Question

: 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 E(X) and Var(X).

Answer 1

Hint : In this question the percentage of people in favour and against a proposal is given where X=0 if the member is against the proposal and X=1 for those who are in favour of the proposal so by using this values we will find the expectation value E(X) and then we will find the variance (X) of the random variable X.

Complete step-by-step answer :
The expectation value of a certain experiment is given by the formula $E\left( X \right) = \sum\nolimits_{i = 1}^n {{x_i}{p_i}}$.
If the member oppose the proposal then $X = 0$
If the member are in favour of the proposal then $X = 1$
Now it is given that 70% of the members are favour of the proposal, hence we can write
$P\left( {X = 1} \right) = 70\% = 0.7$
Also 30% of the members of meeting oppose the proposal, so
$P\left( {X = 0} \right) = 30\% = 0.3$
Hence we get the Probability distribution table as

X	0	1
P(X)	0.3	0.7

Now we know that the expectation value E(X) is given by the formula
$\Rightarrow E\left( X \right) = \sum\nolimits_{i = 1}^n {{x_i}{p_i} - - (i)}$
So by substituting the values in equation (i), we get the expectation value E(X) as
$\Rightarrow E\left( X \right) = \sum\nolimits_{i = 1}^n {{x_i}{p_i} = 0 \times 0.3 + 1 \times 0.7} = 0.7$
Now we will find the variance (X) which is given by the formula
$\Rightarrow Var\left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} - - (ii)$
Where,

$$\Rightarrow E\left( {{X^2}} \right) = \sum\nolimits_{i = 1}^n {x_i^2{p_i}} \\\ = {0^2} \times 0.3 + {1^2} \times 0.7 \\\ = 0.7 \;$$

Now substitute the value of $E\left( {{X^2}} \right)$in equation (ii) where$E\left( X \right) = 0.7$, hence we get

$$\Rightarrow Var\left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} \\\ = 0.7 - {\left( {0.7} \right)^2} \\\ = 0.7\left[ {1 - 0.7} \right] \\\ = 0.7 \times 0.3 \\\ = 0.21 \;$$

Hence the value of $Var\left( X \right) = 0.21$
Therefore the value of
$E\left( X \right) = 0.7$
$Var\left( X \right) = 0.21$
Note : The expectation defines the average value of the probability distribution while the variance is defined as the spread of the distribution from the average value of the probability distribution.