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Worked Examples · Example 11
Q.

Five samples, each of size 5, were drawn from a production process. Their means and ranges were:

Sample12345
Mean Xˉ\bar{X}5052495153
Range RR46537

Using A2=0.577A_2=0.577, D3=0D_3=0, D4=2.115D_4=2.115 (standard factors for n=5n=5), compute the control limits for the Xˉ\bar{X} chart and the RR chart, and state whether the process is in control.

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Step 1 — Compute the grand mean Xˉˉ\bar{\bar{X}} (mean of the five sample means):

Xˉˉ=50+52+49+51+535=2555=51\bar{\bar{X}} = \dfrac{50+52+49+51+53}{5} = \dfrac{255}{5} = 51

Step 2 — Compute the average range Rˉ\bar{R} (mean of the five sample ranges):

Rˉ=4+6+5+3+75=255=5\bar{R} = \dfrac{4+6+5+3+7}{5} = \dfrac{25}{5} = 5

Step 3 — Xˉ\bar{X} chart control limits, using A2=0.577A_2 = 0.577 for n=5n=5:

UCL=Xˉˉ+A2Rˉ=51+(0.577)(5)=51+2.885=53.89UCL = \bar{\bar{X}} + A_2\bar{R} = 51 + (0.577)(5) = 51 + 2.885 = 53.89

LCL=Xˉˉ−A2Rˉ=51−2.885=48.12LCL = \bar{\bar{X}} - A_2\bar{R} = 51 - 2.885 = 48.12

Step 4 — RR chart control limits, using D4=2.115D_4=2.115, D3=0D_3=0:

UCL=D4Rˉ=(2.115)(5)=10.575≈10.58UCL = D_4\bar{R} = (2.115)(5) = 10.575 \approx 10.58

LCL=D3Rˉ=(0)(5)=0LCL = D_3\bar{R} = (0)(5) = 0

Step 5 — Check every sample point against both charts:

SampleXˉ\bar{X}Inside [48.12, 53.89]?RRInside [0, 10.58]?
150Yes4Yes
252Yes6Yes
349Yes5Yes
451Yes3Yes
553Yes7Yes

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