Question 1 of 7: Process Capability and Average-and-Range Control Limits

Question 1: Process Capability and Average-and-Range Control Limits (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A lot of 750 inspected parts whose measured characteristic is normally distributed (part a), and the summary statistics of a variables control-chart study on the same kind of process (part b):

Find. (a) the number of the 750 parts expected to lie inside the three-sigma band of the population, and (b) the upper and lower control limits of the average chart together with the upper control limit of the range chart.

Approach. Part (a) is a direct evaluation of the normal integral between ±3 standard deviations, applied to a finite lot as an expected count; part (b) applies the standard Shewhart factors for n = 4, which convert an average range into three-sigma limits without ever computing a standard deviation explicitly.

1. Part (a) — evaluate the three-sigma area of the normal distribution. For a normal population the proportion falling within k standard deviations of the mean is $$P(\mu-k\sigma \le x \le \mu+k\sigma) \;=\; \operatorname{erf}\!\left(\frac{k}{\sqrt{2}}\right)$$ which at $k = 3$ gives $\operatorname{erf}(2.1213) = 0.997300$, the familiar 99.73 per cent of the empirical rule. The complement is split evenly between the two tails, so each tail carries $(1-0.997300)/2 = 0.001350$, or about 1350 parts per million.
2. Convert the proportion to an expected count of parts. Multiplying the population proportion by the lot size, $$N_{\text{in}} \;=\; N\,P \;=\; 750 \times 0.997300 \;=\; 747.98$$ $$\boxed{N_{\text{in}} \approx 748 \text{ parts within } \pm 3\sigma}$$ The balance, $750 - 747.98 = 2.02$, is the expected number falling outside the band — about two parts, roughly one beyond each tail. Because a count must be an integer, 748 is the expected value rather than a guaranteed result; a second lot of 750 might show one outlier or four.
3. Part (b) — convert the average range into a three-sigma allowance for the subgroup mean. The average chart is centred on the grand average and its limits are placed three standard errors away. Shewhart's factor $A_2$ bundles the conversion from range to standard deviation and the division by $\sqrt{n}$ into a single multiplier: $$A_2\bar{R} \;=\; 0.729 \times 0.008 \;=\; 0.005832 \text{ in}$$
4. Place the control limits of the average chart. Adding and subtracting that allowance from the grand average, $$\mathrm{UCL}_{\bar{x}} = \bar{\bar{x}} + A_2\bar{R} = 8.502 + 0.005832 = 8.507832 \text{ in}$$ $$\mathrm{LCL}_{\bar{x}} = \bar{\bar{x}} - A_2\bar{R} = 8.502 - 0.005832 = 8.496168 \text{ in}$$ $$\boxed{\mathrm{UCL}_{\bar{x}} = 8.5078 \text{ in},\qquad \mathrm{LCL}_{\bar{x}} = 8.4962 \text{ in}}$$ The band is only 0.0117 in wide because it governs the average of four pieces, not an individual piece.
5. Place the limits of the range chart. The range chart is centred on $\bar{R}$ and its limits come from the same table: $$\mathrm{UCL}_R = D_4\bar{R} = 2.282 \times 0.008 = 0.018256 \text{ in}$$ $$\mathrm{LCL}_R = D_3\bar{R} = 0 \times 0.008 = 0$$ $$\boxed{\mathrm{UCL}_R = 0.01826 \text{ in},\qquad \mathrm{LCL}_R = 0}$$ For any subgroup of six or fewer pieces $D_3$ is zero, so the range chart has no usable lower limit: the distribution of the range is too skewed at small n for a three-sigma lower bound to fall above zero.
6. Cross-check the limits against an estimate of the process standard deviation. The average range estimates the process spread through $\hat{\sigma} = \bar{R}/d_2$, with $d_2 = 2.059$ at $n = 4$: $$\hat{\sigma} = \frac{0.008}{2.059} = 0.003885 \text{ in}$$ The standard error of a mean of four is then $\hat{\sigma}/\sqrt{4} = 0.001943$ in, and three of those is 0.005828 in — the same 0.00583 in obtained from $A_2\bar{R}$, which confirms the factor was applied correctly. The natural tolerance of the process, $6\hat{\sigma} = 0.0233$ in, is the figure to compare against the drawing tolerance when judging capability.