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The probability distribution of the random variable X is given by Then, the value of V(X) is equal to |
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Answer» Correct Answer - B `E(X)=Sigmax_iP(x_i)=1/8(1)+1/2(2)+1/8(3)+1/4(4)` `=1/8+1+3/8+1=5/2` Now, `V(X)=E(X^2)-[E(X)]^2` `=Sigmax_i^2.P(x_i)-(5/2)^2` `=1/8(1)^2+1/2(2)^2+1/8(3)^2+1/4(4)^2-(25)/4` `=1/8+2+9/8+4-(25)/4` `=(1+16+9+32-50)/8=(58-50)/8=8/8=1` |
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