The company ships the components in lots of 200. lots containing more than 20 defective components may be returned. find a 95% confidence interval for the proportion of lots that will be returned. use the normal approximation to compute this proportion. round the answers to four decimal places. the 95% confidence interval is

[ 0.0584, 1.0416 ]

Step-by-step explanation:

Data provided in the question:

sample size, n = 200

Number of defective components = 20

thus,

p = = 0.1

Now,

Confidence interval for p is given as:

⇒ p ±

here, z = 1.96 for 95% confidence interval

therefore,

⇒ 0.1 ±

or

⇒ 0.1 ± 0.0416

or

confidence interval = [ 0.1 - 0.0416, 0.1 + 0.0416 ]

or

confidence interval = [ 0.0584, 1.0416 ]

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just sum

10 + 10 = 20