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For the single-sample plan,\(n = 50\), and\(c = 1\), because\(N\)is large enough, average outgoing quality is given by

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

Theorem:

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

\(B(x;n,p) = P(X \le x) = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\)

Using this,\(AOQ\)is

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

The following table contains\(AOQ\)for\(p = 0.01,0.02, \ldots ,0.1\). Below the table there is mentioned graph.

\(\begin{array}{*{20}{c}}p&{AOQ}\\{0.01}&{0.0091}\\{0.02}&{0.0147}\\{0.03}&{0.0167}\\{0.04}&{0.016}\\{0.05}&{0.014}\\{0.06}&{0.0114}\\{0.07}&{0.0089}\\{0.08}&{0.0066}\\{0.09}&{0.0048}\\{0.1}&{0.0034}\\{}&{}\end{array}\)

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. The mentioned graph is below the table.

\(\begin{array}{*{20}{c}}p&{ATI}\\{0.01}&{224.3}\\{0.02}&{565.2}\\{0.03}&{917.2}\\{0.04}&{1219}\\{0.05}&{1455.2}\\{0.06}&{1629.5}\\{0.07}&{1753.3}\\{0.08}&{1838.7}\\{0.09}&{1896.3}\\{0.1}&{1934.1}\\{}&{}\end{array}\)

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

\({p_0} = 0.0318.\)

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

\(AOQL = {p_0} \times P(A) = 0.0167\).