Questions: The intensive care unit lab process has an average turnaround time of 26.2 minutes and a standard deviation of 1.5 minutes. Further, the nominal value from design specification for this service is 25 minutes +/- 5 minutes. What is the capability index of this production process? 0.84 1.48 1.38 0.94

The intensive care unit lab process has an average turnaround time of 26.2 minutes and a standard deviation of 1.5 minutes. Further, the nominal value from design specification for this service is 25 minutes +/- 5 minutes.

What is the capability index of this production process?
0.84
1.48
1.38
0.94

Transcript text: The intensive care unit lab process has an average turnaround time of 26.2 minutes and a standard deviation of 1.5 minutes. Further, the nominal value from design specification for this service is 25 minutes $+/-5$ minutes. What is the capability index of this production process? 0.84 1.48 1.38 0.94

Solution

Solution Steps

To find the capability index (Cp) of the production process, we use the formula:

\[ Cp = \frac{USL - LSL}{6 \sigma} \]

where:

USL (Upper Specification Limit) = 30 minutes (25 + 5)
LSL (Lower Specification Limit) = 20 minutes (25 - 5)
$\sigma$ (standard deviation) = 1.5 minutes

We will plug these values into the formula to calculate Cp.

Step 1: Define the Specification Limits and Standard Deviation

Given the nominal value of the service is 25 minutes with a tolerance of ±5 minutes, we can define the Upper Specification Limit (USL) and Lower Specification Limit (LSL) as follows: \[ USL = 25 + 5 = 30 \text{ minutes} \] \[ LSL = 25 - 5 = 20 \text{ minutes} \] The standard deviation ($\sigma$) is given as: \[ \sigma = 1.5 \text{ minutes} \]

Step 2: Calculate the Capability Index (Cp)

The capability index $Cp$ is calculated using the formula: \[ Cp = \frac{USL - LSL}{6 \sigma} \] Substituting the values we have: \[ Cp = \frac{30 - 20}{6 \times 1.5} = \frac{10}{9} \approx 1.1111 \]