Question

Sample No. 1 2 3 4 5 6 No. Defectives 7 5 20 10 12 7 13 10 5 12 Sample No. 11 12 13 14 15 16 17 18 19 20 No. Defectives 6 6 15 4 12 7 12 3 19 16 Sample No. 21 22 23 24 25 26 27 28 29 30 No. Defectives 17 13 5 7 14 9 13 6 13 3 7 8 9 10 a) Establish 3a upper and lower controllimits. UCL - Center your response as a number between 0 and 1, rounded to throw decimal places). LOL, (enter your response as a number between 0 and 1, rounded to three decimal places).

Answer 1

Solution:

a:

$\bar{c} = \frac{Total\ Defects}{total\ Samples} = \frac{301}{30}$

$\bar{c} = 10.033$

Now, the sigma will be,

30 = 3Vc = 3V10.033

$3\sigma = 9.502$

Hence, the control limits are,

$UCL = \bar{c} + 3\sigma$

$LCL = \bar{c} - 3\sigma$

b: Option A
Since the percent defective cannot be a negative number hence the lower control limits that are negative are considered as zero.

c: Not as strict as
Because the control limit is lower than the one at Birmingham bank, hence, there is more scope within the industry as compared to Birmingham bank for variation.