Complete Question
Complete Question is attached below
Answer:
[tex]UCL= 0.25[/tex]
Step-by-step explanation:
From the question we are told that:
Sample size[tex]n=150[/tex]
Sample Variants [tex]s=7[/tex]
Sigma control limits [tex]Z = 3[/tex]
Therefore
Total number of observations is Given as
[tex]T_o=n*s[/tex]
[tex]T_o=150 *7[/tex]
[tex]T_0=1050[/tex]
Generally
Summation of defectivee
[tex]\sum np=23+34+15+30+25+22+18[/tex]
[tex]\sum np= 167[/tex]
Generally the equation for P-bar is mathematically given by
[tex]P-bar=\frac{\sum np}{T_o}[/tex]
[tex]P-bar=\frac{167}{1050}[/tex]
[tex]P-bar=0.16[/tex]
Therefore
[tex]Sp=\sqrt{\frac{P-bar(1-P-bar)]}{ n}}[/tex]
[tex]Sp=\sqrt{\frac{[0.159(1-0.159)]}{150}}[/tex]
[tex]Sp=0.03[/tex]
Generally the equation for 3-sigma upper control limit of the process is mathematically given by
[tex]UCL = P-bar + Z*Sp[/tex]
[tex]UCL= 0.16 + 3*0.03[/tex]
[tex]UCL= 0.25[/tex]
