CUET CUET Mathematics 2024 - <table class="question-table" border="1"> <thead> <tr> <th> X </th> <th>-2</th> <th>-1</th> <th>0</th> <th>1</th> <th>2</th> </tr> </thead> <tbody> <tr> <td> P(X) </td> <td> 0.2 </td> <td> 0.1 </td> <td> 0.3 </td> <td> 0.2 </td> <td>0.2</td> </tr> </tbody> </table> The variance of X will be : | PYQs + Solutions | AfterBoards
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CUET Mathematics 2024 PYQs

CUET Mathematics 2024

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X X -2 -1 0 1 2
P(X) P(X) 0.2 0.2 0.1 0.1 0.3 0.3 0.2 0.2 0.20.2
The variance of XX will be :

Correct Option: 3
Step 1: Calculate E(X):
E(X)=xiP(X=xi)E(X) = \sum x_i P(X=x_i) \newline E(X)=(2)(0.2)+(1)(0.1)+(0)(0.3)+(1)(0.2)+(2)(0.2)E(X) = (-2)(0.2) + (-1)(0.1) + (0)(0.3) + (1)(0.2) + (2)(0.2) \newline E(X)=0.40.1+0+0.2+0.4=0.1E(X) = -0.4 - 0.1 + 0 + 0.2 + 0.4 = 0.1
Step 2: Calculate E(X²):
E(X2)=xi2P(X=xi)E(X^2) = \sum x_i^2 P(X=x_i) \newline E(X2)=(4)(0.2)+(1)(0.1)+(0)(0.3)+(1)(0.2)+(4)(0.2)E(X^2) = (4)(0.2) + (1)(0.1) + (0)(0.3) + (1)(0.2) + (4)(0.2) \newline E(X2)=0.8+0.1+0+0.2+0.8=1.9E(X^2) = 0.8 + 0.1 + 0 + 0.2 + 0.8 = 1.9
Step 3: Find the variance:
Var(X)=E(X2)[E(X)]2=1.9(0.1)2=1.90.01=1.89\text{Var}(X) = E(X^2) - [E(X)]^2 = 1.9 - (0.1)^2 = 1.9 - 0.01 = 1.89

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