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For the following distribution :
Var(X) is equal to : |
3 10 1 7 |
1 |
The correct answer is Option (3) → 1 $E(X)=1×\frac{1}{10}+2×\frac{1}{5}+3×\frac{3}{10}+4×\frac{2}{5}=3$ $E(X^2)=1^2×\frac{1}{10}+2^2×\frac{1}{5}+3^2×\frac{3}{10}+4^2×\frac{2}{5}=10$ So $Var(X)=E(X^2)-E^2(X)$ $=10-3^2=1$ |
