A random variable X has the following probability distribution :
|
X |
0 |
1 |
2 |
3 |
4 |
|
P(X) |
c |
3c |
4c |
6c |
7c |
The mean of the distribution is:
Answer & explanation
Correct answer: option 1
Given probability distribution
$X: 0,\;1,\;2,\;3,\;4$
$P(X): c,\;3c,\;4c,\;6c,\;7c$
Total probability = 1
$c+3c+4c+6c+7c=21c=1$
$c=\frac{1}{21}$
Mean of distribution
$E(X)=\sum xP(X)$
$E(X)=0\cdot c+1\cdot3c+2\cdot4c+3\cdot6c+4\cdot7c$
$=3c+8c+18c+28c=57c$
$E(X)=57\cdot\frac{1}{21}=\frac{57}{21}=\frac{19}{7}$
The mean of the distribution is $\frac{19}{7}$.