A powder metal manufacturing company is producing sleeves for a locking mechanism.The target (nominal)value for the inside diameter is 1 inch.The inside diameter specifications are 1 ± .005 inches.Assume that the process is in statistical control with $\overline { \overline { X } }$

= 1)0002 inches, $\overline { \mathrm { R } }$
= )003 inches,and subgroup size of 5.What is the sigma level capability for this process?

Question

A powder metal manufacturing company is producing sleeves for a locking mechanism.The target (nominal) value for the inside diameter is 1 inch.The inside diameter specifications are 1 ± .005 inches.Assume that the process is in statistical control with

\overline { \overline { X } }

= 1) 0002 inches,

\overline { \mathrm { R } }

= ) 003 inches,and subgroup size of 5.What is the sigma level capability for this process?

= 1)0002 inches, $\overline { \mathrm { R } }$
= )003 inches,and subgroup size of 5.What is the sigma level capability for this process?

Quizplus · Accepted Answer

The formula to calculate sigma level capability is:

Sigma Level = (USL - LSL) / 6 * Standard Deviation

where USL is the upper specification limit, LSL is the lower specification limit, and Standard Deviation is calculated as $\overline { \mathrm { R } }$
/d2

d2 is a constant. For subgroup size 5, d2 value is 2.326.

USL = 1 + 0.005 = 1.005
LSL = 1 - 0.005 = 0.995
Standard Deviation = 0.003 / 2.326 = 0.00129

Substituting these values in the formula:

Sigma Level = (1.005 - 0.995) / 6 * 0.00129 = 3.72

Therefore, the sigma level capability for this process is 3.72, which is closest to answer choice B.

A Powder Metal Manufacturing Company Is Producing Sleeves for a Locking