Question-11

Suppose that \(P(X=a) = p \hspace{1mm}\text{and} \hspace{1mm} P(X=b) = 1-p\).

Question - 11 (a)

Idenitfy the distribution of \(\dfrac{X-b}{a-b}\).

• Binomial distribution
• Negative Binomial distribution
• Bernoulli distribution
• Cannot determine

Hint

Substitute the values of \(X\) and try to identify the distribution.

Answer

• Binomial distribution
• Negative Binomial distribution
• Bernoulli distribution
• Cannot determine

Solution

Since \(\dfrac{X-b}{a-b}\) will equal 1 with probability \(p\), or 0 with probability \(1-p\).
Therefore, it follows bernoulli distribution.

Question - 11 (b)

Find the value of Var\((X)\).

• \(p(1-p)\)

• \(\dfrac{p(1-p)}{(a-b)^2}\)

• \((a-b)^2 \hspace{0.5mm}p\hspace{0.5mm}(1-p)\)

• \(np(1-p)\)

Hint

Start with the Var\(\left(\dfrac{X-b}{a-b}\right)\).

Answer

• \(p(1-p)\)

• \(\dfrac{p(1-p)}{(a-b)^2}\)

• \((a-b)^2 \hspace{0.5mm}p\hspace{0.5mm}(1-p)\)

• \(np(1-p)\)

Solution

Since \(\dfrac{X-b}{a-b}\) is a bernoulli random variable will take value 1 with probability \(p\), or 0 with probability \(1-p\). Thus, \(p(1-p)=\text{Var}\left(\dfrac{X-b}{a-b}\right) = \dfrac{1}{(a-b)^2} \text{Var}(X-b)=\dfrac{1}{(a-b)^2} \text{Var}(X)\)
\(\implies p(1-p) = \dfrac{1}{(a-b)^2} \text{Var}(X)\)
\(\implies \text{Var}(X) = (a-b)^2p(1-p)\)