91影视

First: Locate and connect desired ARLs on the in-control and out-of-control scales; Second: Solve for\(n\)equation

\({k^\prime } = \frac{{\Delta /2}}{{\sigma /\sqrt n }}\)

and round\(n\)up to the nearest integer.

Third: Use second line to connect the point on the\({k^\prime }\)scale with the point on the in-control ARL scale. Note where this second line crosses\({h^\prime }\)scale. Then\(h\)is

\(h = {h^\prime }\frac{\sigma }{{\sqrt n }}\)

Locate 600 on the in-control ARL scale, and 4 on the out-of-control scale and extending to\({k^\prime }\)scale on the Kemp Nonogram gives\(k = 0.87\).

Solve for\(n\)equation

\({k^\prime } = \frac{{\Delta /2}}{{\sigma /\sqrt n }} = \frac{{0.004/2}}{{0.005/\sqrt n }}\)

which yield

\(n = 4.73 \approx 5.\)

Connect connect\(0.97\)on the\({k^\prime }\)scale to 600 on the in-control ARL scale and extending to\({h^\prime }\)you get\({h^\prime } = 2.8\)(Cusum procedure). Find\(h\)from

\(h = {h^\prime }\frac{\sigma }{{\sqrt n }} \cdot = 2.8 \cdot \frac{{0.005}}{{\sqrt n }} = 0.00626\)

The out-of-control signal results when one of\({d_i},e_i^\prime s\)is larger than\(h\).

\(h = 0.00626.{\rm{ }}\)