Question

Five data entry operators work at the data processing department of the Birmingham Bank. Each day for 30​ days, the number of defective records in a sample of 200 records typed by these operators has been​ noted, as​ follows: Sample No. No. Defectives Sample No. No. Defectives Sample No. No. Defectives 1 6 11 5 21 18 2 4 12 5 22 11 3 18 13 15 23 5 4 9 14 5 24 6 5 10 15 10 25 14 6 9 16 7 26 9 7 12 17 13 27 13 8 9 18 4 28 7 9 7 19 17 29 13 10 13 20 16 30 3 ​a) Establish 3sigma upper and lower control limits. UCL Subscript p ​= . 161 ​(enter your response as a number between 0 and​ 1, rounded to three decimal​ places).

Answer 1

Answer:

UCL = 0.406

LCL = 0.244

Step-by-step explanation:

Data is shown below:

Here ∑np = 293

∑n = 900

Calculating the average of the data:

$\bar{p}=\frac{\sum np}{\sum n}\\\bar{p}=\frac{293}{9000} = 0.325$

$\text{and,} \bar n=\frac{\sum n}{k}\\ \bar n=\frac{9000}{30} = 300$

Now, Calculating Upper and Lower Limit:

$UCL =\bar{p}+ 3\sqrt\frac{\bar p(1- \bar p)}{\bar n} \\UCL = 0.325 + 3\sqrt\frac{0.325(1- 0.325)}{300}\\\Rightarrow UCL = 0.406$

$\text{and,} LCL =\bar{p}- 3\sqrt\frac{\bar p(1- \bar p)}{\bar n} \\LCL = 0.325 - 3\sqrt\frac{0.325(1- 0.325)}{300}\\\Rightarrow LCL = 0.244$