OC Curve and Poisson Distribution Question. Where is the calculation for Pa2 com

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OC Curve and Poisson Distribution Question. Where is the calculation for Pa2 coming from? I don't understand how the sample generated P(x1=2) = 0.216 ; and P (x1=3) = 0.087 ; and P(x=4) = 0.026.

The OC Curve The OC curve also measures the performance of double sampling plans. Calculations for double sampling plans are more involved than for single sampling plans. Let Pai denote the probability of accepting the lot on the first sample and Pa2 the probability of lot acceptance on the second sample. The combined probability of acceptance is given by Pa-Pal +Pa. Let xi and x2 denote the observed number of nonconforming items on the first and second samples, respectively. We have (10-11) In the following example we calculate the probability of lot acceptance for a given proportion nonconforming. The Poisson approximation to the hypergeometric distribution is used here. This procedure can be used to construct the OC curve. Example 10-10 Let's consider a double sampling plan of lot size 3000 given by the following parameters: n,-40. c,-I, r,-5, n,-80. c2=5, r2-6. For a lot proportion nonconforming value of p-0.03, find the probability of accepting such lots. Solution To find the probability of acceptance on the first sample (Pa), we have nip= (40)(0.03)= 1.2. The Poisson cumulative probability distribution tables in Appendix A-2 give Pal = P(Xi 1)= 0.663 We calculate P2 as follows: Note that when calculating P(x2 S 3) and other probabilities dealing with x2, the value of n2P_ (80)(0.03)= 2.4. So Pa2= (0.216)(0.779) + (0.087)(0.570) + (0.026)(0.308) 0.2259 The combined probability of acceptance is Pa-Pal + Pa2=0.663 + 0.2259-0.8889

Explanation / Answer

first of all i think one more term should be added in RHS of calculation of Pa2 as mentioned above.

that is P(x1=5).P(x2=0)

where x1 follows Poisson(1.2)and x2 follows Poisson (2.4)

Complete (A-Z) explanation of the calculation of Pa2

Pa2 : we calculate Pa2 only when first sample doesnot give the conclusion for either selecting or rejecting the lot. this happens only when no. of non conforming units in sample (x1) lie somewhere within the lower specification limit (c1)and upper specification limit (c2).

so if x1 is less than or equal to 1 (=c1)(accept the lot) or more than 5 (=c2)(reject the lot), we do not need to take sample 2 anymore.

with this, now lets talk about the calculation for Pa2

here Pa2 will be calculated only if 2nd sample is taken which is possible only in the following disjoint possible cases

case1. if x1=2, i.e non conforming units in the first sample is 2: AND ALSO

x2 <3, ie non conforming units in second sample not be more than 3.

the require probability of this event is given by P(x1=2).P(x2<=3)

similarly other cases are

case2: if x1 comes out to be 3 AND x2 is less than or equal to 2

probability of this event being P(x1=3).P(x2<=2)

case3:

if x1 comes out to be 4 AND x2 is less than or equal to 1

probability of this event being P(x1=4).P(x2<=1)

case4:

if x1 comes out to be 5 AND x2 is 0

probability of this event being P(x1=5).P(x2=0)

Finally we add probabilities of these 4 cases in order to find Pa2. Thus

Pa2= P(x1=2).P(x2<=3)+P(x1=3).P(x2<=2)+P(x1=4).P(x2<=1)+P(x1=5).P(x2=0)

now explanation of Probability of P(x1=2) = 0.216 ; and P (x1=3) = 0.087 ; and P(x=4) = 0.026.

since x1 follows Poisson Distribution with paramater (n1p)=1.2

hence we use the probability mass function of the Poisson distribution to calculate P(x1= 2 or 3 or 4)

P(x1=a)= [(e)^-1.2]*[(1.2)^a]/a!

*note : above formula is nothing but the pdf of Poisson distribution with parameter 1.2

hence using this formula we get the value of x1= 2 or 3 or 4 by replacing 'a' in the formula by them.

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