Question

If a random variable X has the hypergeometric distribution

with parameters A = 8, B = 20, and n, for what value of n will Var(X) be a maximum?

Short answer

The value of n which maximises the variance is n=14.

Step-by-step solution

Given information

Let X is a random variable that follows hypergeometric distribution with parameters A,B and n. Here, the parameter values are A = 8, B = 20, and n that is, \[X \sim Hyp\left( {A = 8,B = 20,n} \right)\].

Variance formula and substituting the values

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,

\(\begin{array}{c}Var\left( X \right) = \frac{{n\;8 \cdot 20}}{{{{\left( {8 + 20} \right)}^2}}} \cdot \frac{{8 + 20 - n}}{{8 + 20 - 1}}\\ = \frac{{160n}}{{{{\left( {28} \right)}^2}}} \cdot \frac{{28 - n}}{{27}}\\ = \frac{1}{{21168}}\left[ {4480n - 160{n^2}} \right] \ldots \left( 1 \right)\end{array}\)

Finding the desirable value of n

To find the value of n such that the value of Var(X) will be a maximum, one need to differentiate the above equation (1) with respect to n.

\(\frac{{\partial Var\left( X \right)}}{{\partial X}} = \frac{1}{{21168}}\left[ {4480 - 320n} \right] \ldots \left( 2 \right)\)

Equating equation (2) with zero,

\(\)

\(\begin{array}{c}\frac{1}{{21168}}\left[ {4480 - 320n} \right] = 0\\4480 = 320n\\n = 14.\end{array}\)

Therefore, the value of n which maximises the variance is n=14.