The probability distribution of a random variable X is given as:
|
X |
0 |
1 |
2 |
3 |
4 |
|
P(X) |
k |
2k |
4k |
6k |
7k |
The mean of the distribution is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{14}{5}$
$\sum P(X=x)=1 \Rightarrow k+2k+4k+6k+7k=20k=1$
$k=\frac{1}{20}$
$E(X)=\sum xP(X)$
$=0\cdot k+1\cdot2k+2\cdot4k+3\cdot6k+4\cdot7k$
$=2k+8k+18k+28k=56k$
$=56\cdot\frac{1}{20}=\frac{14}{5}$
The mean is $\frac{14}{5}$.