Sampling with replacement in Matlab

[Dustinsfl]

I am trying to simulate the probability of rejecting a good batch for a probability of \(0.94\) using the Binomial Probability Law.

My two cases are
\[
P[k\geq 95] = \sum_{k = 95}^{100}\binom{100}{k}p^k(1 - p)^{100 - k}
\]
and
\[
P[k\geq 98] = \sum_{k = 98}^{100}\binom{100}{k}p^k(1 - p)^{100 - k}
\]

I am not sure with what the question wants me to do so I plotted the both curves from \(p = 0.9\) to \(p = 0.99\).

clear all
close all

p = (0.9:0.001:0.99);                 % probability range
P = zeros(length(p), 1);              % pre-allocating P
PP = zeros(length(p), 1);            % pre-allocating PP
T = zeros(length(p), 1);              % pre-allocating T
TT = zeros(length(p), 1);            % pre-allocating TT

% only allowing 5 defective
for i = 1:length(p)
    for n = 95:100
        T(n, 1) = nchoosek(100, n)*p(i).^n.*(1 - p(i)).^(100 - n);
        P(i, 1) = sum(T); 
    end
end

figure(1)
plot(p, P)
grid on

% only allowing 2 defective
for j = 1:length(p)
    for n = 98:100
        TT(n, 1) = nchoosek(100, n)*p(j).^n.*(1 - p(j)).^(100 - n);
        PP(j, 1) = sum(TT); 
    end
end

figure(2)
plot(p, PP)
grid on

 

[zzephod]

I am trying to simulate the probability of rejecting a good batch for a probability of \(0.94\) using the Binomial Probability Law.

My two cases are
\[
P[k\geq 95] = \sum_{k = 95}^{100}\binom{100}{k}p^k(1 - p)^{100 - k}
\]
and
\[
P[k\geq 98] = \sum_{k = 98}^{100}\binom{100}{k}p^k(1 - p)^{100 - k}
\]

I am not sure with what the question wants me to do so I plotted the both curves from \(p = 0.9\) to \(p = 0.99\).

clear all
close all

p = (0.9:0.001:0.99);                 % probability range
P = zeros(length(p), 1);              % pre-allocating P
PP = zeros(length(p), 1);            % pre-allocating PP
T = zeros(length(p), 1);              % pre-allocating T
TT = zeros(length(p), 1);            % pre-allocating TT

% only allowing 5 defective
for i = 1:length(p)
    for n = 95:100
        T(n, 1) = nchoosek(100, n)*p(i).^n.*(1 - p(i)).^(100 - n);
        P(i, 1) = sum(T); 
    end
end

figure(1)
plot(p, P)
grid on

% only allowing 2 defective
for j = 1:length(p)
    for n = 98:100
        TT(n, 1) = nchoosek(100, n)*p(j).^n.*(1 - p(j)).^(100 - n);
        PP(j, 1) = sum(TT); 
    end
end

figure(2)
plot(p, PP)
grid on

--> p=0.94;  %failure rate
--> N=10000; %number of simulated batches to test
--> ss=rand(N,100); %N samples of batches of size 100
--> sm=sum(ss'>p);  %number of failures in each batch
--> Pest=sum(sm>=5)/N %proportion of batches with 5 or more fails
Pest =
    0.7222
-->

 

[Dustinsfl]

I have two questions then.

1. Can you explain sum(ss'>5)? What does the ' do?
2. This tells us 72% of batches will have 5 or more defects. How does that relate to good batches that are rejected? From my code, we know that 45% of time we accept a a defective batch. So 65% we have good batches. Is the answer then 72 - 65 = 7% of good batches are rejected?

 

[zzephod]

I have two questions then.

1. Can you explain sum(ss'>5)? What does the ' do?
2. This tells us 72% of batches will have 5 or more defects. How does that relate to good batches that are rejected? From my code, we know that 45% of time we accept a a defective batch. So 65% we have good batches. Is the answer then 72 - 65 = 7% of good batches are rejected?

Your question is ambiguous, I read it as the population has 94% good and 6% bad, what is the probability of rejecting a batch of 100 when the rejection criterion is that it have 5 or more defective items.

For questions about what Matlab commands do use the help system of try them out at the console (with small arrays).

.

 

[Dustinsfl]

Your question is ambiguous, I read it as the population has 94% good and 6% bad, what is the probability of rejecting a batch of 100 when the rejection criterion is that it have 5 or more defective items.

For questions about what Matlab commands do use the help system of try them out at the console (with small arrays).

.

the question is what is the probability of rejecting a good batch with that criteria.

 

[zzephod]

the question is what is the probability of rejecting a good batch with that criteria.

What does "good batch" mean?

.

 

[Dustinsfl]

What does "good batch" mean?

.

A batch with 5 or less defects. I am trying to find the probability that we reject a good batch of thousand when we only test a 100.

 

[zzephod]

A batch with 5 or less defects. I am trying to find the probability that we reject a good batch of thousand when we only test a 100.

Please post the original question with the exact wording as asked.

.

 

[Dustinsfl]

Please post the original question with the exact wording as asked.

.

Use a computer to simulation to determine the probability of rejecting a good batch. To simplify your code, assume sampling with replacement. A good batch is defined as one with a probability of obtaining a good chip p=0.95. The two strategies are to accept the batch if 95 or more of the 100 samples are good and if 98 or more are good.

 

[zzephod]

Use a computer to simulation to determine the probability of rejecting a good batch. To simplify your code, assume sampling with replacement. A good batch is defined as one with a probability of obtaining a good chip p=0.95. The two strategies are to accept the batch if 95 or more of the 100 samples are good and if 98 or more are good.

I see no mention of a population or batch of size of thousand, where did that come from?

That is exactly what the original code I posted does (except I was using p=0.94 which is what you posted in your oringinal post).

Doing exactly the same but with p=0.95, and in SciLab rather than Matlab:

-->p=0.95;
-->N=10000;
-->ss=rand(N,100);
-->sm=sum(ss>p,2);size(sm)
 ans  =
    10000.    1.  
 
-->Pest=sum(sm>=5)/N
 Pest  =

    0.5642

I would strongly suggest you practice posting questions as they are asked, and include any side information in the section of the book/lecture pertaining to the question. Otherwise you are just wasting the time of anyone foolish enough to try helping you (but then I knew that before replying didn't I).As I have just installed it, here it is in Octave:

octave:8> p=0.95;N=10000;
octave:9> ss=rand(N,100);
octave:10> sm=sum(ss'>p);
octave:11> Pest=sum(sm>=5)/N
Pest =  0.56080