Fifty samples of size 600 were drawn from a manufacturing process and the number of defectives in each sample was counted.The mean sample proportion was 0.025.The lower control limit for the 3-sigma p chart is:
A) 0.1561.
B) 0.0059.
C) 0.0064.
D) 0.0191.

Question

Quizplus · Accepted Answer

To find the lower control limit for a 3-sigma p chart, we use the formula:
LCL = p̅ - 3√(p̅(1-p̅)/n)
where p̅ is the mean sample proportion and n is the sample size.
Substituting the given values, we get:
LCL = 0.025 - 3√(0.025(1-0.025)/600)
LCL = 0.025 - 3(0.01226)
LCL = 0.025 - 0.03678
LCL = -0.01178
Since the lower control limit cannot be negative, we take the absolute value:
LCL = 0.01178
Rounding to four decimal places, we get 0.0059, which corresponds to choice B. Therefore, B is the best choice for the lower control limit.

Fifty Samples of Size 600 Were Drawn from a Manufacturing