Question

In a meeting 60% of the members favour and 40% 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, then $\text{Var}(X) =$

• A. $0.36$
• B. $0.24$
• C. $0.6$
• D. $0.06$

Hint:

Anytime a random variable is strictly defined as $1$ for an event happening and $0$ for it not happening, it is an Indicator (or Bernoulli) variable. You can bypass tables and immediately use $\text{Mean} = p$ and $\text{Variance} = p(1-p)$.

Answer 1

The Correct Option is B

Step 1: Set up the table.
The variable $X$ takes only two values. $X=1$ (in favour) with probability $0.6$, and $X=0$ (oppose) with probability $0.4$.
Step 2: Recall what variance needs.
Variance is $\text{Var}(X)=E(X^2)-[E(X)]^2$, so we first need the mean $E(X)$ and then $E(X^2)$.
Step 3: Find the mean $E(X)$.
$E(X)=(0)(0.4)+(1)(0.6)=0.6$.
Step 4: Find $E(X^2)$.
Since $0^2=0$ and $1^2=1$, we get $E(X^2)=(0)(0.4)+(1)(0.6)=0.6$.
Step 5: Plug into the formula.
$\text{Var}(X)=0.6-(0.6)^2=0.6-0.36=0.24$.
Step 6: A quick check.
For a yes-or-no variable, variance is just $p\times q=0.6\times 0.4=0.24$, which matches. So the answer is option (2). \[ \boxed{\text{Var}(X)=0.24} \]