A process using a p-chart has pbar = 0.076 and nbar = 4.86. Find the control limits.

For a p-chart the 3-sigma control limits are UCL = p + 3√[p(1–p)/n] LCL = p – 3√[p(1–p)/n] (set LCL = 0 if negative). Using p̄ = 0.076 and n̄ = 486, σp = √[0.076(1–0.076)/486] = √0.000144 ≈ 0.0120 3σp = 0.0360 UCL = 0.076 + 0.036 ≈ 0.112 LCL = 0.076 – 0.036 ≈ 0.040 Thus the limits are 0.040 and 0.112, answer C.

A. Gives limits within ±1σ, not ±3σ; mis-computes standard error.
B. Uses an even smaller span (~±0.004), far less than 3σ; incorrect formula.
D. Essentially equals p̄, ignoring variability; no control limits.
E. One of the listed choices (C) is correct, so “none” is wrong.

1. Montgomery, D. C. “Introduction to Statistical Quality Control,” 8th ed., Wiley, 2020, Sec. 7.2.2, Eq. (7.4), p. 296.

2. ASQ/ANSI Z1.9-2013, “Sampling Procedures and Tables,” Appendix B, p. 37 – p-chart control limit equations.

3. MIT OpenCourseWare 15.764 “Manufacturing Systems,” Lecture 8 Notes, p. 5: derivation of p-chart limits.