Answer:
The answer is "0.142466".
Step-by-step explanation:
Using the p formula for the proportion of nonconforming units through the subgrouping which can vary in sizes:
[tex]p =\frac{np}{n}\\\\[/tex]
[tex]\bar{p}=\frac{\Sigma np}{\Sigma n}\\\\[/tex]
Defects [tex]= \frac{5}{100} \times 50 \\\\[/tex]
[tex]p = \frac{5}{100} \times \frac{50}{50}=0.05\\\\[/tex]
It calculates the controls limits through the p-chart that is:
[tex]UCL_{p},LCL_{p}=\bar{p} \pm \sqrt{\frac{\bar{p}(1-\bar{p})}{\bar{n}}}\\\\[/tex]
So, the upper control limits:
[tex]= 0.05 + 3 \times \sqrt{(\frac{0.05\times(1-0.05)}{50})} \\\\= 0.142466[/tex]
