SOLUTION: Consider the situation of Review Exercise 5.95.
It has been determined that the sampling plan should
be extensive enough that there is a high probability,
say 0.9, that if as ma
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-> SOLUTION: Consider the situation of Review Exercise 5.95.
It has been determined that the sampling plan should
be extensive enough that there is a high probability,
say 0.9, that if as ma
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Question 1187787: Consider the situation of Review Exercise 5.95.
It has been determined that the sampling plan should
be extensive enough that there is a high probability,
say 0.9, that if as many as 2 defectives exist in the lot
of 50 being sampled, at least 1 will be found in the
sampling. With these restrictions, how many of the 50
items should be sampled?
Review Exercise 5.95.
A production process outputs items in lots of 50.
Sampling plans exist in which lots are pulled aside periodically and exposed to a certain type of inspection.
It is usually assumed that the proportion defective is
very small. It is important to the company that lots
containing defectives be a rare event. The current inspection plan is to periodically sample randomly 10 out
of the 50 items in a lot and, if none are defective,
perform no intervention.
(a) Suppose in a lot chosen at random, 2 out of 50 are
defective. What is the probability that at least 1
in the sample of 10 from the lot is defective?
(b) From your answer to part (a), comment on the
quality of this sampling plan.
(c) What is the mean number of defects found out of
10 items sampled?
I tried:
defective probability=0.2 (as many as 2 defective=>so I thought 1 and 2 defective)
no defective=0.1
p(defective)=2/50
p(no defective)=48/50
p(at least 1 defective)=1-p(no defective)=1-(48C10)/(50C10)=1-0.6367=0.3633
and here I get confused.
P(X<=2)=P(2)+P(1)+P(0)=++=
or is 0.9 supposed to be 2/50 and 0.1 supposed to be 48/50 ???
Thank you for your time :) Answer by ikleyn(52802) (Show Source):
