In a meeting, 70 \% of the members favour and 30 \% oppose a certain proposal. A member is selected at random and we take X=0 if he opposed, and X=1 if he is in favour. Find \mathrm{E}(\mathrm{X}) and \operatorname{Var}(\mathrm{X}).
In a meeting, 70 \% of the members favour and 30 \% oppose a certain proposal. A member is selected at random and we take X=0 if he opposed, and X=1 if he is in favour. Find \mathrm{E}(\mathrm{X}) and \operatorname{Var}(\mathrm{X}).

Solution:

If members are opposed, X=0, and if members are in favour, X=1.

\mathrm{P}(\mathrm{X}=0)=30 \%=30 / 100=0.3

P(X=1)=70 \%=70 / 100=0.7

As a result, the necessary probability distribution is,

ieret

E(X) is:

\mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}

=0 \times 0.3+1 \times 0.7

\Rightarrow E(X)=0.7

And E\left(X^{2}\right) is:

\mathrm{E}\left(\mathrm{X}^{2}\right)=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^{2} \cdot \mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right) =(0)^{2} \times 0.3+(1)^{2} \times 0.7

\Rightarrow \mathrm{E}\left(\mathrm{X}^{2}\right)=0.7

Then Variance, \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}

=0.7-(0.7)^{2}

=0.7-0.49

=0.2