Example 12.7

EXAMPLE 12.7 Percent Meeting Specifications

lab

CASE 12.1 Figure 12.13 shows that the distribution of individual turnaround times is approximately Normal. We found in Example 12.2 (pages 605–607) that $\bar{\bar{x}} = 40.62$ . We now need an estimate of the standard deviation $σ$ for the process producing the individual measurements.

If we use $\bar{R}$ as the basis for the standard deviation estimate, then the estimate is given by

$\hat{σ} = \frac{\bar{R}}{d_{2}}$

Table 12.1 (page 603) gives values for $d_{2}$ for sample sizes ranging from 2 to 25. From Example 12.2, we find $\bar{R} = 21.967$ . From Table 12.1, we find $d_{2} = 2.326$ for $n = 5$ , which gives the estimate for $σ$

$\hat{σ} = \frac{21.967}{2.326} = 9.444$

We can now calculate the percent of lab tests that meet the upper specification:

$\begin{array}{l} P (lab time \leq 50) & = & P (Z \leq \frac{50 - 40.62}{9.444}) \\ = & P (Z \leq 0.993) \approx 0.8389 \end{array}$

It is estimated that about 84% of lab tests meet the ER specification of 50 minutes or less turnaround time.