Practicing Success
The probability distribution of a random variable X is
The variance of X is |
\(\frac{31}{64}\) \(\frac{15}{64}\) \(\frac{103}{64}\) 1 |
\(\frac{103}{64}\) |
$E(x^2)=0^2×\frac{1}{4}+(1)^2×\frac{1}{8}+(2)^2×\frac{1}{8}+(3)^2×\frac{1}{2}$ $=0+\frac{1}{8}+\frac{4}{8}+\frac{9}{2×4}⇒\frac{1+4+36}{8}=\frac{41}{8}$ $E(x)=0×\frac{1}{8}+2×\frac{1}{8}+3×\frac{1}{2}=\frac{1}{8}+\frac{2}{8}+\frac{3}{2×4}$ $\frac{1+2+12}{8}=\frac{15}{8}$ Variance(X) = $E[x^2]-E[x]^2=\frac{41}{8}-(\frac{15}{8})^2$ $=\frac{41}{8×8}-\frac{225}{64}=\frac{328-225}{64}$ $=\frac{103}{64}$ Option 3rd is correct. |