Question

Omega Electronics (Jaipur, India) mass produces monitors that are used for medical imaging technology. To monitor the quality of the production process, the company draws samples of 200 tubes each, inspects them, and then classifies them into good or bad units. The inspection data for 15 days of production shown that the average fraction of defective units per day is 0.075. Construct a p-chart with 3$\sigma$ control limits, what is the LCLp?

A) zero

B) between 0 and 0.01

C) between 0.02 and 0.03

D) between 0.01 and 0.02

Answer 1

Answer: D) between 0.01 and 0.02

Explanation:

steps and formulas		Sample size 200
Probability of defects=P=	total defective/total observations	0.0750
Q=	1-P	0.9250
N=	average sample size	200
LCL=	P - 3*Squareroot(P*Q/N)	0.0191
Standard deviation	Squareroot(P*Q/N)	0.0186