Question 10.4: A process has mean μ = 3 and standard deviation σ = 1, Sampl...

We first compute the probability p that a given point plots outside the control limits. Then ARL = 1 / p. The control limits are plotted on the basis of a process that is in control. Therefore they are at $\mu \pm 3 \sigma_{\bar{X}}$, where μ = 3 and $\sigma_{\bar{X}}=\sigma / \sqrt{n}=1 / \sqrt{4}=0.5$. The lower control limit is thus at I .5, and the upper control limit is at4.5. If $\bar{X}$ is the mean of a sample taken after the process mean has shifted, then $\bar{X}\sim N\left(3.5,0.5^2\right)$. The probability that $\bar{X}$ plots outside the control limits is equal to $P(\bar{X}<1.5)+P(\bar{X}>4.5)$. This probability is 0.0228 (see Figure 10.6). The ARL is therefore equal to 110.0228 = 43.9. We will have to observe about 44 samples, on the average, before detecting this shift.