For an [latex]\overline{X}[/latex] chart with control limits at [latex]\mu \pm 3 \sigma_{\bar{X}}[/latex], compute the ARL for a process that is in control.

Let [latex]\bar{X}[/latex] be the mean of a sample. Then [latex]\bar{X}\sim N\left(\mu, \sigma \frac{2}{X}\right)[/latex]. The probability that a point plots outside the control limits is equal to [latex]P\left(\bar{X}<\mu-3 \sigma_{\bar{X}}\right)+P\left(\bar{X}>\mu+3 \sigma_{\bar{X}}\right)[/latex]. This probability is equal to 0.00135 + 0.00135 = 0.0027 (see Figure 10.5). Therefore, on the average, 27 out of every 10,000 points will plot outside the control limits. This is equivalent to l every 10,000/ 27 = 370.4 points. The average run length is therefore equal to 370.4.

Question 10.3: For an X chart with control limits at μ±3σX, compute the ARL...

Principles of Statistics for Engineers and Scientists

For an $\overline{X}$ chart with control limits at $\mu \pm 3 \sigma_{\bar{X}}$ , compute the ARL for a process that is in control.

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Let $\bar{X}$ be the mean of a sample. Then $\bar{X}\sim N\left(\mu, \sigma \frac{2}{X}\right)$ . The probability that a point plots outside the control limits is equal to $P\left(\bar{X}<\mu-3 \sigma_{\bar{X}}\right)+P\left(\bar{X}>\mu+3 \sigma_{\bar{X}}\right)$ . This probability is equal to 0.00135 + 0.00135 = 0.0027 (see Figure 10.5). Therefore, on the average, 27 out of every 10,000 points will plot outside the control limits. This is equivalent to l every 10,000/ 27 = 370.4 points. The average run length is therefore equal to 370.4.