p_bar = 0.025
Sample size, N = 200
Standard error, Sp = √[(p_bar*(1 - p_bar) / N] = √(0.025*(1 - 0.025) / 200) = 0.011
So,
Central line, CL = p_bar = 0.025
UCLp = p_bar + 3 * Sp = 0.025 + 3*0.011 =
0.058
LCLp = Max(0, p_bar - 3 * Sp) = Max(0, 0.025 - 3*0.011) =
0

The process is out of statistical control as sample 2, 3, 7, and 8 are out of UCL.
In a manufacturing operation, the percentage defective averages 2.5 percent and sample size in 200. Compute...
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unit will be examined for nonconformities at each time point. The U
chart is used when a number of units will be sampled at each time
point, and a per unit average number of nonconformities will be
obtained. U Chart Formulas Suppose we have k samples, each of size
ni. Let Di represent the total number of nonconformities in the i
th sample. Formulas for...
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