A company randomly selects 100 light bulbs every day for 40 days from its production process.If 600 defective light bulbs are found in the sampled bulbs then the 3-sigma lower control limit would be
A)0.0357
B)0.0429
C)0.15
D)0.1857

Question

Quizplus · Accepted Answer

Given, n=100 (sample size), N=4000 (total population size), d=600 (number of defective light bulbs in the sample)

The proportion of defective light bulbs in the sample is p = d/n = 0.06

The mean of the proportion of defective light bulbs in the population is = p'=600/4000=0.15

The standard deviation of the proportion p is given by:

σp=√[(p'(1-p'))/n]=√[(0.15*0.85)/100]=0.0387

The 3-sigma lower control limit for a process with a mean of 0.15 and standard deviation of 0.0387 is:

LCL = p'-3σp = 0.15-3(0.0387) = 0.0429

Therefore, the answer is B (0.0429).

A Company Randomly Selects 100 Light Bulbs Every Day for 40