A process shows the following number of defectives. Each sample size for this process is 85.3 8 2 7 7 6 8 8 9 5 Find the control limits.

The data describe an np-chart: ten samples of size n = 85 with counts of defectives 3, 8, 2, 7, 7, 6, 8, 8, 9, 5. Total defectives = 63 ⇒ . p-bar = 63 / (10 × 85) = 0.0741 np-bar = n p-bar = 85 × 0.0741 ≈ 6.3 σnp = √[ n p-bar (1–p-bar) ] = √[85 × 0.0741 × 0.9259 ] ≈ 2.41 Control limits: UCL = np-bar + 3σ ≈ 6.3 + 3(2.41) = 13.5 LCL = np-bar – 3σ ≈ 6.3 – 7.2 < 0, therefore LCL = 0 (“none”). Thus the limits are 0 and 13.5, corresponding to choice A.

B: Gives 12.6/25.2—calculated from incorrect mean and σ; does not match n = 85 or observed data. C: Upper limit 25.2 based on erroneous σ; lower limit should be 0 if that σ were right, but resulting UCL still mis-computed. D: One option (A) exactly matches correct limits, so “none of the above” is false.

1. Montgomery, D.C. Introduction to Statistical Quality Control, 9th ed., Wiley, 2020, Section 6.2 “np Control Charts”, eq. (6.7), pp. 275-276.

2. Vining, G. & Kowalski, S. Statistical Process Monitoring, Chapman & Hall, 2011, Sec. 3.2, pp. 57-58 (formulae for p and np charts).

3. MIT OpenCourseWare, Course ESD.86J “Statistics for Manufacturing”, Lecture 13 slides, pp. 3-4: derivation of np-chart limits.