91Ó°ÊÓ

From the given information, 302 U.S. Postal Service employees (229 men and 73 women) applied for promotion. There were 302 U.S. Postal Service employees (229 men and 73 women) who applied for promotion.

x follows a hypergeometric distribution with$N=302,  n=72  and  r=73$

So, the expected value (mean) of x is

$\begin{array}{rcl}E\left(x\right)& =& \mathrm{μ}\\ & =& \frac{nr}{N}\\ & =& \frac{72×73}{302}\\ & =& 17.403974\\ & ≈& 17\\ & & \end{array}$

The variance of x is

$\begin{array}{rcl}V\left(x\right)& =& {\mathrm{σ}}^2\\ & =& \frac{nr\left(N-r\right)\left(N-n\right)}{N^2\left(N-1\right)}\\ & =& \frac{72×73×\left(302-73\right)×\left(302-72\right)}{{302}^2×\left(302-1\right)}\\ & =& \frac{72×73×229×230}{91204×301}\\ & =& 10.0841\\ & & \end{array}$

The variance of x is 10.0841.

The standard deviation of x is

$\begin{array}{rcl}\mathrm{σ}& =& \sqrt{10.0841}\\ & =& 3.1755\\ & & \end{array}$

The standard deviation of x is 3.1755.

The interval is,$\mathrm{μ}Â\pm 3\mathrm{σ}$

So,

$\begin{array}{rcl}\mathrm{μ}Â\pm 3\mathrm{σ}& =& 17.403974Â\pm \left(3×3.1755\right)\\ & =& 17.403974Â\pm 9.5265\\ & =& \left(17.403974-9.5265,17.403974+9.5265\right)\\ & =& \left(7.8775,26.9305\right)\\ & & \end{array}$

There is no discrimination in promoting females.

Therefore,observed that only five females werepromoted.

So, females were not promoted fairly.