91Ó°ÊÓ

A coin for which the probability of obtaining a head is 0.3, is tossed 15 times. The random variable X denotes the number of heads obtained in 15 tosses.

According to the question:

\(X \sim Binomial\left( {15,0.3} \right).\)

\(\begin{aligned}{}E\left( X \right) = np\\ = 15 \times 0.3\\ = 4.5.\end{aligned}\)

We know that \(M.S.E = E\left( {{{\left( {X - d} \right)}^2}} \right)\) is minimized when \(d = E\left( X \right) = 4.5\).

Hence, for the predicted value of 4.5, X has the smallest M.S.E.

The pdf of X is defined as:

\(\begin{aligned}{}f\left( x \right) = P\left( {X = x} \right)\\ &= \left( {\begin{aligned}{{}{}}n\\x\end{aligned}} \right){p^x}{\left( {1 - p} \right)^{n - x}}\\ &= \left( {\begin{aligned}{{}{}}{15}\\x\end{aligned}} \right){\left( {0.3} \right)^x}{\left( {0.7} \right)^{15 - x}}.\end{aligned}\)

The cdf of X is defined as:

\(\begin{aligned}{}F\left( x \right) &= P\left( {X \le x} \right)\\ &= \sum\limits_{x = 0}^x {P\left( {X = x} \right)} \\ &= \sum\limits_{x = 0}^x {\left( {\begin{aligned}{{}{}}{15}\\x\end{aligned}} \right){{\left( {0.3} \right)}^x}{{\left( {0.7} \right)}^{15 - x}}} .\end{aligned}\)

Create the following table using the formula for the pdf and the cdf of a binomial distribution:

From the above table, it is observed that\(P\left( {X \le 4} \right) \ge 0.5\;{\rm{and}}\;P\left( {X \ge 4} \right) \ge 0.5.\)

Hence, 4 is the median of X.

We know that\(M.A.E = E\left( {\left| {X - d} \right|} \right)\) is minimized when d is the median of X.