer:
The mean is 11/5 and the variance is 14/25, option (B). \[ \boxed{\frac{11}{5},\ \frac{14}{25}} \]