What percentage of output would be expected to be out of tolerance if the process were centered?

In case of ahighspecification limit,

if the process is focused, it will be calculated as follows

Z value x sample standard deviation + revised process average = 2.4

Here, Z =?

Sample standard deviation = 0.2096

The revised process mean = 1.9

Therefore, Z x 0.2096 + 1.9 = 2.4 ……. (1)

0.2096Z = 2.4 - 1.9

0.2096Z = 0.5 …………..… (2)

Z = 0.5 / 0.2096

Z = 2.385.

The probability of Z being 2.385 is 0.9914 from the standard distribution table.

As a result, the fraction of output greater than 2.4 mm is 1 - 0.9914 = 0.0086.

In the case of the lower specification limit, the possibility that the thickness will be maximal up to 1.4 mm if the process is centered is as follows:

Z value x sample standard deviation + revised process average = 1.4.

Here, Z =?

Sample standard deviation = 0.209.

The revised process mean = 1.9.

= Z x 0.2096 + 1.9 = 1.4 ……. (1)

0.2096Z = 1.4 - 1.9

0.2096Z = - 0.5……………… (2)

Z = 0.5 / 0.2096

Z = - 2.385

As a result, the fraction of output greater than 1.4 mm is 0.0086.

The fraction of output is expected to be out of tolerance.

As a result, the fraction of output expected to be more than 1.4 mm = 0.0086.

And the fraction of output expected to be higher than 2.4 mm = 0.0086.

Therefore,

The fraction of output expected to be more than 1.4 mm + the fraction of output expected to be higher than 2.4 mm

= 0.0086 + 0.0086

= 0.0172

= 1.72%.

The fraction of output expected to be out of tolerance is 1.72%.