Question
Calculate mean of the discrete probability distribution $p(x)\left\{\begin{array}{c}\frac{x-1}{6} ; x=2,3 \\ \frac{1}{2} ;\end{array} x=4\right.$
✓
Answer
Here, $p(x)=\frac{x-1}{6}$ is given.
Putting, $x=2,3$
$\mathrm{p}(2)=\frac{2-1}{6}=\frac{1}{6}$
$ \mathrm{P}(3)=\frac{3-1}{6}$
$=\frac{2}{6}$
$=\frac{1}{3}$
and $P(4)=\frac{1}{2}$
So, the probability distribution of $X$ is obtained as follows :
Mean of the distribution:
$\mu=\Sigma x \cdot p(x)$
$ =2\left(\frac{1}{6}\right)+3\left(\frac{1}{3}\right)+4\left(\frac{1}{2}\right)$
$ =\frac{2}{6}+1+2$
$ =\frac{2+6+12}{6}$
$=\frac{20}{6}$
$=\frac{10}{3}$
Hence, the mean of the distribution obtained is $\frac{10}{3}$.
Putting, $x=2,3$
$\mathrm{p}(2)=\frac{2-1}{6}=\frac{1}{6}$
$ \mathrm{P}(3)=\frac{3-1}{6}$
$=\frac{2}{6}$
$=\frac{1}{3}$
and $P(4)=\frac{1}{2}$
So, the probability distribution of $X$ is obtained as follows :
Mean of the distribution:
$\mu=\Sigma x \cdot p(x)$
$ =2\left(\frac{1}{6}\right)+3\left(\frac{1}{3}\right)+4\left(\frac{1}{2}\right)$
$ =\frac{2}{6}+1+2$
$ =\frac{2+6+12}{6}$
$=\frac{20}{6}$
$=\frac{10}{3}$
Hence, the mean of the distribution obtained is $\frac{10}{3}$.
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