Question

A random variable X follows the following distribution
$X=x_i:1,2,3,4$

$p(X=x_i):\frac{2}{6},\frac{3}{6},\frac{0}{6},\frac{1}{6}$
, then the mean and variance are

• A. $1,1$
• B. $1,2$
• C. $2,1$
• D. $2,2$

Answer 1

Answer: (C)

$2,1$

$E\left[X\right]={\sum }_{i=1}^4{x}_i{p}_i=1\ast P(X=1)+2\ast P(X=2)+3\ast P(X=3)+4\ast P(X=4)=\frac{2}{6}+\frac{6}{6}+\frac{0}{6}+\frac{4}{6}=2E\left[X^2\right]={\sum }_{i=0}^3{x}^2{p}_i=\frac{2}{6}+\frac{12}{6}+\frac{0}{6}+\frac{16}{6}=5Variance=E\left[X^2\right]-{E\left[X\right]}^2=5-2^2=1$