*Logs-R-Us Lumber* cuts Douglas fir wooden dowels automatically into nominal lengths of 12 inches. The actual lengths are normally distributed with *μ* = 12.000 inches and *σ* = 0.066 inches.

- If a customer's specifications for the dowels are 12.050 ± 0.150 inches, what percentage
of the dowels will
__not meet__their requirements? - Determine the Capability Index, Cpk for the dowel process and comment
on its capability.

- Assuming it is possible to centre the dowel process on the customer's target requirements, determine the Capability Index, Cp for the process and comment on its capability.
- What standard deviation would
*Logs-R-Us Lumber*need to attain for their process if they wanted to achieve a Cp of 1.5?

Solution:

(a) The production process produces dowels with normally distributed lengths as follows:

The customer will only accept dowels with lengths that are between their LSL, (lower specification limit), of 11.900” and their USL, (upper specification limit), of 12.200” as indicated by the dotted vertical lines. To find the percentage of the dowels that will not meet these requirements, we need to determine the area in the tails of the normal distribution below the LSL and above the USL. |

We begin by finding the corresponding z-values for the LSL and the USL…

For the LSL = 11.900;

From a Normal-Curve Area Table; Area1 = 0.5000 - 0.4357 =0.0643

For the USL = 12.200;

From a Normal-Curve Area Table; Area2 = 0.5000 - 0.4988 = 0.0012

Therefore the total area outside specifications = 0.0643 + 0.0012 = 0.0655

And so the percentage of dowels that will not meet customer's requirements is about 6.6%

(b)

Therefore *Cpk* = min(0.505, 1.01) = 0.505

Since *Cpk* < 1, the process is not capable.

(c)

Since *Cp* < 1, the process is not capable.

(d)

And,

⇒ So *Logs-R-Us Lumber* would need to attain a process standard deviation of at least 0.0333"

to achieve a *Cp* of 1.5.