91Ó°ÊÓ

For the single-sample plan,\(n = 50\), and\(c = 2\), because\(N\)is large enough, average outgoing quality is given by

\(AOQ = \frac{{(N - n)p(A) \times p}}{N}\gg P(A) \times p\)

Theorem:

\(\begin{array}{l}b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{,{\rm{ otherwise}}{\rm{. }}}\end{array}} \right.\\B(x;n,p) = P(X \le x) = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n.\end{array}\)

Using this,\(AOQ\)is

\(AOQ = p \times P(A) = p \times (b(0;50,p) + b(1;50,p) + b(2;50,p)) = p \times B(2;50,p)\)

The following table contains\(AOQ\)for\(p = 0.01,0.02, \ldots ,0.1\). AOQL is maximum for\(p = 0.04\).

\(\begin{array}{*{20}{c}}p&{AOQ}\\{0.01}&{0.01}\\{0.02}&{0.018}\\{0.03}&{0.024}\\{0.04}&{0.027}\\{0.05}&{0.027}\\{0.06}&{0.025}\\{0.07}&{0.022}\\{0.08}&{0.018}\\{0.09}&{0.014}\\{0.1}&{0.011}\\{}&{}\end{array}\)

Using calculus (the algebra won't be shown here), one gets probability\(p\)given by

\({p_0} = 0.0447.\)

From this,\(AOQL\)can be computed as

\(AOQL = {p_0} \times P(A) = 0.0274.\)

ATI can be computed using formula

\(ATI = n \times p(A) + N \times (1 - P(A)).\)

For\(n = 50,N = 2000\), and\(P(A)\)given in\((a)\), one can computed values from the following table.

\(\begin{array}{*{20}{c}}p&{{\rm{ ATI }}}\\{0.01}&{77.3}\\{0.02}&{202.1}\\{0.03}&{418.6}\\{0.04}&{679.9}\\{0.05}&{945.1}\\{0.06}&{1188.8}\\{0.07}&{1393.6}\\{0.08}&{1559.3}\\{0.09}&{1686.1}\\{0.1}&{1781.6}\\{}&{}\end{array}\)

\( a. Use AOQ = p \times P(A); b. {p_0} = 0.0447,AOQL = 0.0274; c. Use ATI = n \times p(A) + N \times (1 - P(A)). \)