Question:
The variance of random variable \(X\), for the following probability mass function (p. m. f.) is ...
\(X=x\) \(1\) \(2\) \(3\) \(P(X=x)\) \(\frac{1}{5}\) \(\frac{2}{5}\) \(\frac{2}{5}\)
The variance of random variable \(X\), for the following probability mass function (p. m. f.) is ...
| \(X=x\) | \(1\) | \(2\) | \(3\) |
| \(P(X=x)\) | \(\frac{1}{5}\) | \(\frac{2}{5}\) | \(\frac{2}{5}\) |
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Compute the mean and the mean of squares from the table and use variance equal to E(X^2) minus the square of E(X).
Updated On: Nov 12, 2026
- \(\frac{14}{25}\)
- \(\frac{11}{5}\)
- \(\frac{\sqrt{14}}{5}\)
- \(\frac{11}{25}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Variance \(=E(X^{2})-[E(X)]^{2}\).
Step 2: Mean
\(E(X)=1\cdot\frac{1}{5}+2\cdot\frac{2}{5}+3\cdot\frac{2}{5}=\frac{1+4+6}{5}=\frac{11}{5}\).
Step 3: Mean of squares
\(E(X^{2})=1\cdot\frac{1}{5}+4\cdot\frac{2}{5}+9\cdot\frac{2}{5}=\frac{1+8+18}{5}=\frac{27}{5}\).
Step 4: Variance
\(\frac{27}{5}-\frac{121}{25}=\frac{135-121}{25}=\frac{14}{25}\).
Final Answer:
The variance is 14/25. \[ \boxed{\text{(A) }\dfrac{14}{25}} \]
Variance \(=E(X^{2})-[E(X)]^{2}\).
Step 2: Mean
\(E(X)=1\cdot\frac{1}{5}+2\cdot\frac{2}{5}+3\cdot\frac{2}{5}=\frac{1+4+6}{5}=\frac{11}{5}\).
Step 3: Mean of squares
\(E(X^{2})=1\cdot\frac{1}{5}+4\cdot\frac{2}{5}+9\cdot\frac{2}{5}=\frac{1+8+18}{5}=\frac{27}{5}\).
Step 4: Variance
\(\frac{27}{5}-\frac{121}{25}=\frac{135-121}{25}=\frac{14}{25}\).
Final Answer:
The variance is 14/25. \[ \boxed{\text{(A) }\dfrac{14}{25}} \]
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