Homework Assignment 6
Problem 3-5.
Solution:
If the producer provides quality steel rods, the percent of defectives equals to
0.1.
Then for this sampling plan:
N=20, n=5, c=1, p=0.1
Let's calculate the probability of acceptance as following:
The probability or rejection is:
Pj = 1 - Pa = 0.08155 = 8.155%
So, even though the producer provides quality steel rods, the probability that his
products are subject to rejection is 8.155%.
Problem 3-6.
Solution:
(1) For this situation:
N=5,000; n=50; c=1; p=0.005;
Let's calculate the probability of acceptance as following:
The probability or rejection is:
Pj = 1 - Pa = 0.0261 = 2.61%
So, even though the producer provides quality steel rods, the probability that
his
products are subject to rejection is 2.61%.
(2) For this situation:
N=5,000; n=50; c=1; p=0.04;
Let's calculate the probability of acceptance as following:
The probability or acceptance is 40.05%.
So, even though the producer provides unqualified steel rods, the probability
that the
customer still accept his is 40.05%.
Problem 3-7.
Solution:
For plan 1:
For plan 2:
For plan 3:
The three OC curves is as following:
According to the OC curves, we can see that the three different sampling plans
provide different protection to the customers and producers. For the customers, the
third plan with N=300, n=30, c=3 provides the best protection. But this plan
provides the worst protection to the producers. The first plan with N=100, n=10, c=3
provides the best production to the producers, but it provides the worst plan to the
customers. The second plan most closely approximates the ideal "square shoulder" OC
curve.
Problem 3-8.
Solution:
(1)
Consider p=0.022
For scheme 1:
For scheme 2:
For scheme 3:
According to the calculation, we can see that probability of acceptance of each of
the three scheme are bigger than 0.1, so all of them can not meet the requirement of
the customer.
We can find another plan which is N=1000, n=180, c=1. The probability of this plan
is calculated as following:
This plan can meet the customer's requirements.
(2)
For scheme 1:
When p=0.01,
When p=0.005,
For scheme 2:
When p=0.01,
When p=0.005,
For scheme 3:
When p=0.01,
When p=0.005,
We can see that the third scheme provides the best protection of the manufacturer.
(3)
No. Since both n and c affected Pa significantly, the acceptance plans with the
acceptance number equal to zero does not necessarily provide the best protection. In
(1), The Pa for Scheme 1 with Ac=0 is 0.111 while the Pa for Scheme3 with Ac=2 is
0.1029. However, Scheme 3 provides better protection.