Question

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 12% 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%. [You may find it useful to reference the z table.]

a. What is the probability that the production machines will be adjusted if the batch consists of 56 items? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

b. What is the probability that the production machines will be adjusted if the batch consists of 112 items? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

Solution

Given that,

p = 0.12

1 - p = 1 - 0.12 = 0.88

a) n = 56

= p = 0.12

=  [p( 1 - p ) / n] = [(0.12 * 0.88) / 56 ] = 0.0434

P( > 0.16 ) = 1 - P( < 0.16 )

= 1 - P(( - ) / < (0.16 - 0.12) / 0.0434)

= 1 - P(z < 0.92)

Using z table

= 1 - 0.8212

= 0.1788

b) n = 112

= p = 0.12

=  [p( 1 - p ) / n] = [(0.12 * 0.88) / 112 ] = 0.0307

P( > 0.16 ) = 1 - P( < 0.16 )

= 1 - P(( - ) / < (0.16 - 0.12) / 0.0307)

= 1 - P(z < 1.30)

Using z table

= 1 - 0.9032

= 0.0968