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3 years ago

13

A company randomly selects 100 light bulbs every day for 40 days from its production process. If 600 defective light bulbs are f

ound in the sampled bulbs then the 3-sigma lower control limit would be

2 answers:

wel3 years ago

4 0

Answer:

3-sigma lower control limit=0.0429

Explanation:

Process average, p = 600 / (100*40)

=600/400

=0.15

Lower Control Limit = p- 3 *sqrt (p*(1-p)/n)

= 0.15 - 3* (0.15*0.85/100)^0.5

=0.15-3*(0.1275/100)^0.5

= 0.0429

Therefore, the 3-sigma lower control =0.0429

lutik1710 [3]3 years ago

3 0

Answer:

3 sigma lower control limit = 0.0429

Explanation:

Given.

n = 100

days = 100

Number of defective bulbs = 600 defective bulbs

Let p = Process Average

p = 600/(100*40)

P = 600/4000

P = 0.15

q = 1 - p

q = 1 - 0.15

q = 0.85

3 sigma lower limit = p - 3*√(pq/n)

Using the above formula

Substitute in the values

3 sigma lower control limit = 0.15 - 3 * √(0.15 * 0.85/100)

3 sigma lower control limit= 0.15 - 3√0.001275

3 sigma lower control limit = 0.15 - 3* 0.035707142142714

3 sigma lower control limit = 0.15 - 0.107121426428142

3 sigma lower control limit = 0.04287857357185

3 sigma lower control limit = 0.0429 ---- approximated

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