Show solutions Fill out the chart below.  It is information needed to construct an OC curve and AOQ curve for the following sampling plan:  N=1900n=125   c=2 DO NOT DRAW THE CURVES. P 100p...

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  1. Fill out the chart below.  It is information needed to construct an OC curve and AOQ curve for the following sampling plan:  N=1900n=125   c=2

DO NOT DRAW THE CURVES.

P

100p

N

np

Pa

AOQ

0.0016

 

125

 

 

 

0.008

 

125

 

 

 

0.012

 

125

 

 

 

 

 

125

1.8

 

 

0.0224

 

125

 

 

 

 

 

125

4.2

 

 

0.0416

 

125

 

 

 

 

 

125

6.0

 

 

0.0512

 

125

 

 

 

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Expert Answers
gsarora17 eNotes educator| Certified Educator

Given: 

Size of lot (N)=1900

Sample Size (n) =125

Acceptance Number (c)=2

Proportion defective (p) can be calculated as :

For np=1.8,

 `p=1.8/125=0.0144`

For np=4.2,

 `p=4.2/125=0.0336`

For np=6.0,

 `p=6.0/125=0.048`

So these values of p completes the Column 1 for p.

Now column 2 can be obtained by multiplying p values with 100.

Now let's fill the column 3 (np) values,

For p=0.0016,

 `np=125*0.0016=0.2`

For p=0.008,

`np=125*0.008=1`

 For p=0.012,

`np=125*0.012=1.5`

For p=0.0224,

`np=125*0.0224=2.8`

For p=0.0416,

`np=125*0.0416=5.2`

For p=0.0512,

`np=125*0.0512=6.4`

Now let's calculate the Probability of acceptance by the cumulative Poisson formula,

`P_a=sum_(x=0)^c(e^(-np)(np)^x)/(x!)`

We have to calculate P_a for each value of p,

For p=0.016

`P_a=P_0+P_1+P_2`

`=(e^(-0.2)(0.2)^0)/(0!)+(e^(-0.2)(0.2)^1)/(1!)+(e^(-0.2)(0.2)^2)/(2!)`

`=0.818730753+0.163746151+0.016374615`

`=0.998851519`

`=~~0.999`

For p=0.008,

`P_a=(e^(-1)(1)^0)/(0!)+(e^(-1)(1)^1)/(1!)+(e^(-1)(1)^2)/(2!)`

`=0.367879441+0.367879441+0.183939721` 

`=0.919698603`

`=~~0.92` 

For p=0.012,

`P_a=(e^(-1.5)(1.5)^0)/(0!)+(e^(-1.5)(1.5)^1)/(1!)+(e^(-1.5)(1.5)^2)/(2!)`

`=0.22313016+0.33469524+0.25102143`

`=0.808846831`

`=~~0.809`

Similarly we can calculate P_a for all values of p, which are shown in the attached image.

Now let' calculate AOQ,

`AOQ=(P_a*p)(N-n)/N`

If `n/N<=0.1`

Then, AOQ `=P_a*p`

Now let's calculate AOQ,

For p=0.0016,

`AOQ=0.0016*0.999=0.0015984`

For p=0.008,

`AOQ=0.008*0.92=0.00736`

Similarly we can calculate AOQ, for other values of p.

Please refer to the attached image, where all calculations are shown.

` `

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