Question

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

Answer 1

To solve this question we should. Know about probability distribution of random variables.
Mean for probability distribution: the mean is equal to the sum over every possible value weighted by the probability of the value; that is, it is computed by taking the product of each possible value of $x$ and $X$ , its probability $P(X)$ and then adding the product together.
In mathematically: $E(X) = \sum {xP(X)}$
Variance: The variance is the expected value of the squared difference between each value and the mean of the distribution. In the finite case, it is simply the average squared difference.
In mathematically: $Var(X) = E({X^2}) - {[E(X)]^2}$

Complete step by step answer:
As given in question;
If,
Member is opposed $X = 0$
Member is favour proposal $X = 1$
Member who is in favour of proposal $= 70\%$
So, $P(X = 1) = 70\% = 0.7$
Member who oppose proposal $= 30\%$
So, $P(X = 0) = 30\% = 0.3$
Hence probability distribution is

$X$	$0$	$1$
$P(X)$	$0.3$	$0.7$

The value of mean of probability distribution is:
$E(X) = \sum {xP(X)}$
Keeping value in it,
$\Rightarrow E(X) = 0 \times 0.3 + 1 \times 0.7$
$\Rightarrow E(X) = 0 + 0.7 = 0.7$
The variance of $x$ will be;
$Var(X) = E({X^2}) - {[E(X)]^2}$
To find;
$E({X^2}) = \sum {{x^2}P(X)}$
By keeping value in it,
$\Rightarrow E({X^2}) = {0^2} \times 0.3 + {1^2} \times 0.7$
$\Rightarrow E({X^2}) = 0 + 0.7$
Now, Keeping value in it,
$Var(X) = 0.7 - {0.7^2}$
$\Rightarrow Var(X) = 0.7(1 - 0.7)$
$\Rightarrow Var(X) = 0.7 \times 0.3 = 0.21$
Hence, mean $E(X) = 0.7$ and variance $Var(X) = 0.21$

Note: Types of probability distribution:
Cumulative probability distribution
Discrete probability distribution
Probability density function is used to define the probability of the random variable coming within a distinct range of values, as opposed to taking anyone values.