Answer and Explanation:
Data provided in the question
defect rate i.e. [tex]\bar p[/tex] = 1.50%
the sample size = n = 200
Now
[tex]S_p = \sqrt{\frac{\bar p (1 - \bar p)}{n} } \\\\= \sqrt{\frac{1.50\% (1 - 1.50\%)}{200} }[/tex]
= 0.008595057
Now the 3 sigma control limits is
UCL_p = [tex]\bar p[/tex] + 35p
= 0.015 + 3 (0.008595057 )
= 0.04078517
LCL_p = [tex]\bar p[/tex] - 35p
= 0.015 - 3 (0.008595057 )
= 0
hence, the 3 sigma control limits are UCL 0.04078517 and LCL 0 respectively