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)0.1500
D)0.1857

Question

Quizplus · Accepted Answer

The proportion of defective light bulbs can be calculated as follows:
p = 600/(100*40) = 0.15
The standard deviation of the proportion can be calculated as follows:
sigma = sqrt(p*(1-p)/n) = sqrt(0.15*0.85/100) = 0.03887
The 3-sigma lower control limit can be calculated as follows:
LCL = p - 3*sigma = 0.15 - 3*0.03887 = 0.0335 (rounded to four decimal places)
The closest option to 0.0335 is option B) 0.0429, which is the best choice.

A Company Randomly Selects 100 Light Bulbs Every Day for 40