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
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
No. of 100 W bulbs, n = 28,539 bulbs
Given:
Power of a single bulb = 100 W
Time period, T = 1 yr = = 31,536,000 s
mass of matter, m = 1 g =
Solution:
According to Eintein's mass-energy equivalence:
E =
Power of a single bulb, P =
P =
P = 28,53,881 W
No. of 100 W bulbs, n =
n =
n = 28,539
a heterogeneous mixture