Acceptance sampling is an important quality control techniqu

Acceptance sampling is an important quality control technique, where a batch of data is tested to determine if the proportion of units having a particular attribute exceeds a given percentage. Suppose that 9% of produced items are known to be nonconforming. Every week a batch of items is evaluated and the production machines are adjusted if the proportion of nonconforming items exceeds 16%.

What is the probability that the production machines will be adjusted if the batch consists of 70 items? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

What is the probability that the production machines will be adjusted if the batch consists of 100 items? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

Solution

Here,

mean = u = 0.09

standard deviation = sqrt(p(1-p)/n) = sqrt(0.09*(1-0.09)/70) = 0.034205263

Thus,

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.16      
u = mean =    0.09      
          
s = standard deviation =    0.034205263      
          
Thus,          
          
z = (x - u) / s =    2.05      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.05   ) =    0.020182215 [ANSWER]

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b)

standard deviation = sqrt(p(1-p)/n) = sqrt(0.09*(1-0.09)/100) = 0.028618176

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.16      
u = mean =    0.09      
          
s = standard deviation =    0.028618176      
          
Thus,          
          
z = (x - u) / s =    2.45      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.45   ) =    0.007142811 [answer]