Question:
For the following probability distribution, the standard deviation of the random variable X is
X 2 3 4 p(X=\(x\)) 0.2 0.5 0.3
For the following probability distribution, the standard deviation of the random variable X is
| X | 2 | 3 | 4 |
| p(X=\(x\)) | 0.2 | 0.5 | 0.3 |
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Compute E(X) and E(X^2), then variance = E(X^2) - [E(X)]^2.
Updated On: Oct 1, 2026
- \(0.5\)
- \(0.6\)
- \(0.61\)
- \(0.7\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
For a discrete random variable, \(\text{Var}(X) = E(X^2) - [E(X)]^2\) and the standard deviation is its square root.
Step 2: Mean
\[ E(X) = 2(0.2) + 3(0.5) + 4(0.3) = 0.4 + 1.5 + 1.2 = 3.1 \]
Step 3: Variance and SD
\[ E(X^2) = 4(0.2) + 9(0.5) + 16(0.3) = 0.8 + 4.5 + 4.8 = 10.1 \]
\[ \text{Var}(X) = 10.1 - (3.1)^2 = 10.1 - 9.61 = 0.49 \]
\[ \sigma = \sqrt{0.49} = 0.7 \]
Option (C), 0.61, is not a value from this calculation, and (B) 0.6 would need variance 0.36.
Final Answer:
The standard deviation is 0.7, option (D). \[ \boxed{0.7} \]
For a discrete random variable, \(\text{Var}(X) = E(X^2) - [E(X)]^2\) and the standard deviation is its square root.
Step 2: Mean
\[ E(X) = 2(0.2) + 3(0.5) + 4(0.3) = 0.4 + 1.5 + 1.2 = 3.1 \]
Step 3: Variance and SD
\[ E(X^2) = 4(0.2) + 9(0.5) + 16(0.3) = 0.8 + 4.5 + 4.8 = 10.1 \]
\[ \text{Var}(X) = 10.1 - (3.1)^2 = 10.1 - 9.61 = 0.49 \]
\[ \sigma = \sqrt{0.49} = 0.7 \]
Option (C), 0.61, is not a value from this calculation, and (B) 0.6 would need variance 0.36.
Final Answer:
The standard deviation is 0.7, option (D). \[ \boxed{0.7} \]
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