A company specifies that the actual contents of packages shall not be less, on average, than the Nominal Quantity (Qn) and that not more than 2.5% of packages may be Non-Standard (NS), i.e. deviate from Qn by more than "total negative error" (TNE). Thus, if Qn=230g and TNE=9g a package is NS if it weighs less than 221g.
Assume a normal distribution and containers are filled independently.
(a) Let Qn=230g and TNE=9g and suppose the company sets the average fill to be u=230g.
(i) What must the standard deviation of the filling operation be so that only
2.5% of containers are NS?
(ii) Suppose the actual standard deviation is 6g, what minimum value should u be so that
the NS regulation is met? Round to whole units.
(b) Suppose u is the value found in (a)(ii) and tha five containers are selected at random
(i) What is the probability that two or more of them are NS?
(ii) What is the probability that the combined weight of the five congtainers is less than
(c) What is the probability that an inspector will need to weigh exactly 6 containers
from the production line to discover a NS container. That is, the first five are standard
and the sixth is NS.
The Attempt at a Solution
I don't have my actual workings with me, but here are my methods.
(a) (i) 2.5% is NS, so I found the z-value, Z, which gives 0.025.
Z = (221-230)/SD
From this I found the standard deviation.
(ii) Z = (221 - Umin)/6
Solved for Umin
(b) (i) P(2 or more are NS) = 1 - P(1 or 0 is not NS)
How do I calculate this probabilty?
(ii) I have no idea here
(c) Do I have to do a binomial distribution?