A random variable X has the following probability distribution
| $x_i$ | -2 | -1 | 0 | 1 | 2 | 3 |
| $p_i$ | 0.1 | k | 0.2 | 2k | 0.3 | k |
Then the mean of X is :
Answer & explanation
Correct answer: option 2
Given probability distribution:
$x_i:\ -2,\ -1,\ 0,\ 1,\ 2,\ 3$
$p_i:\ 0.1,\ k,\ 0.2,\ 2k,\ 0.3,\ k$
Sum of probabilities is $1$:
$0.1+k+0.2+2k+0.3+k=1$
$0.6+4k=1$
$4k=0.4 \Rightarrow k=0.1$
Mean of $X$:
$E(X)=\sum x_ip_i$
$=(-2)(0.1)+(-1)(0.1)+(0)(0.2)+(1)(0.2)+(2)(0.3)+(3)(0.1)$
$=-0.2-0.1+0+0.2+0.6+0.3$
$=0.8$
final answer: mean of $X$ is $0.8$