91Ó°ÊÓ

Let X is a random variable that follows Hypergeometric distribution with parameters A,B and n. Here, the parameter values are A = A, B = T-A, and n=n that is, \[X \sim Hyp\left( {A,T - A,n} \right)\].

From theorem 5.3.4 in the textbook, if X is a hypergeometric variable, then the variance of X is

\(Var\left( X \right) = \frac{{nAB}}{{{{\left( {A + B} \right)}^2}}} \cdot \frac{{A + B - n}}{{A + B - 1}}\)

On substituting the values,

\[\begin{array}{c}Var\left( X \right) = \frac{{n\; \cdot A \cdot \left( {T - A} \right)}}{{{{\left( {A + T - A} \right)}^2}}} \cdot \frac{{\left( {A + T - A} \right) - n}}{{\left( {A + T - A} \right) - 1}}\\ = \frac{{n \cdot A \cdot \left( {T - A} \right)}}{{{T^2}}} \cdot \frac{{T - n}}{{T - 1}}\\ = \frac{1}{{{T^2} \cdot \left( {T - 1} \right)}}\left[ {nA{T^2} - {n^2}AT - n{A^2}T + {n^2}{A^2}} \right] \ldots \left( 1 \right)\end{array}\]

In order to find the value of n for which the value of Var(X) will be a maximum, one need to differentiate the above equation \(\left( 1 \right)\) with respect to n.

\(\frac{{\partial Var\left( X \right)}}{{\partial X}} = \frac{1}{{{T^2} \cdot \left( {T - 1} \right)}}\left[ {A{T^2} - 2nAT - {A^2}T + 2n{A^2}} \right] \ldots \left( 2 \right)\)

Equating \(\left( 2 \right)\) with 0

\(\)

\(\begin{array}{c}\frac{1}{{{T^2} \cdot \left( {T - 1} \right)}}\left[ {A{T^2} - 2nAT - {A^2}T + 2n{A^2}} \right] = 0\\2n\left( {A - T} \right) = T\left( {A - T} \right)\\n = \frac{T}{2}\end{array}\)

But, if T is not an even number, the value of n won’t be an integer. Therefore, we can either use \[\frac{{T - 1}}{2}\;or\;\frac{{T + 1}}{2}\]if T is odd.

Therefore, the answer is:

\[\begin{array}{l}\frac{{T - 1}}{2}\;or\;\frac{{T + 1}}{2}\;if\;T\;is\;odd\\\frac{T}{2}\;if\;T\;is\;even\end{array}\].