91Ó°ÊÓ

See the text for the transformation. Mean value of the given transformation

\(Y = h(X) = {\sin ^{ - 1}}\left( {\sqrt {\frac{X}{n}} } \right)\)

is given by

\({\sin ^{ - 1}}(\sqrt p )\)

and variance

\(\frac{1}{{4n}}.\)

The values using the inverse transformation (arcsin) are

\(i\) \({\sin ^{ - 1}}(\sqrt p )\)

\(\begin{aligned}{*{20}{l}}{1\;\;0.22551}\\{2\;\;\;0.30469}\\{3\;\;\;0.35374}\\{4\;\;\;0.29584}\\{5\;\;\;0.44462}\\{6\;\;0.31332}\\{7\;\;0.18819}\\{8\;\;0.36137}\\{9\;\;0.23673}\\{10\;0.35374}\\{11\;0.39065}\\{12\;0.27741}\end{aligned}\)

\(\begin{aligned}{*{20}{l}}{13\;\;0.28676}\\{14\;\;\;0.32999}\\{15\;\;0.30469}\\{16\;\;0.29584}\\{17\;\;0.27741}\\{18\;\;0.33807}\\{19\;\;0.24747}\\{20\;\;\;0.32175}\\{21\;\;0.29584}\\{22\;\;\;0.30469}\\{23\;\;\;0.24747}\\{24\;\;\;0.35374}\\{25\;\;\;0.39770}\\{26\;\;\;0.28676}\\{27\;\;\;0.23673}\\{28\;\;\;0.32175}\end{aligned}\)

The center line, from the fact that

\(\sum\limits_{i = 1}^k {{{\hat y}_i}} = 9.24373\)

is given by

\(\bar y = \frac{{9.24373}}{{30}} = 0.30812\)

The\(LCL\) is

\(\begin{aligned}{l}LCL = \bar y - 3\sqrt {\frac{1}{{4n}}} = 0.30812 - 3 \cdot \sqrt {\frac{1}{{800}}} \\ = 0.2020\end{aligned}\)

The \(UCL\) is

\(\begin{aligned}{l}UCL = \bar y + 3\sqrt {\frac{1}{{4n}}} = 0.30812 + 3 \cdot \sqrt {\frac{1}{{800}}} \\ = 0.4142\end{aligned}\)

There are two points out of the limits, first one is \(0.18819\) which is lower than the \(LCL\) and\(0.44462\) which is higher than the\(UCL\).