Question 39
<p>A random variable X has the distribution law as given below:</p> <table border="1" cellpadding="0" cellspacing="0" style="width: 250px;"> <tbody> <tr> <td> <p style="text-align: center;"><strong>X</strong></p> </td> <td> <p style="text-align: center;">1</p> </td> <td> <p style="text-align: center;">2</p> </td> <td> <p style="text-align: center;">3</p> </td> </tr> <tr> <td> <p style="text-align: center;"><strong>P(X = x)</strong></p> </td> <td> <p style="text-align: center;">0.3</p> </td> <td> <p style="text-align: center;">0.4</p> </td> <td> <p style="text-align: center;">0.3</p> </td> </tr> </tbody> </table> <p><br /> The variance of the distribution is:</p>
📘 Step-by-Step Solution & Detailed Explanation:
Step 1: Compute the Expected Value (Mean), \( \mu = E[X] \) \[ \mu = \sum x \cdot P(X=x) = 1 \cdot 0.3 + 2 \cdot 0.4 + 3 \cdot 0.3 \] \[ \mu = 0.3 + 0.8 + 0.9 = 2.0 \] Step 2: Compute the Variance, \( \text{Var}(X) = E[X^2] - (E[X])^2 \) \[\] First, compute \( E[X^2] \): \[ E[X^2] = \sum x^2 \cdot P(X=x) = 1^2 \cdot 0.3 + 2^2 \cdot 0.4 + 3^2 \cdot 0.3 \] \[ E[X^2] = 1 \cdot 0.3 + 4 \cdot 0.4 + 9 \cdot 0.3 = 0.3 + 1.6 + 2.7 = 4.6 \] Now, \[ \text{Var}(X) = E[X^2] - \mu^2 = 4.6 - 2^2 = 4.6 - 4 = \boxed{0.6} \] Final Answer: 0.6
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