Question

If the number of observations for each sample is 150 units, what is the 3-sigma upper control limit of the process

Answer 1

Complete Question

Complete Question is attached below

Answer:

$UCL= 0.25$

Step-by-step explanation:

From the question we are told that:

Sample size$n=150$

Sample Variants $s=7$

Sigma control limits  $Z = 3$

Therefore

Total number of observations is Given as

$T_o=n*s$

$T_o=150 *7$

$T_0=1050$

Generally

Summation of defectivee

$\sum np=23+34+15+30+25+22+18$

$\sum np= 167$

Generally the equation for P-bar is mathematically given by

$P-bar=\frac{\sum np}{T_o}$

$P-bar=\frac{167}{1050}$

$P-bar=0.16$

Therefore

$Sp=\sqrt{\frac{P-bar(1-P-bar)]}{ n}}$

$Sp=\sqrt{\frac{[0.159(1-0.159)]}{150}}$

$Sp=0.03$

Generally the equation for 3-sigma upper control limit of the process is mathematically given by

$UCL = P-bar + Z*Sp$

$UCL= 0.16 + 3*0.03$

$UCL= 0.25$