Question

An inspection of 10 samples of size 400 each from 10 lots revealed the following no.
of defective units: 17, 15, 14, 26, 9, 4, 19, 12, 9 and 15. Calculate control limits for
the no. of defective units and state your conclusion.​

Answer 1

The control limit for the number of defective units is 25.03 and 2.97 when number of defective units were discovered of 10 samples.

Given that,

The following number of defective units were discovered during an examination of 10 samples, each of size 400, from 10 lots: 17, 15, 14, 26, 9, 4, 19, 12, 9 and 15.

We have to determine the control limits for the number of defective pieces and present your findings.​

We know that,

P = 17, 15, 14, 26, 9, 4, 19, 12, 9 and 15.

k = 10 and n=400

$\bar{P}$ = $\frac{\frac{17+15+14+26++4+19+12+9+15}{400} }{10}$

$\bar{P}$ = $\frac{\frac{140}{400} }{10}$

$\bar{P}$ = $\frac{0.35 }{10}$

$\bar{P}$ = 0.035

$1-\bar{P}$ = 1-0.035 = 0.965

n$\bar{P}$ = 400×0.035 = 14 (central line)

Upper control limit = n$\bar{P}$ + 3$\sqrt{n\bar{P}(1-\bar{P})}$

= 14 +  3$\sqrt{400\times 0.035\times 0.965}$

= 14 + 3 × 3.6756

= 14 + 11.0268

= 25.0268

= 25.03

Lower control limit = n$\bar{P}$ - 3$\sqrt{n\bar{P}(1-\bar{P})}$

= 14 -  3$\sqrt{400\times 0.035\times 0.965}$

= 14 - 3 × 3.6756

= 14 - 11.0268

= 2.97

Therefore, The control limit for the number of defective units is 25.03 and 2.97.

To know more, visit:

https://brainly.in/question/30395598

https://brainly.in/question/1942704

#SPJ1