91Ó°ÊÓ

\(\begin{aligned}{}X \sim Binomial\left( {7,\frac{1}{4}} \right)\\Y \sim Binomial\left( {5,\frac{1}{2}} \right)\end{aligned}\)

\(\begin{aligned}{}E\left( X \right) &= np\\ &= 7 \times \frac{1}{4}\\ &= \frac{7}{4}.\end{aligned}\)

We know that \(M.S.E = E\left( {{{\left( {X - d} \right)}^2}} \right)\) is minimized when \(d = E\left( X \right) = \frac{7}{4}\).

\(\begin{aligned}{}Now,E\left( {{{\left( {X - d} \right)}^2}} \right) = Var\left( X \right)\\ = np\left( {1 - p} \right)\\ = 7 \times \frac{1}{4} \times \frac{3}{4}\\ = \frac{{21}}{{16}}\\ = 1.3125.\end{aligned}\)

Therefore, the M.S.E of X is 1.3125.

\(\begin{aligned}{}Similarly,E\left( Y \right) = np\\ = 5 \times \frac{1}{2}\\ = \frac{5}{2}.\end{aligned}\)

We know that \(M.S.E = E\left( {{{\left( {Y - d} \right)}^2}} \right)\) is minimized when \(d = E\left( Y \right) = \frac{5}{2}\).

\(\begin{aligned}{}Now,\,\,E\left( {{{\left( {Y - d} \right)}^2}} \right) = Var\left( Y \right)\\ = np\left( {1 - p} \right)\\ = 5 \times \frac{1}{2} \times \frac{1}{2}\\ = \frac{5}{4}\\ = 1.25.\end{aligned}\)

Therefore, the M.S.E of Y is 1.25.

Hence, the random variable Y has a smaller M.S.E