Question

The thickness (in millimeters) of the coating applied to disk drives is a characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness \((x)\) has a normal distribution with a mean of \(3 \mathrm{~mm}\) and a standard deviation of \(0.05\) \(\mathrm{mm}\). Suppose that the process will be monitored by selecting a random sample of 16 drives from each shift's production and determining \(\bar{x}\), the mean coating thickness for the sample. a. Describe the sampling distribution of \(\bar{x}\) (for a sample of size 16 ). b. When no unusual circumstances are present, we expect \(\bar{x}\) to be within \(3 \sigma_{\bar{x}}\) of \(3 \mathrm{~mm}\), the desired value. An \(\bar{x}\) value farther from 3 than \(3 \sigma_{\bar{x}}\) is interpreted as an indication of a problem that needs attention. Compute \(3 \pm 3 \sigma_{\bar{x}}\). (A plot over time of \(\bar{x}\) values with horizontal lines drawn at the limits \(\mu \pm 3 \sigma_{\bar{x}}\) is called a process control chart.) c. Referring to Part (b), what is the probability that a sample mean will be outside \(3 \pm 3 \sigma_{\bar{x}}\) just by chance (i.e., when there are no unusual circumstances)? d. Suppose that a machine used to apply the coating is out of adjustment, resulting in a mean coating thickness of \(3.05 \mathrm{~mm}\). What is the probability that a problem will be detected when the next sample is taken? (Hint: This will occur if \(\bar{x}>3+3 \sigma_{\bar{x}}\) or \(\bar{x}<3-3 \sigma_{\bar{x}}\) when \(\mu=\) 3.05.) b. When no unusual circumstances are present, we expect \(\bar{x}\) to be within \(3 \sigma_{\bar{x}}\) of \(3 \mathrm{~mm}\), the desired value. An \(\bar{x}\) value farther from 3 than \(3 \sigma_{\bar{x}}\) is interpreted as an indication of a problem that needs attention. Compute \(3 \pm 3 \sigma_{\bar{x}}\). (A plot over time of \(\bar{x}\) values with horizontal lines drawn at the limits \(\mu \pm 3 \sigma_{\bar{x}}\) is called a process control chart.)

Short answer

The sampling distribution is \(N(3, 0.0125^2)\). The control chart limits are from 2.9625 mm to 3.0375 mm. The probability of a sample mean landing outside these limits due to random chance is 0.27 %. The probability of problem detection when the machine is out of adjustment needs to be calculated using CDF for Normal distribution.

Step-by-step solution

Describe the sampling distribution

On the condition of no unusual circumstances, the thickness of the coating \(x\) is a normal distribution \(N(3, 0.05^2)\), when the sample size \(n\) is 16, so the distribution of \(\bar{x}\) will also be a normal distribution. The expected value of \(\bar{x}\) will be the population mean \(\mu\) and the standard deviation will be \(\sigma_{\bar{x}} = \sigma / \sqrt{n} = 0.05 / \sqrt{16} = 0.0125 \mathrm{mm}\). Hence, the sampling distribution of \(\bar{x}\) is \(N(3, 0.0125^2)\).

Calculate control chart limits

The limits for the control chart are given by \(3 \pm 3 \sigma_{\bar{x}}\). Thus, we have \(3 \pm 3 * 0.0125 = 3 \pm 0.0375\). So, the control chart limits are FROM 2.9625 mm TO 3.0375 mm.

Find out the probability for a sample mean outside control limits

Since we have a normal distribution, the \(3 \pm 3 \sigma_{\bar{x}}\) range covers approximately 99.73% of the probability, thus the probability of a sample mean landing outside this range due to random chance is about \(1 - 0.9973 = 0.0027\) or 0.27%.

Calculate the probability of detecting a problem

If the machine is out of adjustment and the mean coating thickness becomes 3.05 mm, a problem will be detected when the sample mean is outside the previously calculated control limits. To calculate this probability we'll be using cumulative distribution function (CDF) for normal distribution. We need to find out the CDF values for 2.9625 and 3.0375 when \(N(3.05, 0.0125^2)\), calculate probabilities for the \(x\) being less or greater than these values, and finally add these probabilities together to get the final probability of detection.