If $\overline { \overline { X } }$

= 16.1, $\bar { R }$
= )03,and n = 6,and if the specifications are [15.9,16.3],is the process capable?
A)Yes
B)No
$$\mathrm { NTL } = 16.1 \pm 3 \left( \frac { 0.03 } { 2.534 } \right) = 16.065 \text { to } 16.135$$

Question

If $\overline { \overline { X } }$

= 16.1, $\bar { R }$
= )03,and n = 6,and if the specifications are [15.9,16.3],is the process capable?
A)Yes
B)No
$$\mathrm { NTL } = 16.1 \pm 3 \left( \frac { 0.03 } { 2.534 } \right) = 16.065 \text { to } 16.135$$

Quizplus · Accepted Answer

The process is capable because the entire NTL (New Tolerance Limits) range falls within the specification limits of [15.9, 16.3].

If X‾‾\overline { \overline { X } }X = 161 Rˉ\bar { R }Rˉ

If $\overline { \overline { X } }$ = 161 $\bar { R }$